NCERT Solutions for Class 10 Maths Chapter 9 Some Applications of Trigonometry

Updated on Jan 13, 2021 - 11:33 a.m. IST by Dinesh Goyal #CBSE Class 10th

NCERT Solutions for Class 10 Maths Chapter 9 - Some Applications of Trigonometry is the 9th chapter of Maths which deals with the applications of trigonometry in our daily life. NCERT Class 10 maths solutions for chapter 9 consist of the answers to each question given in the 9th chapter of NCERT class 10 maths book. There is only one exercise in chapter 9 of NCERT class 10 maths solutions which contain 16 questions. In these NCERT solutions for class 10 maths chapter 9, students will learn how trigonometry is used for finding the distances and heights and various objects, without actually measuring them.
NCERT solutions for class 10 maths chapter 9 are designed by the subject experts to help students in their preparation of board exams. Also, check NCERT solutions for classes 6 to 12.

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NCERT Solutions for Class 10 Maths Chapter 9 - All Exercises

NCERT Solutions for Class 10 Maths Chapter 9 Some Applications of Trigonometry Excercise: 9.1

Q1 A circus artist is climbing a 20 m long rope, which is tightly stretched and tied from the top of a vertical pole to the ground. Find the height of the pole, if the angle made by the rope with the ground level is 30° (see Fig. 9.11).

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Answer:

Given that,
The length of the rope (AC) = 20 m. and $\angle ACB$ $=30^o$
Let the height of the pole (AB) be $h$

So, in the right triangle $\Delta ABC$

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By using the Sin rule
$\sin \theta = \frac{P}{H} =\frac{AB}{AC}$
$\sin 30^o =\frac{h}{20}$
$h =10$ m.
Hence the height of the pole is 10 m.

Q2 A tree breaks due to storm and the broken part bends so that the top of the tree touches the ground making an angle 30° with it. The distance between the foot of the tree to the point where the top touches the ground is 8 m. Find the height of the tree.

Answer:

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Suppose DB is a tree and the AD is the broken height of the tree which touches the ground at C.
GIven that,
$\angle ACB = 30^o$ , BC = 8 m
let AB = $x$ m and AD = $y$ m
So, AD+AB = DB = $x+y$

In right angle triangle $\Delta ABC$ ,
$\tan \theta = \frac{P}{B}=\frac{x}{8}$
$\tan 30^o =\frac{x}{8}=\frac{1}{\sqrt{3}}$
So, the value of $x$ = $8/\sqrt{3}$

Similarily,
$\cos 30^o = \frac{BC}{AC} = \frac{8}{y}$
the value of $y$ is $16/\sqrt{3}$

So, the total height of the tree is-

$x+y=\frac{24}{\sqrt{3}}=8\sqrt{3}$

= 8 (1.732) = 13.856 m (approx)

Q3 A contractor plans to install two slides for the children to play in a park. For the children below the age of 5 years, she prefers to have a slide whose top is at a height of 1.5 m, and is inclined at an angle of 30° to the ground, whereas for elder children, she wants to have a steep slide at a height of 3m, and inclined at an angle of 60° to the ground. What should be the length of the slide in each case?

Answer:

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Suppose $x$ m is the length of slides for children below 5 years and the length of slides for elders children be $y$ m.

Given that,
AF = 1.5 m, BC = 3 m, $\angle AEF = 30^o$ and $\angle BDC = 60^o$

In triangle $\Delta$ EAF,
$\sin \theta = \frac{AF}{EF} = \frac{1.5}{x}$
$\sin 30^o = \frac{1.5}{x}$
The value of $x$ is 3 m.

Similarily in $\Delta$ CDB,
$\sin 60^o = \frac{3}{y}$
$\frac{\sqrt{3}}{2}= \frac{3}{y}$
the value of $y$ is $2\sqrt{3}$ = 2(1.732) = 3.468

Hence the length of the slide for children below 5 yrs. is 3 m and for the elder children is 3.468 m.

Q4 The angle of elevation of the top of a tower from a point on the ground, which is 30 m away from the foot of the tower, is 30°. Find the height of the tower.

Answer:

1594290279361
Let the height of the tower AB is $h$ and the angle of elevation from the ground at point C is $\angle ACB = 30^o$
According to question,
In the right triangle $\Delta ABC$ ,
$\tan \theta = \frac{AB}{BC} = \frac{h}{30}$
$\tan 30^o =\frac{1}{\sqrt{3}}=\frac{h}{30}$
the value of $h$ is $10\sqrt{3}$ = 10(1.732) = 17.32 m
Thus the height of the tower is 17.32 m

Q5 A kite is flying at a height of 60 m above the ground. The string attached to the kite is temporarily tied to a point on the ground. The inclination of the string with the ground is 60°. Find the length of the string, assuming that there is no slack in the string.

Answer:

A
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Given that,
The length of AB = 60 m and the inclination of the string with the ground at point C is $\angle ACB = 60^o$ .
Let the length of the string AC be $l$ .
According to question,
In right triangle $\Delta$ CBA,
$\sin 60^{o} = \frac{AB}{AC} = \frac{60}{l}$
$\frac{\sqrt{3}}{2} = \frac{60}{l}$
The value of length of the string ( $l$ ) is $40\sqrt{3}$ = 40(1.732) = 69.28 m
Hence the length of the string is 69.28 m.

Q6 A 1.5 m tall boy is standing at some distance from a 30 m tall building. The angle of elevation from his eyes to the top of the building increases from 30° to 60° as he walks towards the building. Find the distance he walked towards the building.

Answer:

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Given that,
The height of the tallboy (DC) is 1.5 m and the height of the building (AB) is 30 m.
$\angle ADF = 30^o$ and $\angle AEF = 60^o$

According to question,
In right triangle AFD,
$\\\Rightarrow \tan 30^o=\frac{AF}{DF} = \frac{28.5}{DF}\\\\\Rightarrow \frac{1}{\sqrt{3}} = \frac{28.5}{DF}$
So, DF = $(28.5)\sqrt{3}$

In right angle triangle $\Delta AFE$ ,
$\tan 60^o =\frac{AF}{FE}=\frac{28.5}{EF}$
$\sqrt{3}=\frac{28.5}{EF}$
EF = $9.5\sqrt{3}$

So, distance walked by the boy towards the building = DF - EF = $19\sqrt{3}$

Q7 From a point on the ground, the angles of elevation of the bottom and the top of a transmission tower fixed at the top of a 20 m high building are 45° and 60° respectively. Find the height of the tower.

Answer:

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Suppose BC = $h$ is the height of transmission tower and the AB be the height of the building and AD is the distance between the building and the observer point (D).
We have,
AB = 20 m, BC = $h$ m and AD = $x$ m
$\angle CDA = 60^o$ and $\angle BDA = 45^o$

According to question,
In triangle $\Delta$ BDA,
$\tan 45^o = \frac{AB}{AD}=\frac{20}{x}$
So, $x$ = 20 m

Again,
In triangle $\Delta$ CAD,

$\\\Rightarrow \tan 60^o = \frac{AB+BC}{AD}=\frac{20+h}{20}\\\\\Rightarrow \sqrt{3}= 1+\frac{h}{20}\\\\\Rightarrow h=20(\sqrt{3}-1)\\\\\Rightarrow 20(0.732) = 14.64 m$

Answer- the height of the tower is 14.64 m

Q8 A statue, 1.6 m tall, stands on the top of a pedestal. From a point on the ground, the angle of elevation of the top of the statue is 60° and from the same point, the angle of elevation of the top of the pedestal is 45°. Find the height of the pedestal.

Answer:

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Let the height of the pedestal be $h$ m. and the height of the statue is 1.6 m.
the angle of elevation of the top of the statue and top of the pedestal is( $\angle DCB=$ $60^o$ )and( $\angle ACB=$ $45^o$ ) respectively.

Now,
In triangle $\Delta ABC$ ,
$\tan 45^o =1 =\frac{AB}{BC}=\frac{h}{BC}$
therefore, BC = $h$ m

In triangle $\Delta CBD$ ,
$\\\Rightarrow \tan 60^o = \frac{BD}{BC}=\frac{h+1.6}{h}\\\\\Rightarrow \sqrt{3}= 1+\frac{1.6}{h}$
the value of $h$ is $0.8(\sqrt{3}+1)$ m
Hence the height of the pedestal is $0.8(\sqrt{3}+1)$ m

Q9 The angle of elevation of the top of a building from the foot of the tower is 30° and the angle of elevation of the top of the tower from the foot of the building is 60°. If the tower is 50 m high, find the height of the building.

Answer:

1594290421503
It is given that, the height of the tower (AB) is 50 m. $\angle AQB = 30^o$ and $\angle PBQ = 60^o$
Let the height of the building be $h$ m

According to question,
In triangle PBQ,
$\tan 60^o = \frac{PQ}{BQ} = \frac{50}{BQ}$
$\\\sqrt{3} = \frac{50}{BQ}\\ BQ = \frac{50}{\sqrt{3}}$ .......................(i)

In triangle ABQ,

$\tan 30^o = \frac{h}{BQ}$
${BQ}=h\sqrt{3}$ .........................(ii)
On equating the eq(i) and (ii) we get,

$\frac{50}{\sqrt{3}}=h\sqrt{3}$
therefore, $h$ = 50/3 = 16.66 m = height of the building.

Q10 Two poles of equal heights are standing opposite each other on either side of the road, which is 80 m wide. From a point between them on the road, the angles of elevation of the top of the poles are 60° and 30°, respectively. Find the height of the poles and the distances of the point from the poles.

Answer:

1594290440253
Given that,
The height of both poles are equal DC = AB. The angle of elevation of of the top of the poles are $\angle DEC=30^o$ and $\angle AEB=60^o$ resp.
Let the height of the poles be $h$ m and CE = $x$ and BE = 80 - $x$

According to question,
In triangle DEC,

$\\\Rightarrow \tan 30^o = \frac{DC}{CE} = \frac{h}{x}\\\\\Rightarrow \frac{1}{\sqrt{3}}= \frac{h}{x}\\\\\Rightarrow x=h\sqrt{3}$ ..............(i)

In triangle AEB,
$\\\Rightarrow \tan 60^o = \frac{AB}{BE}=\frac{h}{80-x}\\\\\Rightarrow \sqrt{3}=\frac{h}{80-x}\\\\\Rightarrow x=80 - \frac{h}{\sqrt{3}}$ ..................(ii)
On equating eq (i) and eq (ii), we get

$\sqrt{3}h=80 - \frac{h}{\sqrt{3}}$
$\frac{h}{\sqrt{3}}=20$
$h=20\sqrt{3}$ m
So, $x$ = 60 m

Hence the height of both poles is ( $h=20\sqrt{3}$ )m and the position of the point is at 60 m from the pole CD and 20 m from the pole AB.

Q11 A TV tower stands vertically on a bank of a canal. From a point on the other bank directly opposite the tower, the angle of elevation of the top of the tower is 60°. From another point 20 m away from this point on the line joining this point to the foot of the tower, the angle of elevation of the top of the tower is 30° (see Fig. 9.12). Find the height of the tower and the width of the canal.

1594290461635

Answer:

1594290476491
Suppose the $h$ is the height of the tower AB and BC = $x$ m
It is given that, the width of CD is 20 m,
According to question,

In triangle $\Delta ADB$ ,
$\\\Rightarrow \tan 30^o = \frac{AB}{20+x}=\frac{h}{20+x}\\\\\Rightarrow \frac{1}{\sqrt{3}}=\frac{h}{20+x}\\\\\Rightarrow 20+x = h\sqrt{3}\\\\\Rightarrow x = h\sqrt{3}-20$ ............(i)

In triangle ACB,
$\\\Rightarrow \tan 60^o = \frac{h}{x}=\sqrt{3}\\\\\Rightarrow x= \frac{h}{\sqrt{3}}$ .............(ii)

On equating eq (i) and (ii) we get:

$h\sqrt{3}-20= \frac{h}{\sqrt{3}}$
from here we can calculate the value of $h=10\sqrt{3}= 10 (1.732) = 17.32\: m$ and the width of the canal is 10 m.

Q12 From the top of a 7 m high building, the angle of elevation of the top of a cable tower is 60° and the angle of depression of its foot is 45°. Determine the height of the tower.

Answer:

1594290494703
Let the height of the cable tower be (AB = $h+7$ )m
Given,
The height of the building is 7 m and angle of elevation of the top of the tower $\angle ACE = 60^o$ , angle of depression of its foot $\angle BCE = 45^o$ .

According to question,

In triangle $\Delta DBC$ ,
$\\\tan 45^o = \frac{CD}{BD} = \frac{7}{BD} = 1\\ BD =7\ m$
since DB = CE = 7 m

In triangle $\Delta ACE$ ,

$\\\tan 60^o = \frac{h}{CE}=\frac{h}{7}=\sqrt{3}\\ \therefore h = 7\sqrt{3}\ m$

Thus, the total height of the tower equal to $h+7$ $=7(1+\sqrt{3}\) m$

Q13 As observed from the top of a 75 m high lighthouse from the sea-level, the angles of depression of two ships are 30° and 45°. If one ship is exactly behind the other on the same side of the lighthouse, find the distance between the two ships.

Answer:

Given that,
The height of the lighthouse (AB) is 75 m from the sea level. And the angle of depression of two different ships are $\angle ADB = 30^0$ and $\angle ACB = 45^0$ respectively
1594290544799
Let the distance between both the ships be $x$ m.
According to question,

In triangle $\Delta ADB$ ,

$\tan 30^0 = \frac{AB}{BD}=\frac{75}{x+y} = \frac{1}{\sqrt{3}}$
$\therefore x+y = 75 \sqrt{3}$ .............(i)

In triangle $\Delta ACB$ ,

$\tan 45^0 = 1 =\frac{75}{BC}=\frac{75}{y}$
$\therefore y =75\ m$ .............(ii)

From equation (i) and (ii) we get;
$x = 75(\sqrt{3}-1)=75(0.732)$
$x = 54.9\simeq 55\ m$

Hence, the distance between the two ships is approx 55 m.

Q14 A 1.2 m tall girl spots a balloon moving with the wind in a horizontal line at a height of 88.2 m from the ground. The angle of elevation of the balloon from the eyes of the girl at any instant is 60°. After some time, the angle of elevation reduces to 30° (see Fig. 9.13). Find the distance traveled by the balloon during the interval.

1594290601110

Answer:

Given that,
The height of the girl is 1.2 m. The height of the balloon from the ground is 88.2 m and the angle of elevation of the balloon from the eye of the girl at any instant is ( $\angle ACB =60^0$ ) and after some time $\angle DCE =30^0$ .
1594290642418
Let the $x$ distance travelled by the balloon from position A to position D during the interval.
AB = ED = 88.2 - 1.2 =87 m

Now, In triangle $\Delta BCA$ ,

$\\\tan 60^0 = \sqrt{3}=\frac{AB}{BC}=\frac{87}{BC}\\ \therefore BC = 29\sqrt{3}$

In triangle $\Delta DCE$ ,

$\\\tan 30^0 = \frac{1}{\sqrt{3}}=\frac{DE}{CE}=\frac{87}{CE}\\ \therefore CE = 87\sqrt{3}$

Thus, distance traveled by the balloon from position A to D

$= CE - BC =87\sqrt{3}-29\sqrt{3} = 58\sqrt{3}$ m

Q15 A straight highway leads to the foot of a tower. A man standing at the top of the tower observes a car at an angle of depression of 30°, which is approaching the foot of the tower with a uniform speed. Six seconds later, the angle of depression of the car is found to be 60°. Find the time taken by the car to reach the foot of the tower from this point.

Answer:

1594290668622
Let $h$ be the height of the tower (DC) and the speed of the car be $x\ ms^{-1}$ . Therefore, the distance (AB)covered by the car in 6 seconds is 6 $x$ m. Let $t$ time required to reach the foot of the tower. So, BC = $x$ $t$

According to question,
In triangle $\Delta DAC$ ,
$\\\tan 30^0 = \frac{1}{\sqrt{3}}=\frac{h}{6x+xt}\\ x(6+t) = h\sqrt{3}$ ..........................(i)

In triangle $\Delta BCD$ ,

$\\\tan 60^0 = \sqrt{3} = \frac{h}{xt}\\ \therefore h = 3.xt$ ...................(ii)

Put the value of $h$ in equation (i) we get,
$\\x(6+t) = (\sqrt{3}.\sqrt{3})xt\\ 6x +xt = 3xt\\ 6x = 2xt$
$t = 3$

Hence, from point B car take 3 sec to reach the foot of the tower.

Q16 The angles of elevation of the top of a tower from two points at a distance of 4 m and 9 m from the base of the tower and in the same straight line with it are complementary. Prove that the height of the tower is 6 m.

Answer:

1594290687445
Let the height of the tower be $h$ m.
we have PB = 4m and QB = 9 m
Suppose $\angle BQA = \theta$ , so $\angle APB =90- \theta$

According to question,

In triangle $\Delta ABQ$ ,

$\\\tan \theta = \frac{h}{9}\\ \therefore h = 9 \tan \theta$ ..............(i)

In triangle $\Delta ABP$ ,

$\\\tan (90-\theta)=\cot \theta = \frac{h}{4}\\ \therefore h = 4\cot \theta$ .....................(ii)

multiply the equation (i) and (ii), we get

$\\h^2 = 36\\ \Rightarrow h = 6 m$

Hence the height of the tower is 6 m.

NCERT Solutions for Class 10 Maths for Other Chapters

Chapter No.	Chapter Name
Chapter 1	NCERT solutions for class 10 maths chapter 1 Real Numbers
Chapter 2	NCERT solutions for class 10 maths chapter 2 Polynomials
Chapter 3	NCERT solutions for class 10 maths chapter 3 Pair of Linear Equations in Two Variables
Chapter 4	NCERT solutions for class 10 maths chapter 4 Quadratic Equations
Chapter 5	NCERT solutions for class 10 chapter 5 Arithmetic Progressions
Chapter 6	NCERT solutions for class 10 maths chapter 6 Triangles
Chapter 7	NCERT solutions for class 10 maths chapter 7 Coordinate Geometry
Chapter 8	NCERT solutions for class 10 maths chapter 8 Introduction to Trigonometry
Chapter 9	NCERT solutions for class 10 maths chapter 9 Some Applications of Trigonometry
Chapter 10	NCERT solutions class 10 maths chapter 10 Circles
Chapter 11	NCERT solutions for class 10 maths chapter 11 Constructions
Chapter 12	NCERT solutions for class 10 chapter maths chapter 12 Areas Related to Circles
Chapter 13	NCERT solutions class 10 maths chapter 13 Surface Areas and Volumes
Chapter 14	NCERT solutions for class 10 maths chapter 14 Statistics
Chapter 15	NCERT solutions for class 10 maths chapter 15 Probability

Know more about NCERT Solutions for Class 10 Maths Chapter 9

As per NCERT Class 10 Chapter 9 Maths solutions, Trigonometry is one of the most ancient topics studied by scholars all over the world. This chapter introduces new terms like the line of sight, angle of depression and angle of elevation. NCERT solutions for class 10 maths chapter 9 will give you assistance while practising the questions of the practice exercises including optional exercise.

Let's understand a few terms used in this chapter through the diagram.

The line of sight- It is an imaginary line drawn from the observer's eye to the object viewed by the observer.
The angle of elevation- The angle of elevation of an object viewed by an observer, is the angle formed by the line of sight with the horizontal level when the object is above the horizontal level,

The angle of elevation

The angle of depression- The angle of depression of an object viewed by an observer, is the angle formed by the line of sight with the horizontal level when the object is below the horizontal level.

The angle of depression

How to use NCERT Solutions for Class 10 Maths Chapter 9?

Before coming to this chapter, please ensure that you have completed the previous chapter.
Read the conceptual theory given in the NCERT textbook and have a look over some examples present in the textbook for the deeper understanding of the topics.
Once you complete the above points then you can jump to the practice exercises available in the book.
While practising the questions in practice exercises, you can use NCERT solutions for class 10 maths chapter 9 to know how to solve, what is the final answer and so on.
When you have done the practice exercise then the best thing to make your concepts strong is the last 5 years CBSE class 10 question papers.

Keep working hard & happy learning!

NCERT Solutions for Class 10 for other subjects.

Solutions

Frequently Asked Question (FAQs) - NCERT Solutions for Class 10 Maths Chapter 9 Some Applications of Trigonometry

Question: Does CBSE provides the solutions of NCERT class 10 ?

Answer:

No, CBSE doesn’t provided NCERT solutions for any class or subject.

Question: Where can I find the complete solutions of NCERT class 10 maths ?

Answer:

Here you will get the detailed NCERT solutions for class 10 maths by clicking on the link.

Question: How does the NCERT solutions are helpful in CBSE board exam?

Answer:

In CBSE board exam most of the questions are directly asked from NCERT textbook, so must know NCERT very well. In NCERT solutions, you will get are detailed solutions provided by the experts who knows how best to answer in the board exam in order to get good marks.

Question: What is the weightage of the chapter application of trigonometry in CBSE board exam ?

Answer:

Two chapters trigonometry and applications of trigonometry has 12 marks weightage in 10 board final exam.

Question: How many chapters are there in the class 10 maths ?

Answer:

There are 15 chapters in the class 10 maths NCERT.

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Questions related to CBSE Class 10th

Showing 31 out of 31 Questions

382 Views

Board cbse class 10 Is cancelled and promote all student please sir because of corona do not study

Aadarsh Gupta 9th Apr, 2021

Hello aspirant,

No CBSE and ICSE board didn't not cancelled the 10th and 12th exam.

As they had declared that 10th and 12th class exam will not be cancelled and students will have to give it in offline mode in this pandemic also.

Hope this helps you

All the best for your future

52 Views

CBSE class 10th malayalam can someone please tell me a good one

Ashish Kumar Verma 8th Apr, 2021

Hello Dear,

Malayalam is the mother tongue of those who have their roots in Kerala. It is one of the Dravidian languages that has similarities in words with its other contemporary languages such as Tamil and Telugu. People who speak Malayalam are referred to as Mallus. You can download the previous year papers from the link below:

https://school.careers360.com/download/sample-papers/cbse-10th-class-malayalam-solved-sample-paper-2021

Good Luck

64 Views

When will the Karnataka NTSE Stage 1 Merit list(i.e people who have cleared stage-1) and the cut off come out?

Souhrid Sarkar 23rd Mar, 2021

Hi Shreyas,

DSERT, Karnataka has released the NTSE Karnataka 2021 Result stage 1 on March 9, 2021 on their official website ( dsert.kar.nic.in ). It is available in the form of district-wise marks lists , consisting of registration numbers and marks of all students.

Along with the NTSE stage 1 result, cutoff and merit list of selected candidates is also released.

The students those who have cleared the stage 1 exam are eligible to seat for the stage 2 exam which is going to be held on June 13, 2021 . The scholarship will be given to the students who will pass the stage 2 examination.

To know more about NTSE Karnataka Important Dates, Cut-off, Scholarship visit :

https://www.google.com/amp/s/school.careers360.com/articles/ntse-karnataka-result/amp

Hope it helps
Thank You !!!

715 Views

Age criteria to appear in class 10 CBSE exam

Manisha 31st Jan, 2021

Hello Bishal

According to guidelines of CBSE, minimum age to appear for class x must be 14 years. There is no upper limit to appear for class x cbse board.

A candidate can appear for maximum three attempts.

Some candidates give private exams or sometimes students fail in standard ix then they privately appear for class x then their age must be more than 14 years. Sometimes students appear for x class after one year gap of passing class ix then also their age would be 15 or 16 as there is no upper limit age.

154 Views

My child is supposed to make another attempt at compartment exam of 10th cbse Missed last date for applying. since the boards are postponed to May 4th can I pay her fees and apply now.

Yash Dev Student Expert 15th Jan, 2021

Hello sir I'm sorry to inform you but now you're not eligible to fill the registration form as last date to fill the registration Form was 9th December 2020. Yes examination gets postponed to May but portal to fill the registration form is not open , in case if the registration form portal will open again you can fill the registration form and make the payment.

Feel free to comment if you've any doubt

Good luck