Exercise-6.2

Q1. In Fig. 6.23, if AB || CD, CD || EF and y : z = 3 : 7, find x.

Solution:

Given $$\begin{aligned}AB\left| \right| CD\\ CD\left| \right| EF\\ \Rightarrow AB\left| \right| EF\\ y:z=3:7\end{aligned}$$ To find \(x\) $$y+z=180^{\circ }$$ (Angles same side of the transversal) $$\begin{aligned}y:z&=3:7\\ y&=\dfrac{3}{10}\times 180^{\circ}\\ &=54^{\circ}\\ z&=\dfrac{7}{10}\times 180^\circ\\ &=126^{\circ}\end{aligned}$$ $$x+y=180^{\circ }$$ (Angles same side of transversal line) $$\begin{aligned}x+54^{\circ}&=180^{\circ}\\ x&=180^{\circ}-54^{\circ}\\ &=126^{\circ}\end{aligned}$$

Q2. In Fig. 6.24, if AB || CD, EF ⊥ CD and ∠ GED = 126°, find ∠ AGE, ∠ GEF and ∠ FGE.

Solution:

Given $$\begin{aligned} AB\parallel CD\\ EF\bot CD\\ \angle GED=126^\circ\end{aligned}$$ To find

\(\angle GEF,\; \angle FGE \text{ and}\;\angle AGE\\\)

$$\begin{aligned} \angle GED&=\angle GEF+\angle FED\\ 126^\circ&=\angle GEF+90^{\circ}\\ \angle GEF&=126^{\circ}-90^{\circ}=36^{\circ}\\ \angle GED+\angle EFG+\angle FGE&=180^{\circ}\end{aligned}$$ (Sum of Internal Angles of Δ GEF) $$\begin{aligned}36^{\circ}+90^{\circ}+\angle FGE&=180^{\circ }\\ \angle FGE&=180^{\circ}-126^{\circ}\\ &=54^{\circ}\\ \angle AGE+\angle FGE&=180\end{aligned}$$ (linear pair) $$\begin{aligned}\angle AGE+54^{\circ }&=180^{\circ}\\ \angle AGE&=180^{\circ}-54^{\circ}\\ &=126^{\circ}\end{aligned}$$

Q3. In Fig. 6.25, if PQ || ST, ∠ PQR = 110° and ∠ RST = 130°, find ∠ QRS.

Solution:

Given $$\begin{aligned}PQ\left| \right| ST\\ \angle PQR=110^{\circ }\\ \angle RST=130^{\circ }\end{aligned}$$ To find $$\angle QRS$$ Construction: Draw a line AB || ST $$\angle RST=\angle SRA$$ (Alternate angle) $$\begin{align}\angle SRA=130^{\circ}\\ \Rightarrow \angle ARQ+\angle QRS=130^{\circ }\tag{1}\\ \angle ARQ+\angle PQR=180^{\circ}\end{align}$$ (Sum of angles on same side of transversal) $$\begin{aligned}\angle ARQ+110^{\circ}&=180^{\circ}\\ \angle ARQ&=180^{\circ }-110^{\circ}\\ \angle ARQ&=70^{\circ}\end{aligned}$$ Substituting value of \(\angle ARQ\) in Equation (1) $$\begin{aligned}\angle ARQ+\angle QRS&=130^{\circ }\\ \angle QRS&=130^{\circ }-\angle ARQ\\ &=130^{0}-70^{0}\\ &=60^{\circ }\end{aligned}$$

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

Solution:

Given $$\begin{aligned} AB\parallel CD\\ \angle APQ=50^\circ\\ \angle PRD=127^\circ\\ AB\parallel CD\end{aligned}$$ therefore $$\angle APQ=\angle PQR$$ (Alternate angles) $$\begin{aligned}2APQ&=x\\ 50^{\circ }&=x\\ \Rightarrow x&=50^{\circ}\end{aligned}$$ PR stands on line CD therefore $$\angle QRP+\angle PRD=180^{\circ }$$ (linear pair) $$\begin{aligned}\angle QRP&=180^{\circ}-127^{\circ}\\ &=53^{\circ}\\ \angle PQR+\angle QRP+2RPQ&=180^{\circ}\end{aligned}$$ (sum of internal angles of ΔPQR) $$\begin{aligned}50^{\circ}+53^{\circ}+\angle RPQ&=180^{\circ }\\ 103^{\circ}+y&=180^{\circ }\\ y&=180^{\circ}-103^{\circ}\\ &=77^{\circ}\\ x&=50^{\circ}\\ y&=77^{\circ}\end{aligned}$$

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:

Given $$PQ\parallel RR$$ Ray AB reflected from point B Ray BC reflected from point E
To prove
AB||CD
Constructions
line BE is drawn from point B to E such that BE ⟂ RS line CF is drawn from point a to F such that CF ⟂ PQ
Proof $$\angle ABE=\angle EBC\tag{1}$$ (Angle of incidence = Angle of reflection)
Similarly
$$\begin{align}\angle BCF&=\angle FCD\tag{2}\\ BE\parallel CF\\ \therefore \angle EBC&=\angle BCF\tag{3}\end{align}$$ (Alternate angles to transversal BC) Adding equations (1) and (3) $$\begin{align}\angle ABE+\angle EBC&=\angle EBC+\angle FCD\\ \angle ABC&=\angle EBC+\angle FCD\tag{4}\end{align}$$ Adding equations (2) and (3) $$\begin{align}\angle BCF+\angle EBC&=\angle FCD+\angle BCF\\ &= \angle BCD\tag{5}\end{align}$$ From equations (4) and (5) $$\angle ABC=\angle BCD$$ Alternate angles to the transversal BC are equal, therefore $$ AB\parallel CD$$