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Imagine building tall structures or measuring the distance of tall buildings. Isn’t that fascinating? The answer lies in the chapter on Trigonometry. Trigonometry and Its Equations- NCERT Exemplar Class 10 introduces students to important trigonometric ratios, which are important for problems involving angles, heights, and distances. Students gain a foundation of concepts, such as the angle of elevation and depression, and learn how to calculate unknown values by applying these ratios to right-angled triangles.
Regular practice of exercises and following the CBSE Syllabus for Class 10 improve students' problem-solving skills and comprehension of the different kinds of exam questions. An in-depth understanding of the subject offers a solid basis for advanced mathematics and practical applications in the fields of engineering, architecture, and navigation. For students preparing for both academic advancement and competitive exams, this chapter is crucial.
Class 10 Maths Chapter 8 Exemplar Solutions Exercise: 8.1 Page number: 89-91 Total questions: 15 |
Question:1
If cos A =4/5 , then the value of tan A is
1.$\frac{3}{5}$ (b) $\frac{3}{4}$ (c) $\frac{4}{3}$ (d) $\frac{5}{3}$
Answer:
Answer. [B]
Solution. It is given that cos A = 4/5
$\text{We know that cos}\theta=\frac{Base}{Hypotenuse}$
$\therefore$ value of base = 4
Hypotenuse = 5
Use Pythagoras theorem in $\bigtriangleup$ABC
(Hypotenuse)2 = (Base)2 + (perpendicular)
$\left ( AC \right )^{2}= \left ( BC \right )^{2}+\left ( AB \right )^{2}$
$\left ( 5 \right )^{2}= \left ( 4 \right )^{2}+\left ( AB \right )^{2}$
$25-16= \left ( AB \right )^{2}$
$9= \left ( AB \right )^{2}$
$\sqrt{9}= AB$
$3= AB$
The value of perpendicular is 3
$\text{Also we know that tan}\theta=\frac{perpendicular}{base}$
$\therefore \tan \, A= \frac{3}{4}$
Hence, option (B) is correct.
Question:2
If sin A =1/2, then the value of cot A is
a) $\sqrt{3}$ (b) $\frac{1}{\sqrt{3}}$ (c) $\frac{\sqrt{3}}{2}$ (d) 1
Answer:
$\\sinA=\frac{1}{2}\text{ for angle }30^0\\cosA=\frac{\sqrt{3}}{2}\\cotA=\frac{cosA}{sinA}=\frac{\frac{\sqrt{3}}{2}}{\frac{1}{2}}=\sqrt{3}$
Hence, option (a) is correct.
Question:3
The value of the expression [cosec (75° + θ) – sec (15° – θ) – tan (55° + θ) + cot (35° – θ)] is
(A) – 1 (B) 0 (C) 1 (D)$\frac{3}{2}$
Answer:
Answer. $\quad[B]$
Solution. Given expression is :
$
\begin{aligned}
& {\left[\operatorname{cosec}\left(75^{\circ}+\theta\right)-\sec \left(15^{\circ}-\theta\right)-\tan \left(55^{\circ}+\theta\right)+\cot \left(35^{\circ}-\theta\right)\right]} \\
& {\left[\operatorname{cosec}(90-(15-\theta))-\sec \left(15^{\circ}-\theta\right)-\tan (90-(35-\theta))+\cot \left(35^{\circ}-\theta\right)\right]} \\
& \{\because \text { we can write }(75+\theta)=(90-(15-\theta)) \text { and }(55+\theta)=(90-(35-\theta))\} \\
& \{\because \operatorname{cosec}(90-q)=\sec q \text { and tan }(90-q)=\cot q\} \\
& {[\sec (15-\theta)-\sec (15-\theta)-\cot (35-\theta)+\cot (35-\theta)]=0}
\end{aligned}
$
Hence, option (B) is correct
Question:4
Given that sinθ =a/b, then cosθ is equal to
(A) $\frac{b}{\sqrt{b^{2}-a^{2}}}$ (B) $\frac{b}{a}$ (C) $\frac{\sqrt{b^{2}-a^{2}}}{b}$ (D)$\frac{a}{\sqrt{b^{2}-a^{2}}}$
Answer:
Answer. [C]
Solution. It is given that sin$\theta$ = a/b
$cos\theta=\sqrt{1-sin^2\theta}=\sqrt{1-\frac{a^2}{b^2}}$
$\therefore \cos \theta = \frac{\sqrt{b^{2}-a^{2}}}{b}$
Hence option (C) is correct.
Question:5
If cos (α + β) = 0, then sin (α – β) can be reduced to
(A) cos β (B) cos 2β (C) sin α (D) sin 2α
Answer:
$\\\text{Given that }\cos(\alpha+\beta)=0\\\Rightarrow (\alpha+\beta)=90^0...........(1)$
$\text{wehave to find }\sin(\alpha-\beta)$
$ \text{from(1) we can write }\alpha=90-\beta$
$ \therefore \ \sin(\alpha-\beta)=\sin(90-\beta-\beta)$
$=\sin(90-2\beta)=\cos2\beta $
$ \text{since sin(90-x)=cosx})$
Hence, option B is correct.
Question:6
The value of (tan1° tan2° tan3° ... tan89°) is
(A) 0 (B) 1 (C) 2 (D)1/2
Answer:
Answer. [B]
Solution. Given :-tan1° tan2° tan3° ... tan89°
tan1° tan2° tan3° ... tan89°tan87° tan 88° tan89° …(1)
We can also write equation (1) in the form of
[tan (900 – 890). tan (900 – 880). tan (900 – 870) …… tan 87°. tan 88° tan 89°]
[$\because$ we can write tan 1° in the form of tan (900 – 890) similarly, we can write other values]
[cot 890. cot 880. cot 870 …. tan 870. tan 880. tan 890]
$\because$ [tan (902 – $\theta$) = cot$\theta$ ]
Also
$\left [ \frac{1}{\tan 89^{\circ}}\frac{1}{\tan 88^{\circ}} \frac{1}{\tan 87^{\circ}}\cdots \tan 87^{\circ}.\tan 88^{\circ}\tan 89^{\circ} \right ]$
$\because$ Throughout all terms are cancelled by each other and remaining will be tan45
Hence, the value is 1
$\because$ option B is correct.
Question:7
If cos 9α = sinα and 9α < 90° , then the value of tan5α is
(A) $\frac{1}{\sqrt{3}}$
(B) $\sqrt{3}$
(C) $1$
(D) $0$
Answer:
Answer. [C]
Solution. Given :- cos 9$\alpha$ = sin$\alpha$
cos9 $\alpha$ = cos(90 – $\alpha$)
$\because$ (cos (90 – $\alpha$) = sin$\alpha$)
9$\alpha$ = 90 – $\alpha$
9$\alpha$+ $\alpha$ = 90
10 $\alpha$= 90
$\alpha = \frac{90}{10}$
Now tan 5$\alpha$ is
Put $\alpha$ = 9 we get
tan 5$\times$ (9)
tan 450
= 1
{$\because$ from the table of trigonometric ratios of angles we know that tan 450 = 1}
Hence, option C is correct.
Question:8
If ΔABC is right-angled at C, then the value of cos (A+B) is
$(A) 0 \ \ \ \ \ (B) 1 \ \ \ \ \ (C)1/2 \ \ \ \ \ (D)\frac{\sqrt{3}}{2}$
Answer:
Answer. [A]
Solution. It is given that $\angle$C = 90°
In $\bigtriangleup$ABC
$\angle$A +$\angle$B +$\angle$C = 180 [$\because$ sum of interior angles of triangle is 180°]
$\angle$A + $\angle$B + 90o = 180 [$\because$ C = 90° (given)]
$\angle$A + $\angle$B = 1800 – 900
$\angle$A + $\angle$B = 90° …(1)
cos($\angle$A +$\angle$B) = cos (90°)
cos(90°) = 0
[$\because$ from the table of trigonometric ratios of angles we know that cos 90° = 0]
Hence option A is correct.
Question:9
If sinA + sin2A = 1, then the value of the expression (cos2A+ cos4A) is
(A) 1 (B)1/2 (C) 2 (D) 3
Answer:
Answer. [A]
Solution. It is given that sinA + sin2A = 1 …(*)
sinA = 1 – sin2AsinA = cos2A …(1) ( $\because$ 1 – sin2A = cos2A)
Squaring both sides we get
sin2 A = cos4A …(2)
Hence cos2A + cos4A =
= sinA + sin2A {using (1) and (2)}
= sinA + sin2A = 1 (given)
Hence option (A) is correct.
Question:10
Given that sinα =1/2 and cosβ =1/2 , then the value of (α + β) is
(A) 0° (B) 30° (C) 60° (D) 90°
Answer:
Answer. [D]
Solution.
$\\\sin \alpha=1/2\Rightarrow \alpha=30^0\\\cos \beta=1/2\Rightarrow \beta=60^0\\\therefore \alpha+\beta=90^0$
Hence option (D) is correct.
Question:11
The value of the expression
$\left [ \frac{\sin ^{2}22^{\circ}+\sin ^{2}68^{\circ}}{\cos ^{2}22^{\circ}+\cos ^{2}68^{\circ}}+\sin ^{2}63^{\circ}+\cos ^{2}63^{\circ}\sin 27^{\circ} \right ]\ is$
(A) 3 (B) 2 (C) 1 (D) 0
Answer:
Answer. [B]
Solution.
$\left [ \frac{\sin ^{2}22^{\circ}+\sin ^{2}68^{\circ}}{\cos ^{2}22^{\circ}+\cos ^{2}68^{\circ}}+\sin ^{2}63^{\circ}+\cos ^{2}63^{\circ}\sin 27^{\circ} \right ]$
$\frac{\sin ^{2}22^{\circ}+\sin ^{2}68^{\circ}}{\cos ^{2}22^{\circ}+\cos ^{2}68^{\circ}}+\sin ^{2}63^{\circ}+\cos ^{2}63^{\circ}\sin \left ( 90^{\circ}-63^{\circ} \right )$
$\left [ \because \sin \left ( 90^{\circ}-\theta \right )= \cos \theta \right ]$
$\frac{\sin ^{2}22^{\circ}+\sin ^{2}68^{\circ}}{\cos ^{2}22^{\circ}+\cos ^{2}68^{\circ}}+\sin ^{2}63^{\circ}+\cos 63^{\circ}\times \cos 63^{\circ}$
$= \frac{\sin ^{2}22^{\circ}+\sin ^{2}\left (90^{\circ}-22^{\circ} \right )}{\cos ^{2}\left ( 90^{\circ}-68^{\circ} \right )+\cos ^{2}68^{\circ}}+\sin ^{2}63^{\circ}+\cos ^{2}63^{\circ}$
$\begin{Bmatrix} \because 68^{\circ}= \left ( 90^{\circ}-22^{\circ} \right ) & \\ 22^{\circ}= \left ( 90^{\circ}-68^{\circ} \right )& \end{Bmatrix}$
$= \frac{\sin ^{2}22^{\circ}+\cos ^{2}22^{\circ}}{\sin ^{2}68^{\circ}+\cos ^{2}68^{\circ}}+\sin ^{2}63^{\circ}+\cos ^{2}63^{\circ}$
$\begin{Bmatrix} \because \sin \left ( 90^{\circ}-\theta \right ) = \cos \theta & \\ \cos \left ( 90^{\circ} -\theta \right )= \sin \theta & \end{Bmatrix}$
$= \frac{1}{1}+1\; \left [ \because \sin ^{2}\theta +\cos ^{2}\theta = 1 \right ]$
= 1+ 1 =2
Hence option (B) is correct.
Question:12
If $4 \tan\theta= 3$ then $\left ( \frac{4\sin \theta -\cos \theta }{4\sin \theta +\cos \theta } \right )$ is equal to
$(A)\frac{2}{3}$ $(B)\frac{1}{3}$ $(C)\frac{1}{2}$ $(D)\frac{3}{4}$
Answer:
Hence option (C) is correct.
$\\\left ( \frac{4\sin \theta -\cos \theta }{4\sin \theta +\cos \theta } \right )\text{ Divide numerator and denominator by and rewriting the given expression}\\\\\Rightarrow ( \frac{4\sin \theta -\cos \theta }{4\sin \theta +\cos \theta })=\frac{4\frac{\sin\theta}{cos\theta}-\frac{cos\theta}{cos\theta}}{4\frac{\sin\theta}{cos\theta}+\frac{cos\theta}{cos\theta}}\\\\=\frac{4\tan\theta-1}{4\tan\theta+1}=\frac{3-1}{3+1}=\frac{2}{4}=\frac{1}{2}\ (\text{Given }4\tan\theta=3)$
Question:13
If sinθ – cosθ = 0, then the value of (sin4θ + cos4θ) is
(A) 1 (B)3/4 (C) 1/2 (D) 1/4
Answer:
$\sin\theta-\cos\theta=0$
squaring both sides we get
$(\sin\theta-\cos\theta)^2=0$
$\sin^2\theta+\cos^2\theta-2\sin\theta\cos\theta=0$
($\therefore$ (a – b)2 = a2 + b2 – 2ab)
$\sin^2\theta+\cos^2\theta=2\sin\theta\cos\theta.......(1)$
$1=2\sin\theta\cos\theta\because (\sin^2\theta+\cos^2\theta=1)$
$\frac{1}{2}=\sin \theta \cos \theta$
Squaring both sides we get
$\frac{1}{4}=\sin^{2} \theta \cos^{2} \theta$ …(2)
Now squaring both side of equation (1) we get
$\left ( \sin ^{2}\theta +\cos ^{2}\theta \right )^{2}= \left ( 2\sin \theta \cos \theta \right )^{2}$
$\left ( \sin ^{2}\theta \right )^{2}+\left ( \cos ^{2}\theta \right )^{2}+2\sin ^{2}\theta \cdot \cos ^{2}\theta = 4\sin ^{2}\theta \cos ^{2}\theta$
$\left [ \because \left ( a+b \right )^{2} = a^{2}+b^{2}+2ab\right ]$
$\left ( \sin ^{4}\theta \right )+\left ( \cos ^{4}\theta \right )= 4\sin ^{2}\theta \cos ^{2}\theta -2\sin ^{2}\theta \cos ^{2}\theta$
$\sin ^{4}\theta +\cos ^{4}\theta = 2\sin ^{2}\theta \cos ^{2}\theta$
(Use equation (2))
$\sin ^{4}\theta +\cos ^{4}\theta = 2\left ( \frac{1}{4} \right )$
$\sin ^{4}\theta +\cos ^{4}\theta =\frac{1}{2}$
Hence option (C) is correct.
Question:14
sin (45° + θ) – cos (45° – θ) is equal to
(A) 2cosθ (B) 0 (C) 2sinθ (D) 1
Answer:
Answer. [B]
Solution. Here
:$\sin \left ( 45^{\circ}+\theta \right )-\cos \left ( 45^{\circ}-\theta \right )$
Sin[90° - (45°- θ)] – cos(45°- θ)
$\left [ \because \left ( 45^{\circ} +\theta \right ) = \left ( 90^{\circ}-\left ( 45-\theta \right ) \right )\right ]$
Cos(45°- θ) – cos(45°- θ) [$\because$ sin (90 – θ) = cosθ]
= 0
Hence option (B) is correct
Question:15
A pole 6 m high casts a shadow 2$\sqrt{3}$ m long on the ground, then the Sun’s elevation is
(A) 60° (B) 45° (C) 30° (D) 90°
Answer:
Answer. [A]
Solution. Given :
height pole = 6 m
Shadow of pole = $2\sqrt{3}m$
Now make figure according to given condition
Let angle of elevation is $\alpha$
$\therefore \tan \alpha = \frac{Perpendicular}{base}$
$\tan \alpha = \frac{6}{2\sqrt{3}}$
$\tan \alpha = \frac{3}{\sqrt{3}}= \sqrt{3}$
$\tan \alpha = \tan 60^{\circ}$ $\left [ \because \tan 60^{\circ}=\sqrt{3} \right ]$
a = 60°
Hence the Sun's elevation is 60°.
Class 10 Maths Chapter 8 exemplar solutions Exercise: 8.2 Page number: 93 Total questions: 12 |
Question:1
Answer:
$\frac{\tan 47^{\circ}}{\cot 43^{\circ}}= 1$
Taking L.H.S.
$\frac{\tan 47^{\circ}}{\cot 43^{\circ}}$
$\frac{\tan \left ( 90^{\circ} -43^{\circ}\right )}{\cot 43^{\circ}}$ $\left ( \because 47= \left ( 90-43 \right ) \right )$
$\frac{\cot 43^{\circ}}{\cot 43^{\circ}}= 1$ $\left ( \because \tan \left ( 90-\theta \right )= \cot \theta \right )$
Hence L.H.S. = R.H.S.
So, the given expression is true.
Question:2
The value of the expression (cos2 23° – sin2 67°) is positive.
Answer:
Answer. [False]
Solution. (cos2 23° – sin2 67°)
= $\left ( \cos ^{2}\left ( 90^{\circ}-67^{\circ} \right )-\sin ^{2}67^{\circ} \right )$ $\left ( \because 23^{\circ}= 90^{\circ}-67^{\circ} \right )$
= Sin267° - sin267° (cos(90-θ) = sin θ)
= 0
Hence the value of the expression is neutral
So, the given statement is false.
Question:3
The value of the expression (sin 80° – cos80°) is negative.
Answer:
Answer. [False]
Solution. (sin 80° – cos 80°)
We know that from 0 to 90° sin$\theta$ and cos$\theta$ both are positive i.e.
$0< \sin \theta \leq 90^{\circ}$ (always positive)
$0< \cos \theta \leq 90^{\circ}$ (always positive)
At 45° both the values of sin$\theta$ and cos$\theta$ are the same but after 45° to 90° value of sin is greater than the value of cos$\theta$ Hence sin 80° > cos 80°.
If we subtract a smaller term from bigger than the result is positive.
Hence (sin 80° – cos80°) > 0
So, the given statement is false
Question:4
Prove that $\sqrt{\left ( 1-\cos ^{2}\theta \right )\sec ^{2}\theta }= \tan \theta$
Answer:
L.H.S
$\sqrt{\left ( 1-\cos ^{2}\theta \right )\sec ^{2}\theta } \cdots \left ( 1 \right )$
We know that
$1-\cos ^{2}\theta = \sin ^{2}\theta\text{ in (1)}$
$\sqrt{\sin ^{2}\theta \cdot \sec ^{2}\theta }$
$=\sqrt{\sin ^{2}\theta \times \frac{1}{\cos ^{2}\theta } }$ $\left ( \because \sec \theta = \frac{1}{\cos \theta } \right )$
$=\sqrt{\frac{\sin ^{2}\theta }{\cos ^{2}\theta }}$
$=\sqrt{\tan ^{2}\theta }= \tan \theta$ $\left ( \because \frac{\sin \theta }{\cos \theta }= \tan \theta \right )$
L.H.S. = R.H.S.
Hence the given expression is true.
Question:5
If cosA + cos2A = 1, then sin2A + sin4A = 1.
Answer:
Given
cosA + cos2A = 1 …(1) cos A = 1 – cos2A
cosA = sin2A …(2) ($\because$ sin2$\theta$ = 1 – cos2$\theta$)
$\sin ^{2} A+\sin ^{4}A= 1$
L.H.S.
$\sin ^{2} A+\sin ^{4}A$
$\sin ^{2} A+ \left ( \sin ^{2}A \right )^{2}$
$\cos A+\left ( \cos A \right )^{2}$ (from (2))
cosA + cos2A
= 1 (R.H.S.) (from (1))
Hence sin2A + sin4A = 1
So, the given statement is true.
Question:6
(tanθ + 2) (2 tan θ + 1) = 5 tan θ + sec2θ.
Answer:
(tanθ + 2) (2 tan θ + 1) = 5 tan θ + sec2θ
Taking L.H.S.
(tanθ + 2) (2 tan θ + 1)
tanθ.(2tan θ+1) + 2(2tan θ +1)
$2\tan ^{2}\theta +\tan \theta +4\tan \theta+2$
$2\tan ^{2}\theta +5\tan \theta+2$
$2\left ( \tan ^{2}\theta +1 \right )+5\tan \theta\: \cdots \left ( 1 \right )$
We know that
$\sec ^{2 }\theta-\tan ^{2}\theta= 1$
$\left ( 1+\tan ^{2}\theta= \sec ^{2}\theta \right )$
Put the above value in (1)we get
$2\sec ^{2}\theta +5\tan \theta \neq 5\tan \theta +\sec ^{2}\theta$
L.H.S. $\neq$ R.H.S.
Hence the given expression is false.
Question:7
Answer:
Answer. [False]
Solution. Let us take 2 cases.
Case 1:
Case 2 :
$\theta _{1}$ is the angle when length is small and $\theta _{2}$ is the angle when shadow length is increased.
for finding $\theta _{1}$, $\theta _{2}$ find tan $\theta$
$\tan \theta_{1} = \frac{Perpendicular}{Base}= \frac{height\, of\, tower}{length\, of\, shadow}$
$\tan \theta_{2} = \frac{height\, of\, tower}{length\, of\, shadow}$
In both the case height of the tower is the same but in case 2 length of the shadow is increased and if the length of shadow increased value of $\theta _{2}$ decreased.
Hence the given statement is false.
Question:8
Answer:
Answer. [False]
Solution. According to question.
In the figure $\theta _{1}$ is the angle of elevation and $\theta _{2}$ is the angle of depression of the cloud.
For finding $\theta _{1}$find tanq in $\bigtriangleup$ECD
$\tan \theta _{1}= \frac{perpendicular}{Base}= \frac{DC}{EC}= \frac{H}{\iota }$
Find tan$\theta$ in $\bigtriangleup$ECB for $\theta _{2}$
$\tan \theta _{2}= \frac{BC}{EC}= \frac{3}{\iota }$
Here we found that $\theta _{1}$ and $\theta _{2}$ both are different.
Hence the given statement is false.
Question:9
The value of 2sinθ can be a+1/a , where a is a positive number, and a ≠ 1.
Answer:
Answer. [False]
Solution. We know that
-1≤ sin θ ≤ 1
Multiply by 2.
-2≤ 2 sin θ ≤ 2
Here we found that value of 2 sin $\theta$ is lies from – 2 to 2.
But if we take a > 0 and a $\neq$ 1 then
$a+\frac{1}{a}> 2$
For example a = 3
3 + 1/3 = 3.33
Hence $a+\frac{1}{a}$ is always greater than 2 in case of positive number except 1
But value of 2 sin $\theta$ is not greater than 2
Hence the given statement is false
Question:10
$\cos \theta = \frac{a^{2}+b^{2}}{2ab}$ where a and b are two distinct numbers such that ab> 0.
Answer:
We know that
$-1\leq \cos \theta \leq 1$
We also know that
$\left ( a-b \right )^{2}= a^{2}+b^{2}-2ab$
$\text{Since }$ $\left ( a-b \right )^{2}$ $\text{is a square term hence it is always positive }$
$\left ( a-b \right )^{2}> 0$
a2 + b2 – 2ab > 0
$a^{2}+b^{2}> 2ab$
We observe that $a^{2}+b^{2}$ is always greater than 2ab.
Hence,
$\frac{a^{2}+b^{2}}{2ab}> 1$
Because if we divide a big term by small then the result is always greater than 1.
cos$\theta$ is always less than or equal to 1
Hence the given statement is false.
Question:11
Answer:
Answer. [False]
Solution. According to question
Case: 1
Here BC is the tower.
Let the height of the tower is H and distance AB = a
In $\bigtriangleup$ABC
$\tan\theta=\frac{Perpendicular}{Base}$
$\tan 30^{\circ}= \frac{H}{a}$ $\left ( \because \theta = 30^{\circ} \right )$
$\frac{1}{\sqrt{3}}= \frac{H}{a}\; \cdots \left ( 1 \right )$ $\left ( \because \tan 30^{\circ} = \frac{1}{\sqrt{3}}\right )$
Case :2 When height is doubled
Here ED = a
In $\bigtriangleup DEF$
$\tan \theta = \frac{2H}{a}$
$\tan \theta = \frac{2}{3}$ (from (1))
But $\tan 60^{\circ}= \sqrt{3}$ (If the angle is double)
$\sqrt{3}\neq \frac{2}{\sqrt{3}}$
hence the given statement is false.
Question:12
Answer:
Answer. [True]
Solution. According to question
In case-1. Height is H and observation distance is a.
In case-2, both height and observation distance is increased by 10%.
In case -1
$\tan \theta _{1}= \frac{H}{a}$ $\left ( \because \tan \theta = \frac{perpendicular}{Base} \right )$ .....(1)
In case -2
$\tan \theta _{2}= \frac{H+\frac{H}{10}}{a+\frac{a}{10}}$
$= \frac{\frac{11H}{10}}{\frac{11a}{10}}= \frac{11H}{10}\times \frac{10}{11a}= \frac{H}{a}$
$\tan \theta _{2}= \frac{H}{a} \cdots \left ( 2 \right )$
from equation (1) and (2) we observe that $\theta _{1}= \theta _{2}$
Hence the given statement is true.
Class 10 Maths Chapter 8 exemplar solutions Exercise: 8.3 Page number: 95 Total questions: 15 |
Question:1
Answer:
$\frac{\sin \theta }{1+\cos \theta }+\frac{1+\cos \theta}{\sin \theta}= 2\cos ec\theta$
Taking L.H.S.
$= \frac{\sin \theta }{1+\cos \theta }+\frac{1+\cos \theta}{\sin \theta}$
Taking LCM
$= \frac{\sin ^{2}\theta +\left ( 1+\cos \theta \right )^{2}}{\left ( 1+\cos \theta \right )\sin \theta }$
$= \frac{\sin ^{2}\theta +\cos ^{2}\theta +2\cos \theta }{\left ( 1+\cos \theta \right )\sin \theta }$ $\left ( \because \left ( a+b \right )^{2} = a^{2}+b^{2}+2ab\right )$
$= \frac{1+1+2\cos \theta }{\left ( 1+\cos \theta \right )\sin \theta }$ $\left ( \because \sin ^{2}\theta +\cos ^{2}\theta = 1 \right )$
$= \frac{2\left ( 1+\cos \theta \right )}{\left ( 1+\cos \theta \right )\sin \theta }$
$= \frac{2}{\sin \theta }$
$= 2 \cos ec\theta$ $\left ( \because \frac{1}{\sin \theta } = \cos ec \theta \right )$
L.H.S. = R.H.S.
Hence proved.
Question:2
Prove the following :
$\frac{tan A}{1+\sec A}-\frac{\tan A}{1-\sec A}= 2cosecA$
Answer:
Solution.
$\frac{tan A}{1+\sec A}-\frac{\tan A}{1-\sec A} = 2cosecA$
Taking L.H.S.
$= \frac{\tan A}{1+\sec A}-\frac{\tan A}{1-\sec A}$
Taking L.C.M.
$=\frac{\tan A\left ( 1-\sec A \right )-\tan A\left ( 1+\sec A \right )}{\left ( 1+\sec A \right )\left ( 1-\sec A \right )}$
$=\frac{\tan A-\tan A\\sec A-\tan A-\tan A\sec A}{1-\sec ^{2}A}$ $\left ( \because \left ( a-b \right )\left ( a+b \right ) = a^{2}-b^{2}\right )$
$= \frac{-2\tan A\sec A}{-\tan ^{2}A}$ $\left ( \because \sec ^{2}A-\tan ^{2}A= 1 \right )$
$= \frac{2\sec A}{\tan A}$
$= \frac{2}{\cos A}\times \frac{\cos A}{\sin A}$ $\begin{pmatrix} \because \sec \theta = \frac{1}{\cos \theta } & \\ \because \tan \theta =\frac{\sin \theta }{\cos \theta } & \end{pmatrix}$
$= \frac{2}{\sin A}= 2\cos ec A$ $\left ( \because \frac{1}{\sin \theta }= \cos ec\theta \right )$
L.H.S. = R.H.S.
Hence proved
Question:3
If tan A = 3/4 , then show that sinAcos A = 12/25 .
Answer:
Given:-
$\tan A= \frac{3}{4}$
To prove:-
$sin A\cos A= \frac{12}{25}...........(1)$
we know that
$\tan \theta = \frac{P}{B}$
$\frac{P}{B}= \frac{3}{4}$
P=3, B = 4
Using Pythagoras theorem
$H^{2}= B^{2}+P^{2}$
$H^{2}= \left ( 4^{2} \right )+\left ( 3^{2} \right )$
$H^{2}= 9+16$
$H^{2}= 25$
$H= \sqrt{25}= 5$
H = 5
we know that
$\sin \theta = \frac{P}{H},\cos \theta = \frac{B}{H}$
Hence
$\sin A= \frac{3}{5},\cos A= \frac{4}{5}$
Put the value of sinA and cosA in equation (1)
$\frac{3}{5}\times \frac{4}{5}= \frac{12}{25}$
L.H.S. = R.H.S.
Hence proved.
Question:4
Prove the following : (sin α + cos α) (tan α + cot α) = sec α + cosec α
Answer:
Solution.
(sin α + cos α) (tan α + cot α) = sec α + cosec α
Taking L.H.S.
$\left ( \sin \alpha +\cos \alpha \right )\left ( \tan \alpha +\cot \alpha \right )$
$= \left ( \sin \alpha +\cos \alpha \right )\left ( \frac{\sin \alpha }{\cos \alpha } +\frac{\cos \alpha}{\sin \alpha } \right )\left ( \because \tan \theta = \frac{\sin \theta }{\cos \theta },\cot \theta = \frac{\cos \theta }{\sin \theta } \right )$
Taking L.C.M.
$= \frac{\left ( \sin \alpha +\cos \alpha \right )\left ( \sin ^{2}\alpha +\cos^{2} \alpha \right )}{\sin \alpha +\cos \alpha }$
$= \frac{\sin \alpha }{\sin \alpha \cos \alpha }+\frac{\cos \alpha }{\sin \alpha +\cos \alpha }$ $\left ( \because \sin^{2} \theta +\cos^{2} \theta = 1 \right )$
by separately divide
$= \frac{\sin \alpha }{\sin \alpha \cos \alpha }+\frac{\cos \alpha }{\sin \alpha +\cos \alpha }$
$= \frac{1}{\cos \alpha }+\frac{1}{\sin \alpha }$
$= \sec \alpha +\cos ec \, \alpha$ $\left ( \because \frac{1}{\cos \theta }= \sec \theta ,\frac{1}{\sin \theta }= \cos ec\theta \right )$
L.H.S. = R.H.S.
Hence proved.
Question:5
$\left ( \sqrt{3}+1 \right )\left ( 3-\cot 30^{\circ} \right )= \tan ^{3}60^{\circ}-2\sin 60^{\circ}$
Taking L.H.S
$\left ( \sqrt{3}+1 \right )\left ( 3-\cot 30^{\circ} \right )$
$= \sqrt{3}\left ( 3-\cot 30^{\circ} \right )+1\left ( 3-\cot 30^{\circ} \right )$
$=3 \sqrt{3}-\sqrt{3}\cot 30^{\circ}+3-\cot 30^{\circ}$
We know that
$\cot 30^{\circ}= \sqrt{3}$
$= 3\sqrt{3}-\sqrt{3}\left (\sqrt{3} \right )+3-\sqrt{3}$
$= 3\sqrt{3}-3+3-\sqrt{3}$
$= 3\sqrt{3}-\sqrt{3}$
$= 2\sqrt{3}$
R.H.S.
$\tan ^{3}60^{\circ}-2\sin 60^{\circ}$
We know that
$tan 60^o = \sqrt{3}$
$\tan ^{3}60^{\circ}-2\sin 60^{\circ}= \left ( \sqrt{3} \right )^{3}-2\left ( \frac{\sqrt{3}}{2} \right )$
$= 3\sqrt{3}-\sqrt{3}= 2\sqrt{3}$
Hence proved
Question:6
Prove the following :
$1+\frac{\cot ^{2}\alpha }{1+\cos ec\alpha }= \cos ec\, \alpha$
Answer:
$1+\frac{\cot ^{2}\alpha }{1+\cos ec\alpha }= \cos ec\, \alpha$
Taking L.H.S.
$= 1+\frac{\cot ^{2}\alpha }{1+\cos ec\, \alpha }$
$= \frac{1+\cos ec\, \alpha +\cot ^{2}\alpha }{1+\cos ec\, \alpha }$
$= \frac{\cos ec^{2}\alpha-\cot ^{2}\alpha+ \cos ec\, \alpha+\cot ^{2}\alpha \, \, }{1+\cos ec\, \alpha }$ $\left ( \because \cos ec^{2}\theta -\cot ^{2}\theta = 1 \right )$
$= \frac{\cos ec^{2}\alpha +\cos ec\, \alpha }{1+\cos ec\, \alpha }$
$= \frac{\cos ec\, \alpha \left ( \cos ec\, \alpha+1 \right ) }{\left ( 1+\cos ec\, \alpha \right )}$
$= \cos ec\, \alpha$
L.H.S. = R.H.S.
Hence proved.
Question:7
Prove the following: tan θ + tan (90° – θ) = sec θ sec (90° – θ)
Answer:
Solution.
tan θ + tan (90° – θ) = sec θ sec (90° – θ)
Taking L.H.S.
= tan θ + tan (90° – θ)
= tanθ + cotθ ($\because$ tan (90 – θ) = cot θ)
$=\frac{\sin \theta }{\cos \theta }+\frac{\cos \theta }{\sin \theta }$ $\left ( \because \tan \theta = \frac{\sin \theta }{\cos \theta },\cot \theta = \frac{\cos \theta }{\sin \theta } \right )$
Taking L.C.M.
$\frac{\sin ^{2}\theta +\cos ^{2}\theta }{\sin \theta \cos \theta }$
$= \frac{1}{\cos \theta \cdot \sin \theta }$ $\left ( \because \sin ^{2}\theta +\cos ^{2}\theta = 1 \right )$
$= \frac{1}{\cos \theta }\times \frac{1}{\sin \theta }$
$= \sec \theta \times \cos ec\, \theta$ $\left ( \because \frac{1}{\cos \theta } = \sec \theta ,\frac{1}{\sin \theta }= \cos ec\, \theta \right )$
$= \sec \theta \times \sec \left ( 90^{\circ}-\theta \right )$ $\left ( \because \sec \left ( 90-\theta \right ) = \cos ec\, \theta \right )$
L.H.S. = R.H.S.
Hence proved.
Question:8
Answer:
Answer. [30°]
Solution. According to question
Here BC is the height of the pole i.e. h meters and AB is the length of shadow i.e. $\sqrt{3}h$ .
For finding angle q we have to find tanq in $\bigtriangleup$ABC
$\tan \theta = \frac{Perpendicular}{Base}$
$= \tan \theta = \frac{h}{\sqrt{3}h}$
$\theta = 30^{\circ}$ $\left ( \because \tan 30^{\circ}= \frac{1}{\sqrt{3}} \right )$
Hence angle of elevation is 30°.
Question:9
If $\sqrt{3}$ tan θ = 1, then find the value of sin2θ – cos2θ.
Answer:
Given :
$\sqrt{3}\tan\theta=1$
$\tan \theta = \frac{1}{\sqrt{3}}$
$\theta = 30^{\circ}$ $\left ( \because \tan30^{\circ}= \frac{1}{\sqrt{3}} \right )$
$\sin ^{2}\theta -\cos ^{2}\theta = \sin ^{2}30^{\circ}-\cos ^{2}30^{\circ}$
(Because θ = 300)
$= \left ( \frac{1}{2} \right )^{2}-\left ( \frac{\sqrt{3}}{2} \right )^{2}$ $\begin{bmatrix} \because \sin 30^{\circ}= \frac{1}{2} & \\ \cos 30^{\circ}= \frac{\sqrt{3}}{2}& \end{bmatrix}$
$= \frac{1}{4}-\frac{3}{4}$
Taking L.C.M.
$= \frac{1-3}{4}= \frac{-2}{4}$
$\sin ^{2}\theta -\cos ^{2}\theta = \frac{-1}{2}$
Question:10
Answer:
Length of ladder = 15 m
The angle between wall and ladder = 60°
Let the height of wall = H
In $\bigtriangleup$ABC $\angle$C = 60°, $\angle$B = 90°
We know that
$\angle$A + $\angle$B + $\angle$C = 180° (Sum of interior angles of a triangle is 180)
$\angle$A + 90 + 60 = 180°
$\angle$A = 30°
In $\bigtriangleup$ABC
$\sin 30^{\circ} = \frac{H}{15}$
$H= \sin 30^{\circ} \times 15$
$H= \frac{1}{2} \times 15$ $\left ( \because \sin 30^{\circ} = \frac{1}{2}\right )$
$H= 7\cdot 5\, m$
Hence the height of the wall is 7.5 m
Question:11
$\text{Simplify} (1+tan^2\theta)(1-sin\theta)(1+sin\theta)$
Answer:
$(1+tan^2\theta)(1-sin\theta)(1+sin\theta)$
$= \left ( \sec ^{2}\theta \right )\left ( \left ( 1 \right )^{2} -\left ( \sin \theta \right )^{2}\right )$ $\left ( \because \left ( a-b \right )\left ( a+b \right ) = a^{2}-b^{2}\right )$
$= \left ( \sec ^{2}\theta \right )\left ( 1-\sin ^{2}\theta \right )$
$= \left ( \sec ^{2}\theta \right )\left ( \cos ^{2}\theta \right )$ $\left ( \because \sin ^{2}\theta + \cos ^{2}\theta= 1 \right )$
$= \frac{1}{ \cos ^{2}\theta}\times \cos ^{2}\theta$ $\left ( \because \sec ^{2}\theta = \frac{1}{\cos \theta } \right )$
= 1
Question:12
If 2sin2θ – cos2θ = 2, then find the value of θ.
Answer:
2sin2θ – cos2θ = 2
$2\left ( 1-\cos ^{2}\theta \right )-\cos ^{2}\theta = 2$ $\left ( \because \sin ^{2}\theta +\cos ^{2}\theta= 1 \right )$
$2-2\cos ^{2}\theta-\cos ^{2}\theta= 2$
$2-3\cos ^{2}\theta-2= 0$
$-3\cos ^{2}\theta= 0$
$\cos ^{2}\theta= 0$
$\cos \theta= 0$
$\theta= 90^{\circ}$ $\left ( \because \cos 90^{\circ}= 0 \right )$
Hence value of $\theta$ is 90°
Question:13
Answer:
L.H.S
$= \frac{\cos ^{2}\left ( 90-\left ( 45-\theta ^{\circ} \right ) \right )+\cos ^{2}\left ( 45-\theta ^{\circ} \right ) }{\tan \left ( 60^{\circ}+\theta \right )\tan \left ( 30^{\circ}-\theta \right )}$ $\left ( \because \sin \left ( 90^{\circ}-\theta \right ) = \cos \theta \right )$
$= \frac{\sin ^{2}\left ( 45^{\circ}-\theta \right )+\cos ^{2}\left ( 45^{\circ} -\theta\right )}{\tan \left ( 90^{\circ}-\left (30^{\circ} -\theta \right ) \right )\tan \left ( 30^{\circ}-\theta \right )}$ $\left ( \because \tan \left ( 90-\theta \right )= \cot \theta \right )$
$= \frac{\sin ^{2}\left ( 45-\theta \right )+\cos ^{2}\left ( 45-\theta \right )}{\cot \left ( 30^{\circ}-\theta \right )\tan \left ( 30^{\circ}-\theta \right )}$ $\left ( \because \sin ^{2}\theta +\cos ^{2}\theta = 1 \right )$
$= \frac{1}{\frac{1}{\tan \left ( 30^{\circ}-\theta \right )}\times \tan \left ( 30-\theta \right )}$ $\left ( \because \cot \theta= \frac{1}{\tan \theta} \right )$
$= \frac{1}{1}=1$
L.H.S. = R.H.S.
Hence proved.
Question:14
Answer:
Answer. [45°]
Solution. According to the question.
In $\bigtriangleup$EDC EC = 20.5 m
DC = 20.5 m
To find angle $\theta$ in $\bigtriangleup$EDC we need to find tan$\theta$.
$\tan \theta = \frac{P}{B}$
$\tan \theta = \frac{EC}{DC}$
$\tan \theta = \frac{20\cdot 5}{20\cdot 5}$
$\tan \theta = 1$
$\theta = 45^{\circ}$ $\left ( \because \tan 45^{\circ}= 1 \right )$
Hence the angle of elevation is 45°.
Question:15
Show that tan4θ + tan2θ = sec4θ – sec2θ.
Answer:
Taking L.H.S.
tan4θ + tan2θ
(tan2θ) + tan2θ…(1)
We know that sec2θ – tan2θ = 1
Put
$\tan ^{2}\theta = \sec ^{2}\theta -1$ in (1)
$= \left ( \sec ^{2}\theta -1 \right )^{2}+\sec ^{2}\theta -1$
$= \left ( \sec ^{2}\theta \right )^{2}+\left ( 1 \right )^{2}-2\left ( \sec ^{2}\theta \right )\left ( 1 \right )+\sec ^{2}\theta -1$
$\left ( \because \left ( a-b \right )^{2} = a^{2}+b^{2}-2ab\right )$
$= \sec ^{4}\theta +1-2\sec ^{2}\theta +\sec ^{2}\theta -1$
$= \sec ^{4}\theta -\sec ^{2}\theta$
LHS = RHS
Hence proved
Class 10 Maths Chapter 13 exemplar solutions Exercise: 8.4 Page number: 99-100 Total questions: 18 |
Question:1
If cosecθ + cotθ = p, then prove that
$cos\theta=\frac{p^{2}-1}{p^{2}+1}$ .
Answer:
Given: cosecθ + cotθ = p …(1)
Taking right hand side.
$\frac{p^{2}-1}{p^{2}+1}$
Put value of p from equation (1) we get
$= \frac{\left ( \cos ec\, \theta +\cot \theta \right )^{2}-1}{\left ( \cos ec\, \theta +\cot \theta \right )^{2}+1}$
$= \frac{ \cos ec\,^{2} \theta +\cot^{2} \theta+2 \cos ec\, \theta \cot \theta-1}{ \cos ec\,^{2} \theta +\cot^{2} \theta+2 \cos ec\, \theta \cot \theta+1}$
$\left [ \because \left ( a+b \right )^{2} = a^{2}+b^{2}+2ab \right ]$
$= \frac{\frac{1}{\sin ^{2}\theta }+\frac{\cos ^{2}\theta }{\sin ^{2}\theta}+\frac{2\cos \theta}{\sin ^{2}\theta}-1}{\frac{1}{\sin ^{2}\theta }+\frac{\cos ^{2}\theta }{\sin ^{2}\theta}+\frac{2\cos \theta}{\sin ^{2}\theta}+1}$
$\left [ \because \cos ec\, \theta = \frac{1}{\sin \theta } ,\cot \theta = \frac{\cos \theta }{\sin \theta }\right ]$
$= \frac{\frac{1+\cos ^{2}\theta +2\cos \theta -\sin ^{2}\theta }{\sin ^{2}\theta}}{\frac{1+\cos ^{2}\theta +2\cos \theta +\sin ^{2}\theta }{\sin ^{2}\theta}}$ [by taking LCM]
$= \frac{1+\cos ^{2}\theta +2\cos \theta -\sin ^{2}\theta }{1+\cos ^{2}\theta +2\cos \theta +\sin ^{2}\theta}$
$= \frac{1-\sin ^{2}\theta +\cos \theta +2\cos \theta }{1+\sin ^{2}\theta +\cos \theta +2\cos\theta}$
$= \frac{\cos ^{2}\theta +\cos ^{2}\theta+2\cos \theta }{1+1+2\cos \theta }$ $\begin{bmatrix} \because 1-\sin ^{2\theta }= \cos ^{2} \theta & \\ \sin ^{2 }+\cos ^{2}\theta = 1 & \end{bmatrix}$
$= \frac{2\cos ^{2}\theta +2\cos \theta }{2+2\cos \theta}$
$= \frac{2\cos \theta \left ( \cos \theta +1 \right )}{2 \left ( \cos \theta +1 \right )}$
$= \frac{2\cos \theta }{2}$
$= \cos \theta$
which is equal to the eft-hand side
Hence proved.
Question:2
Prove that $\sqrt{\sec ^{2}\theta +\cos ec^{2}\theta }= \tan \theta +\cot \theta$
Answer:
Taking left-hand side
$\sqrt{\sec ^{2}\theta +\cos ec^{2}\theta }$
$= \sqrt{\frac{1}{\cos ^{2}\theta }+\frac{1}{\sin ^{2}\theta }}$ $\left [ \because \sec \theta = \frac{1}{\cos \theta } ,\cos ec= \frac{1}{\sin \theta }\right ]$
$= \sqrt{\frac{\sin ^{2}\theta +\cos ^{2}\theta }{\cos ^{2}\theta \cdot \sin ^{2}\theta }}$
$= \sqrt{\frac{1}{\cos ^{2}\theta\cdot \sin ^{2} \theta }}$ $\left [ \because \sin ^{2}\theta +\cos ^{2}\theta = 1 \right ]$
$\frac{1}{\cos \theta \sin \theta }=\frac{sin^2\theta+cos^2\theta}{cos \theta sin\theta}=\frac{sin^2\theta}{cos \theta sin\theta}+\frac{cos^2\theta}{cos \theta sin\theta}\\\\=\frac{sin\theta}{cos\theta}+\frac{cos\theta}{sin\theta}=tan\theta+cot\theta=RHS\\\text{Hence proved}$
Question:3
Answer:
Answer. [27.322 m]
Solution.
The angle of elevation of the top of a tower AB from certain point C is 30°
Let observer moves from C to D that is CD = 20m
Now angle of elevation increased by 15° that is 45° on point D
In $\bigtriangleup$ABD
$\tan 45^{\circ}= \frac{AB}{BD}$ $\left [ \because \tan \theta = \frac{Perpendicular}{Base} \right ]$
$1= \frac{AB}{BD}$
BD = AB …(1)
In $\bigtriangleup$ABC
$\tan 30^{\circ}= \frac{AB}{BC}$ $\left [ \because \tan \theta = \frac{Perpendicular}{Base} \right ]$
$\frac{1}{\sqrt{3}}= \frac{AB}{BD+DC}$ $\begin{bmatrix} \because \tan 30= \frac{1}{\sqrt{3}} & \\ BC= BD+DC & \end{bmatrix}$
$\frac{1}{\sqrt{3}}= \frac{AB}{BD+20}$ $\left ( \because DC= 20 \right )$
By cross multiplication we get
$BD+20= \sqrt{3}AB$
Now put the value of BD from equation (1) we have
$AB+20= \sqrt{3}AB$
$20= \sqrt{3}AB-AB$
$20= AB\left ( \sqrt{3}-1 \right )$
$AB= \frac{20}{\sqrt{3}-1}$
$AB= \frac{20}{1\cdot 732-1}= \frac{20}{0\cdot 732}$
AB = 27.322
Hence the height of the tower is 27.322 m
Question:4
If 1 + sin2θ = 3sinθ cosθ, then prove that tanθ = 1 or1/2
Answer:
Solution. Given : 1 + sin2θ = 3sinθ cosθ
To Prove - tanθ = 1 or 1/2
Dividing both side by sinθ we get
$\frac{1+\sin ^{2}\theta }{\sin ^{2}\theta}= \frac{3\cos \theta }{\sin \theta }$
$\frac{1}{\sin ^{2}\theta}+\frac{\sin ^{2}\theta}{\sin ^{2}\theta}= 3\cot \theta$ $\left ( \because \frac{\cos \theta }{\sin \theta } = \cot \theta \right )$
$\cos ec^{2}\theta +1= 3\cot \theta$ $\left ( \because \frac{1}{\sin ^{2}\theta}= \cos ec^{2}\theta \right )$
$1+\cot ^{2}\theta +1= 3\cot \theta$ $\left ( \because \cos ec^{2}\theta = 1+\cot ^{2} \theta \right )$
$2+\cot ^{2} \theta+3\cot \theta= 0$
$\cot ^{2} \theta+3\cot \theta+2= 0$
$\cot ^{2} \theta-2\cot \theta-\cot \theta +2= 0$
$\cot \theta\left ( \cot \theta-2 \right )-1\left ( cot \theta-2\right )= 0$
$\left ( \cot \theta-2 \right )\left ( \cot \theta-1 \right )= 0$
$\cot \theta = 1,2$
We know that
$\tan \theta = \frac{1}{\cot \theta }$
$\therefore \tan \theta = 1,\frac{1}{2}$
Hence proved.
Question:5
Given that sinθ + 2cosθ = 1, then prove that 2sinθ – cosθ = 2.
Answer:
Solution. Given:- sinθ + 2cosθ = 1
squaring both sides we have
$\left ( \sin \theta +2\cos \theta \right )^{2}= 1^{2}$
$\sin ^{2}\theta +4\cos ^{2}\theta +4\sin \theta \cos \theta = 1$
$\left ( \because \left ( a+b \right )^{2} = a^{2}+b^{2}+2ab\right )$
$\sin ^{2}\theta +4\cos ^{2}\theta = 1-4\sin \theta \cos \theta \cdots \left ( 1 \right )$
To prove :
$2\sin \theta -\cos \theta = 2$
Taking the left-hand side
$2\sin \theta -\cos \theta \cdots \left ( 2 \right )$
On squaring equation (2) we get
$\left (2 \sin \theta -\cos \theta \right )^{2}$
$= 4\sin ^{2}\theta +\cos ^{2}\theta -4\sin \theta \cos \theta$
$\left( \because (a - b)^{2} = a^{2} + b^{2} - 2ab \right)$
$= 3\sin ^{2}\theta +\sin ^{2}\theta+\cos ^{2}\theta -4\sin \theta \cos \theta$
$= 3\sin ^{2}\theta +1-4\sin \theta \cos \theta$ $\left [ \because \sin ^{2} \theta +\cos ^{2} \theta= 1\right ]$
$= 3\sin ^{2} \theta+\sin^{2} \theta+4\cos ^{2} \theta$ [use equation (1)]
$= 4\sin ^{2}\theta +4\cos ^{2}\theta$
$= 4\left ( \sin ^{2}\theta +\cos ^{2}\theta \right )$
= 4
$\left [ \because \sin ^{2} \theta +\cos ^{2} \theta= 1\right ]$
So here we get the value of (2sin$\theta$ – cos$\theta$)2 is 4
$\left ( 2\sin \theta -\cos \theta \right )^{2}= 4$
$2\sin \theta -\cos \theta = \sqrt{4}$
$2\sin \theta -\cos \theta = 2$
Hence proved
Question:6
Answer:
Solution. According to question
Let the height of tower = h
the distance of the first point from its foot = s
the distance of the second point from its foot = t
$\tan \theta = \frac{h}{s}\cdots \left ( 1 \right )$ $\left ( \because \tan \theta = \frac{Perpendicular}{Base} \right )$
$\tan \left ( 90-\theta \right )= \frac{h}{t}$
$\cot \theta = \frac{h}{t}\cdots \left ( 2 \right )$
Multiply equation (1) and (2) we get
$\tan \theta \times \cot \theta= \frac{h}{t}\times \frac{h}{s}$
$\tan \theta \frac{1}{ \tan \theta }= \frac{h^{2}}{st}$ $\left ( \because \cot \theta = \frac{1}{\tan \theta } \right )$
$1= \frac{h^{2}}{st}$
$st= h^{2}$
$h= \sqrt{st}$
Hence proved.
Question:7
Answer:
Solution. According to question
Let the height of tower = h
$\tan 60^{\circ}= \frac{h}{BD}$ $\left [ \because \tan \theta = \frac{Perpendicular}{Base} \right ]$
$\sqrt{3}= \frac{h}{BD}$ $\left [ \because \tan 60^{\circ}= \sqrt{3} \right ]$
$BD= \frac{h}{\sqrt{3}}\cdots \left ( 1 \right )$
$\tan 30^{\circ}= \frac{h}{BC}= \frac{h}{BD+DC}$ $\left [ \because BC= BD+DC \right ]$
$\frac{1}{\sqrt{3}}= \frac{h}{BD+50}$
$BD+50= \sqrt{3h}$
[by cross multiplication]
$\frac{h}{\sqrt{3}}+50= \sqrt{3h}$
[from equation (1)]
$\frac{h}{\sqrt{3}}= \sqrt{3h}-50$
$h= 3h-50\sqrt{3}$
$h= \frac{50\sqrt{3}}{2}$
$h= 25\sqrt{3}m$
Question:8
Answer:
According to question
Here $h$ is the height of Flagstaff AD.
Let $l$ is the height of the tower
$\alpha$ and $\beta$ be the angle of elevation of the bottom and the top of the flagstaff.
In $\triangle B D C$
$
\tan \alpha=\frac{l}{B C}\left[\because \tan \theta=\frac{\text { Perpendicular }}{\text { Base }}\right]
$
$
B C=\frac{l}{\tan \alpha} \cdots
$
In $\triangle \mathrm{ABC}$
$
\tan \beta=\frac{A B}{B C}
$
$
\tan \beta=\frac{h+l}{B C}
$
$
B C=\frac{h+l}{\tan \beta} \cdots
$
Equate equation (1) and (2) we get
$
\begin{aligned}
& \frac{l}{\tan \alpha}=\frac{h+l}{\tan \beta} \\
& l \tan \beta=h \tan \alpha+l \tan \alpha[\text { by cross multiplication] } \\
& l \tan \beta-l \tan \alpha=h \tan \alpha \\
& l(l \tan \beta-\tan \alpha)=h \tan \alpha \\
& l=\frac{h \tan \alpha}{\tan \beta-\tan \alpha}
\end{aligned}
$
Hence Proved
Question:9
If tanθ + secθ =$l$, then prove that
$sec\theta=\frac{l ^{2}+1}{2l }$
Answer:
Solution. Given : $\tan \theta+\sec \theta=1$
There fore
$
\begin{aligned}
& \frac{l^2+1}{2 l}=\frac{(\tan \theta+\sec \theta)^2+1}{2(\tan \theta+\sec \theta)} \\
& \frac{\tan ^2 \theta+\sec ^2 \theta+2 \tan \theta \sec \theta+1}{2(\tan \theta)+2 \sec \theta} \\
& \frac{\sec ^2 \theta+\sec ^2 \theta+2 \tan \theta \sec \theta}{2(\tan \theta+\sec \theta)} \quad\left[\because 1+\tan ^2 \theta=\sec ^2 \theta\right] \\
& \frac{2 \sec ^2 \theta+2 \tan \theta \sec \theta}{2(\tan \theta+2 \sec \theta)} \\
& \frac{2 \sec \theta(\sec \theta+\tan \theta)}{2(\tan \theta+\sec \theta)} \\
& =\sec \theta(\text { R.H.S) }
\end{aligned}
$
Hence proved
Question:10
If sinθ + cosθ = p and secθ + cosecθ = q, then prove that q (p2 – 1) = 2p.
Answer:
Solution. Given :-sinθ + cosθ = p
and secθ + cosecθ = q
To prove :-q (p2 – 1) = 2p
Taking left hand side
q.(p2– 1) =
Put value of q and p we get
$\left ( \sec \theta +\cos ec\theta \right )\left [ \left ( \sin \theta +\cos \theta \right )^{2}-1 \right ]$
$\left ( \because \left ( a+b \right )^{2} = a^{2}+b^{2}+2ab\right )$
= $\left ( \frac{1}{\cos \theta +\frac{1}{\sin \theta }} \right )\left [ \left ( \sin^{2} \theta+\cos^{2} \theta+2 \sin \theta \cos \theta \right ) -1\right ]$
$\left ( \because \sec \theta = \frac{1}{\cos \theta } ,\cos ec\theta =\frac{1}{\sin \theta } \right )$
$= \frac{1}{\cos \theta }\left [ \sin ^{2}\theta +\cos ^{2}\theta +2\sin \theta \cos \theta -1 \right ]$
$+ \frac{1}{\sin \theta }\left [ \sin ^{2}\theta +\cos ^{2}\theta +2\sin \theta \cdot \cos \theta -1 \right ]$
$= \frac{\sin ^{2}}{\cos \theta }+\cos \theta+2\sin \theta -\frac{1}{\cos \theta}+\sin \theta+\frac{\cos^{2} \theta}{\sin \theta}+2\cos \theta-\frac{1}{\sin \theta}$
$= 3\cos \theta +3\sin \theta -\frac{1}{\cos \theta}+\frac{\left ( 1+\cos ^{2}\theta \right )}{\cos \theta }+\frac{\left ( 1+\sin ^{2}\theta \right )}{\sin \theta }-\frac{1}{\sin \theta }$
$\left ( \because \sin ^{2}\theta = 1-\cos ^{2}\theta \right )$
$\left ( \because \cos ^{2}\theta = 1-\sin ^{2}\theta \right )$
$= 3\cos \theta +3\sin \theta+\frac{1}{\cos \theta}\times \left ( -1+1-\cos^{2} \theta \right )+\frac{1}{\sin \theta }\times \left ( 1-\sin^{2} \theta-1 \right )$
$\left ( \because \sin^{2} \theta+\cos^{2} \theta= 1 \right )$
$= 3\cos \theta -\cos \theta +3\sin \theta -\sin \theta$
$= 2\cos \theta +2\sin \theta$
$= 2\left ( \cos \theta +\sin \theta \right )$
2p (R.H.S)
Hence proved.
Question:11
If a sinθ + b cosθ = c, then prove that a cosθ – b sinθ =$\sqrt{a^{2}+b^{2}-c^{2}}$
Answer:
Solution. Given:- asinθ + b cosθ = c
squaring both side we get
$\left ( a\sin \theta +b\cos \theta \right )^{2}= c^{2}$
$a^{2}\sin ^{2}\theta +b^{2}\cos ^{2}\theta+2ab\sin \theta \cos \theta = c^{2}$
$\left ( \because \left ( a+b \right )^{2} = a^{2}+b^{2}+2ab\right )$
$\Rightarrow 2ab\sin \theta \cos \theta = c^{2}-a^{2}\sin ^{2}\theta -b^{2}\cos ^{2}\theta \cdots \left ( 1 \right )$
To prove : acosθ – b sinθ =$\sqrt{a^{2}+b^{2}-c^{2}}$
Taking left hand side : a cosθ – b sinθ and square it we get
$\left ( a\cos \theta -b \sin \theta \right )^{2}$
$= a^{2}\cos ^{2}\theta +b^{2}\sin ^{2}\theta -2ab\cos \theta \sin \theta$
$\left [ \because \left ( a-b \right )^{2} = a^{2}+b^{2}-2ab\right ]$
$= a^{2}\cos ^{2}\theta +b^{2}\sin ^{2}\theta -\left ( c^{2}-a^{2}\sin ^{2}\theta -b^{2}\cos ^{2}\theta \right )$ $\text{[Using (1)]}$
$= a^{2}\cos ^{2}\theta +b^{2}\sin ^{2}\theta - c^{2}+a^{2}\sin ^{2}\theta +b^{2}\cos ^{2}\theta$
$= a^{2}\left ( \cos ^{2}\theta+\sin ^{2}\theta \right )+b^{2}\left ( \sin ^{2}\theta+ \cos ^{2}\right )-c^{2}$
$= a^{2}+b^{2} - c^{2}$ $\left ( \because \sin ^{2}\theta +\cos ^{2}\theta = 1 \right )$
Hence $\left ( a\cos \theta -b\sin \theta \right )^{2}= a^{2}+b^{2}-c^{2}$
$\Rightarrow \left ( a\cos \theta -b\sin \theta \right )= \sqrt{a^{2}+b^{2}-c^{2}}$
Hence proved.
Question:12
Answer:
Solution To prove :- $\frac{1+\sec \theta -\tan \theta }{1+\sec \theta +\tan \theta }= \frac{1-\sin \theta }{\cos \theta }$
Taking left hand side
$\frac{1+\sec \theta -\tan \theta }{1+\sec \theta +\tan \theta }$
$=\frac{1+\frac{1}{\cos \theta }-\frac{\sin \theta }{\cos \theta }}{1+\frac{1}{\cos \theta }+\frac{\sin \theta }{\cos \theta }}$ $\begin{bmatrix} \because \because \sec \theta = \frac{1}{\cos \theta } & \\ \tan \theta = \frac{\sin \theta }{\cos \theta }& \end{bmatrix}$
$=\frac{\frac{\cos \theta+1-\sin \theta}{\cos \theta}}{\frac{\cos +1+\sin \theta}{\cos \theta}}$
$\frac{\cos \theta+1-\sin \theta}{\cos \theta+1+\sin \theta}$
Multiply nominator and denominator by (1 – sin $\theta$)
$=\frac{\left ( \cos \theta +1-\sin \theta \right )\left ( 1-\sin \theta \right )}{\left ( \cos \theta +1+\sin \theta \right )\left ( 1-\sin \theta \right )}$
$=\frac{\cos \theta-\cos \theta\sin \theta+1-\sin \theta-\sin \theta+\sin^{2} \theta}{\cos \theta-\sin \theta\cos \theta+1-\sin \theta+\sin \theta-\sin^{2} \theta}$
$=\frac{\cos ec\left ( 1-\sin \theta \right )+\left ( 1-\sin \theta \right )-\sin \left ( 1-\sin \theta \right )}{\cos \theta -\sin \theta \cos \theta+1-\sin ^{2} \theta}$
$=\frac{\left ( 1-\sin \theta \right )\left ( \cos \theta +1-\sin \theta \right )}{\cos \theta-\sin \theta\cos \theta+\cos^{2} \theta}$ $\left ( \because 1-\sin ^{2}\theta = \cos ^{2}\theta \right )$
$=\frac{\left ( 1-\sin \theta \right )\left ( \cos \theta +1-\sin \theta \right )}{\cos \theta\left ( \cos \theta+1-\sin \theta \right )}$
$= \frac{1-\sin \theta}{\cos \theta}$
Hence proved
Question:13
Answer:
Solution. According to the question
Here 30 m is the length of tower AB.
Let h is the height of tower DC
Let the distance between them is x
In $\bigtriangleup$ABC
$\tan 60^{\circ}= \frac{30}{x}$ $\left [ \tan \theta = \frac{Perpendicular}{Base} \right ]$
$\sqrt{3}= \frac{30}{x}$
$x= \frac{30}{\sqrt{3}}\cdots \left ( 1 \right )$
In $\bigtriangleup$BDC
$\tan 30^{\circ}= \frac{h}{x}$
$\frac{1}{\sqrt{3}}= \frac{h}{30}\times \sqrt{3}$ (using (1))
$\frac{30}{\sqrt{3}}=h\sqrt{3}$
$\frac{30}{\sqrt{3}\times\sqrt{3} }= h$
$\frac{30}{3}= h$
h = 10m
Hence the height of the second tower is 10
Distance between them
$=\frac{30}{\sqrt{3}}m$ .
Question:14
Answer:
According to question
Let x and y are two objects and $\beta$ and $\alpha$ use the angles of depression of two objects.
In $\bigtriangleup$AOX
$\tan \beta = \frac{h}{Ox}$ $\because \tan \theta = \frac{Perpendicular}{Base}$
$Ox= \frac{h}{\tan \beta }$
$Ox= h\cot \beta \cdots \left ( 1 \right )$
In $\bigtriangleup$AOY
$\tan \alpha = \frac{h}{Oy}= \frac{h}{Ox+xy}\; \; \left ( \because Oy= Ox+xy \right )$
$Ox+xy= \frac{h}{\tan \alpha }$
$Ox+xy= h\cot \alpha$
$xy= h\cot \alpha- Ox$
$xy= h\cot \alpha- h\cot \beta$ (from equation (1))
$xy= h\left ( \cot \alpha- \cot \beta \right )$
Hence the distance between two objects is
$h\left ( \cot \alpha- \cot \beta \right )$ .
Question:15
Answer:
Solution. According to the question:-
Here a and b be the angles of indication when the ladder at rest and when it pulled away from the wall
In $\bigtriangleup$AOB
$\cos \alpha = \frac{OB}{AB}$ $cos \theta = \frac{Base}{Hypotenuse}$
$OB= AB\cos \alpha \cdots \left ( 1 \right )$
$\sin \alpha = \frac{AO}{AB}$ $\sin \theta = \frac{Perpendicular}{hypotenuse}$
$AO= AB\sin \alpha \cdots \left ( 2 \right )$
Similarly In $\bigtriangleup$DOC
$\cos \beta = \frac{OC}{DC}$
$OC= DC\cos \beta \cdots \left ( 3 \right )$
$\sin \beta = \frac{OD}{DC}$
$OD= DC\sin \beta \cdots \left ( 4 \right )$
Now subtract equation (1) from (3) we get
OC – OB = DC cos$\beta$ – AB cos$\alpha$
Here OC – OB = P
and DC = AB because length of ladder remains $\Rightarrow$ P = AB cos
$\beta$ – AB cos$\alpha$
P = AB (cos$\beta$ – cos$\alpha$) …(5)
Subtract equation (4) from (2) we get
AO – OD = AB sina – DC sin $\beta$
Here AO – OD = q
and AB = DC because length of ladder remains same
$\Rightarrow$ q = AB sin $\alpha$ – AB sin$\beta$
q = AB (sin $\alpha$ – sin $\beta$) …(6)
on dividing equation (5) and (6) we get
$\frac{p}{q}= \frac{AB\left ( \cos \beta -\cos \alpha \right )}{AB\left ( \sin \alpha -\sin \beta \right )}$
$\frac{p}{q}= \frac{\cos \beta -\cos \alpha}{\sin \alpha -\sin \beta}$
Hence proved
Question:16
Answer:
Solution. According to question
Let h is the height of the tower
In $\bigtriangleup$ABE
$\tan \theta = \frac{P}{B}$
$\tan 60^{\circ} = \frac{h}{AB}$ $\left ( \because \theta = 60^{\circ} \right )$
$\sqrt{3}= \frac{h}{AB}$ $\left ( \because \tan 60^{\circ}= \sqrt{3} \right )$
$AB= \frac{h}{\sqrt{3}}\: \cdots \left ( 1 \right )$
In $\bigtriangleup$EDC
$\tan \theta = \frac{P}{B}$
$\tan 45^{\circ} = \frac{h-10}{DC}$
$1= \frac{h-10}{AB}$ $\left ( \because AB-DC,\tan 45^{\circ}= 1 \right )$
$AB= h-10\: \cdots \left ( 2 \right )$
from equation (1) and (2)
$h-10= \frac{h}{\sqrt{3}}$
$\sqrt{3}h-\sqrt{3}10= h$
$\sqrt{3}h-h= \sqrt{3}\times 10$
$h\left ( \sqrt{3}-1 \right )= 10\sqrt{3}$
$h= \frac{10\sqrt{3}}{\sqrt{3}-1}$
Hence the height of the tower
$h= \frac{10\sqrt{3}}{\sqrt{3}-1}m$
Question:17
Answer:
Solution. According to the question :
Let the height of the other house is X.
In $\bigtriangleup$DCF.
$\tan \alpha = \frac{P}{B}= \frac{EC}{DC}$
$\tan \alpha =\frac{x-h}{DC}$
$DC= \frac{x-h}{\tan \alpha }\cdots \left ( 1 \right )$
In $\bigtriangleup$DAB
$\tan \beta = \frac{P}{B}= \frac{DA}{AB}$
$\tan \beta = \frac{h}{AB}= \frac{h}{DC}$ $\left ( \because AB= DC \right )$
$\tan \beta = \frac{h}{DC}$
$DC= \frac{h}{\tan \beta }\cdots \left ( 2 \right )$
from equation (1) and (2)
$\frac{X-h}{\tan \alpha}= \frac{h}{\tan \beta}$
$X-h= \frac{\tan \alpha h}{\tan \beta }$
$X= \frac{\tan \alpha h+\tan \beta h}{\tan \beta}$
$X= \frac{h\left ( \tan \alpha +\tan \beta \right )}{\tan \beta}$
separately divide
$X= h\left ( \frac{\tan \alpha }{\tan \beta }+1 \right )$
$X= h\left ( 1+\tan \alpha \cot \beta \right )$ $\left ( \because \frac{1}{\tan \theta }= \cot \theta \right )$
$X= h\left ( 1+\tan \alpha \cot \beta \right )$
Hence proved
Question:18
Answer:
Solution
Let y be the height of the balloon from the second window
In $\bigtriangleup$AOB
$\tan 30= \frac{y}{d} \; \; \left ( \because \tan \theta = \frac{Perpendicular}{Base} \right )$
$d= y\sqrt{3}\; \cdots \left ( 1 \right )$ $\left ( \because \tan 30= 1\sqrt{3} \right )$
In $\bigtriangleup$OCD
$\tan 60= \frac{4+y}{d}$
$\sqrt{3}d= 4+y$
$d= \frac{4+y}{\sqrt{3}}\cdots \left ( 2 \right )$
Equating equation (1) & (2) we get
$y\sqrt{3}= \frac{4+y}{\sqrt{3}}$
$y\sqrt{3}\times \sqrt{3}= 4+y$
$3y-y= 4$
$2y= 4$
$y= \frac{4}{2}$
y = 2
Height of balloon = 2 + 4 + y
= 2 + 4 + 2
= 8m
NCERT Exemplar Class 10 Maths Solutions Chapter 8 pdf downloads are available through online tools for the students to access this content in an offline version, so that no breaks in continuity are faced while practising NCERT Exemplar Class 10 Maths Chapter 8.
These Class 10 Maths NCERT exemplar chapter 8 solutions provide a basic knowledge of Trigonometric ratios, which has great importance in higher classes.
The questions based on Trigonometric ratios can be practised in a better way, along with these solutions.
The NCERT exemplar Class 10 Maths chapter 8 solution Introductions to Trigonometry and Its Equations has a good amount of problems for practice and is sufficient for a student to easily sail through other books such as NCERT Class 10 Maths, RD Sharma Class 10 Maths, and RS Aggarwal Class 10 Maths.
Here are the subject-wise links for the NCERT solutions of class 10:
Given below are the subject-wise NCERT Notes of class 10 :
Here are some useful links for NCERT books and NCERT syllabus for class 10:
Given below are the subject-wise exemplar solutions of class 10 NCERT:
We know that sine is the ratio of perpendicular and hypotenuse.
We know that cos is the ratio of base and hypotenuse.
Therefore, we can say that for complementary angles sine and cosine will give the same values.
We know that sine is the ratio of perpendicular and hypotenuse. Hypotenuses cannot have a smaller length on the perpendicular side; hence, the maximum possible value can be one.
The chapter Introduction to Trigonometry & Its Equations is quite important for Board exams as it carries around 8-10% weightage of the whole paper.
Generally, MCQs, Very short, Short, and Long answers type of questions are asked in the board examinations and NCERT exemplar Class 10 Maths solutions chapter 8 are adequate to score well in this chapter.
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Hello
Since you are a domicile of Karnataka and have studied under the Karnataka State Board for 11th and 12th , you are eligible for Karnataka State Quota for admission to various colleges in the state.
1. KCET (Karnataka Common Entrance Test): You must appear for the KCET exam, which is required for admission to undergraduate professional courses like engineering, medical, and other streams. Your exam score and rank will determine your eligibility for counseling.
2. Minority Income under 5 Lakh : If you are from a minority community and your family's income is below 5 lakh, you may be eligible for fee concessions or other benefits depending on the specific institution. Some colleges offer reservations or other advantages for students in this category.
3. Counseling and Seat Allocation:
After the KCET exam, you will need to participate in online counseling.
You need to select your preferred colleges and courses.
Seat allocation will be based on your rank , the availability of seats in your chosen colleges and your preferences.
4. Required Documents :
Domicile Certificate (proof that you are a resident of Karnataka).
Income Certificate (for minority category benefits).
Marksheets (11th and 12th from the Karnataka State Board).
KCET Admit Card and Scorecard.
This process will allow you to secure a seat based on your KCET performance and your category .
check link for more details
https://medicine.careers360.com/neet-college-predictor
Hope this helps you .
Hello Aspirant, Hope your doing great, your question was incomplete and regarding what exam your asking.
Yes, scoring above 80% in ICSE Class 10 exams typically meets the requirements to get into the Commerce stream in Class 11th under the CBSE board . Admission criteria can vary between schools, so it is advisable to check the specific requirements of the intended CBSE school. Generally, a good academic record with a score above 80% in ICSE 10th result is considered strong for such transitions.
hello Zaid,
Yes, you can apply for 12th grade as a private candidate .You will need to follow the registration process and fulfill the eligibility criteria set by CBSE for private candidates.If you haven't given the 11th grade exam ,you would be able to appear for the 12th exam directly without having passed 11th grade. you will need to give certain tests in the school you are getting addmission to prove your eligibilty.
best of luck!
According to cbse norms candidates who have completed class 10th, class 11th, have a gap year or have failed class 12th can appear for admission in 12th class.for admission in cbse board you need to clear your 11th class first and you must have studied from CBSE board or any other recognized and equivalent board/school.
You are not eligible for cbse board but you can still do 12th from nios which allow candidates to take admission in 12th class as a private student without completing 11th.
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