NCERT Solutions for Class 9 Maths Chapter 6 Exercise 6.2 - Lines and Angles

Q2. In Fig. 6.24, if AB $||$ CD, EF $\bot$ CD and $\angle$ GED = 126°, find $\angle$ AGE, $\angle$ GEF and $\angle$ FGE.

Solution:

Given,
AB || CD, EF $\perp$ CD and $\angle$ GED = $126^0$
In the above figure,
GE is transversal. So, that $\angle$ AGE = $\angle$ GED = $126^0$ [Alternate interior angles]
Also, $\angle$ GEF = $\angle$ GED - $\angle$ FED
$\angle$ GEF = $126^0-90^0 = 36^0$
$\angle GEF = 36^0$
Since AB is a straight line
Therefore, $\angle$ AGE + $\angle$ FGE = $180^0$
So, $\angle$ FGE = $180^0-\angle AGE
$\angle$ FGE = 180^0 - 126^0$
$\Rightarrow \angle FGE =54^0$

Q3. In Fig. 6.25, if PQ $||$ ST, $\angle$ PQR = 110° and $\angle$ RST = 130°, find $\angle$ QRS. [ Hint : Draw a line parallel to ST through point R.]

Solution:

Draw a line EF parallel to the ST through R.
Since PQ || ST and ST || EF
$\Rightarrow$ EF || PQ

$\angle$ PQR = $\angle$ QRF = $110^0$ (Alternate interior angles)$\angle$ QRF = $\angle$ QRS + $\angle$ SRF .............(i)
Again, $\angle$ RST + $\angle$ SRF = $180^0$ (Interior angles of two parallels ST and RF)
$\Rightarrow \angle SRF =180^0-130^0 = 50^0$ ( $\angle$ RST = $130^0$ , given)
Thus, $\angle$ QRS = $110^0-50^0 = 60^0$ (From equation (i))

Q4. In Fig. 6.26, if AB $||$ CD, $\angle$ APQ = 50° and $\angle$ PRD = 127°, find x and y .

Solution:
Given,
AB || CD, $\angle$ APQ = $50^0$ and $\angle$ PRD = $127^0$
PQ is a transversal. So,
$\angle$ APQ = $\angle$ PQR= $50^0$ (Alternate interior angles)
$\therefore x = 50^0$
Again, PR is a transversal. So,
$\angle$ y + $50^0$ = $127^0$ (Alternate interior angles)
$\Rightarrow y = 77^0$

Q5. In Fig. 6.27, PQ and RS are two mirrors placed parallel to each other. An incident ray AB strikes the mirror PQ at B, the reflected ray moves along the path BC and strikes the mirror RS at C and again reflects back along CD. Prove that AB $||$ CD.

Solution:

Draw a ray BL $\perp$ PQ and CM $\perp$ RSSince PQ || RS (Given)
So, BL || CM and BC is a transversal
$\therefore$ $\angle$ LBC = $\angle$ MCB (Alternate interior angles).............(i)
It is known that, angle of incidence = the angle of reflection
So, $\angle$ ABL = $\angle$ LBC and $\angle$ MCB = $\angle$ MCD
$\Rightarrow \angle ABL =\angle MCD$ ..................(ii)
Adding eq (i) and eq (ii), we get
$\angle$ ABC = $\angle$ DCB
Both the interior angles are equal
Hence AB || CD