Exercise-6.1

Q1. In Fig. 6.1, lines AB and CD intersect at O. If \(\angle AOC + \angle BOE = 70^\circ\) and \(\angle BOD = 40^\circ\), find \(\angle BOE\) and reflex \(\angle COE\).

Solution:

Given: $$\begin{aligned}\angle AOC+\angle BOE=70^{\circ }\\ \angle BOD=40^{\circ}\end{aligned}$$ To find and \(\angle BOE\) and reflex \(\angle COE\)
Solution: $$\angle AOC+\angle COB=180^{\circ }\tag{1}$$ (Linear pair)
But $$\angle COB=\angle COE+\angle BOE$$

Substituting value of \(\angle COB\) in eqn (1)

$$\begin{aligned}\angle AOC+\angle COB&=180^{\circ}\\ \angle AOC+\angle COE+\angle BOE&=180^{\circ}\\ \angle AOC+\angle BOE+\angle COE&=180^{\circ}\\ 70^{0}+\angle COE&=180^{\circ}\\ \angle COE&=180^{0}-70^{\circ}\\ &=110^{\circ}\end{aligned}$$ Ray EO stand on line CD $$\therefore \angle OE+\angle EOD=180^{\circ}$$ (linear pair) $$\begin{aligned}110^\circ+\angle EOD=180^{\circ}\\ \angle EOD=180^{\circ}-110^{\circ}\\ \angle EOD=70^{\circ}\\ \angle EOD=\angle BOE+\angle BOD\\ 70^\circ=\angle BOE+40^{\circ}\\ \Rightarrow \angle BOE=70^{0}-40^{\circ }\\ =30^{0}\end{aligned}$$ $$\angle COE=110^{\circ }$$

Reflex of \(\angle COE\)

$$\begin{aligned}\angle COE=360^{\circ}-110^{\circ}\\ =250^{\circ}\end{aligned}$$

Q2. In Fig. 6.2, lines XY and MN intersect at O. If \(\angle POY = 90°\) and \(a : b = 2 : 3\), find \(c\).

Solution:

Line XY and line MN intersect each other at point \(O\) $$\therefore \angle XON=\angle MOY$$ vertically opposite Anges $$\angle MOY+\angle XOM=180^{\circ }$$ (linear pair) $$\small\begin{aligned}\angle MOY&= \angle MOP+\angle POY\\ \angle MOP + \angle POY+\angle XOM&=180^{\circ}\\ \angle MOP + \angle XOM&=180^{\circ}-90^{\circ}\\ \angle MOP + \angle XOM&=90^{\circ}\\ a+b&=90^{\circ}\\ a:b&=2:3\\ a&=\dfrac{2}{5}\times 90^{\circ}\\ &=36^{\circ}\\ b&=\dfrac{3}{5}\times 90^{\circ}\\ &=54\\ \angle XON&=\angle MOP+\angle POY\\ &=a+90^{\circ}\\ &=36^{\circ}+90^{\circ}\\ &=126^{\circ}\end{aligned}$$

Q3. In Fig. 6.3, ∠ PQR = ∠ PRQ, then prove that \(\angle PQS = \angle PRT\)

Solution:

Given that $$\angle PQR=\angle PRQ$$ To Prove: $$\angle PQS=\angle PRT$$ Line PQ stand upon line SR $$\therefore \angle PQS\ + \angle PQR=180^{\circ}\tag{1}$$ (linear Angle Pair) Line PR stand up on line SR $$\therefore \angle PRT+\angle PRQ=180^{\circ}\tag{2}$$ Equating Equation (1) and (2) $$\angle PQS+\angle PQR=\angle PRT+\angle PRQ$$ But $$\begin{aligned}\angle PQR&=\angle PRQ\\ \angle PQS+\angle PQR&=\angle PRT+\angle PQR\\ \therefore \angle PQS&=\angle PRT\end{aligned}$$ Hence Proved.

Q4. In Fig. 6.4, if \(x + y = w + z\), then prove that AOB is a line.

Solution:

Given: $$x+y=w+z$$ To prove:
AB is a line, we need to prove that $$x+y=180^{\circ }$$ Sum of angles around point \(O = 360^\circ\) $$\begin{aligned}x+y+w+z&=360^{\circ }\\ \text{Let } x+y=w+z&=k\\ k+k&=360^{0}\\ 2k&=360^{0}\\ k&=180^{0}\\ \Rightarrow x+y&=180^{0}\\ w+z&=180^{0}\end{aligned}$$ Hence line AB is a straight line
Proved.

Q5. In Fig. 6.5, POQ is a line. Ray OR is perpendicular to line PQ. OS is another ray lying between rays OP and OR. Prove that \(\angle ROS = \frac{1}{2}( \angle QOS – \angle POS)\)

Solution:

Given:
POQ is a line $$OR\,\bot\, PQ$$ To prove $$\angle ROS=\dfrac{1}{2}\left( \angle QOS-\angle POS\right) $$ $$\angle ROQ+\angle ROP=180^{\circ }$$ (Linear Pair) $$\begin{align}\angle ROQ&=180-90^{\circ}\\ &=90^{\circ }\\ \angle ROS+\angle POS&=90^{\circ}\tag{1}\end{align}$$ Ray OS stands upon line PQ $$\therefore \angle POS+\angle QOS=180^{\circ}$$ But $$\begin{align}\angle QOS&=\angle ROS+LROQ\\ \angle QOS&=\angle ROS+90^{\circ }\\ \Rightarrow \angle QOS-\angle ROS&=90^{\circ}\tag{2}\end{align}$$ Substituting (1) in (2) $$\begin{aligned}\angle ROS+\angle POS&=\angle QOS-\angle ROS\\ 2\angle ROS&=\angle QOS-\angle POS\\ \angle ROS&=\dfrac{1}{2}\left[ \angle QOS-\angle POS\right] \end{aligned}$$ Proved.

Q6. It is given that \(\angle XYZ = 64^\circ\) and \(XY\) is produced to point \(P\). Draw a figure from the given information. If ray \(YQ\) bisects \(\angle ZYP\), find \(\angle XYQ\) and reflex \(\angle QYP.\)

Solution:

Given $$\angle XYZ=64^{\circ }$$

Construction: \(XY\) produced to point \(P\)
Ray \(YQ\) bisect \(\angle ZYP\)

$$\therefore \angle PYZ=\angle QYZ=x\\$$

Line \(YZ\) stand upon \(XP\)

$$\therefore \angle ZYP+\angle XYZ=180^{0}\\ \left( LinearPair\right)$$ $$\small\begin{aligned}\therefore \angle ZYP+64^{\circ}&=180^{\circ}\\ \angle ZYP&=180^{\circ}-64^{\circ}\\ &=116^{\circ}\\ \Rightarrow 2x&=116^{\circ}\\ x&=58^{\circ}\\\\ \angle ZYQ&=58^{\circ}\\ \angle XYP&=\angle XYZ+\angle ZYQ\\ &=64^{\circ}+58^{\circ}\\ &=112^{\circ}\\ \angle QYP&=x\\ &=58^{\circ}\end{aligned}$$ Reflex of $$\begin{aligned}\angle QYP &=360^{\circ}-58^{\circ}\\ &=302^{\circ}\end{aligned}$$