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Q1

In Fig. 6.13, lines $AB$ and $CD$ intersect at $O$. If $\mathrm{\angle }AOC+\mathrm{\angle }BOE={70}^{\circ }$ and $\mathrm{\angle }BOD={40}^{\circ }$, find $\mathrm{\angle }BOE$ and reflex $\mathrm{\angle }COE$.

Solution. When two lines intersect, vertically opposite angles are equal. So

$$\mathrm{\angle }AOC=\mathrm{\angle }BOD={40}^{\circ }\mathrm{.}$$

Given $\mathrm{\angle }AOC+\mathrm{\angle }BOE={70}^{\circ }$, we get

$${40}^{\circ }+\mathrm{\angle }BOE={70}^{\circ }⟹\mathrm{\angle }BOE={30}^{\circ }\mathrm{}$$

Now $\mathrm{\angle }BOC$ is a linear pair with $\mathrm{\angle }AOC$, so

$$\mathrm{\angle }BOC={180}^{\circ }-\mathrm{\angle }AOC={180}^{\circ }-{40}^{\circ }={140}^{\circ }\mathrm{}$$

$\mathrm{\angle }COE$ (the smaller angle between $CO$ and $OE$) equals $\mathrm{\angle }COB+\mathrm{\angle }BOE={140}^{\circ }+{30}^{\circ }={170}^{\circ }$
Therefore the reflex $\mathrm{\angle }COE={360}^{\circ }-{170}^{\circ }={\mathbf{190}}^{\circ }\mathrm{}$

Answers: $\mathrm{\angle }BOE={\mathbf{30}}^{\circ },\text{ reflex }\mathrm{\angle }COE={\mathbf{190}}^{\circ }\mathrm{<span\; class="base"\; style="font-size:\; revert;\; font-family:\; var(--wp--preset--font-family--manrope);\; letter-spacing:\; -0.1px;"><span\; class="mpunct">,</span><span\; class="mspace"> </span><span\; class="mord\; text"><span\; class="mord">reflex </span></span><span\; class="mord">\angle </span><span\; class="mord\; mathnormal">COE</span><span\; class="mrel">=</span></span><span\; class="base"\; style="font-size:\; revert;\; font-family:\; var(--wp--preset--font-family--manrope);\; letter-spacing:\; -0.1px;"><span\; class="mord"><span\; class="mord\; mathbf">19</span><span\; class="mord\; mathbf">0</span><span\; class="msupsub"><span\; class="vlist-t"><span\; class="vlist-r"><span\; class="vlist"><span\; class="sizing\; reset-size6\; size3\; mtight"><span\; class="mbin\; mtight">\circ </span></span></span></span></span></span></span><span\; class="mord">.</span></span>}$

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Q2

In Fig. 6.14, lines $XY$ and $MN$ intersect at $O$. If $\mathrm{\angle }POY={90}^{\circ }$ and $a:b=2:3$, find $c$.

Solution. From the figure the rays (in order around $O$) are $OP,OY,ON,OX,OM$. Note that $OP$ to $OX$ (via $OM$) is the right angle ${90}^{\circ }$; more directly,

$$\mathrm{\angle }POX=\mathrm{\angle }POM+\mathrm{\angle }MOX=a+b={90}^{\circ }\mathrm{.}$$

Given $a:b=2:3$, write $a=2k,b=3k$. Then $2k+3k=5k={90}^{\circ }\Rightarrow k={18}^{\circ }$. Thus

$$b=3k={54}^{\circ }\mathrm{<span\; style="font-size:\; revert;\; font-family:\; var(--wp--preset--font-family--manrope);\; letter-spacing:\; -0.1px;">.</span>}$$

Since $OM$ and $ON$ are opposite rays (they lie on the same straight line $MN$), we have $b+c={180}^{\circ }$. Hence

$$c={180}^{\circ }-b={180}^{\circ }-{54}^{\circ }={\mathbf{126}}^{\circ }\mathrm{.}$$

Answer: $\boxed{{\displaystyle c={126}^{\circ }}}$.

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Q3

In Fig. 6.15, $\mathrm{\angle }PQR=\mathrm{\angle }PRQ$. Prove that $\mathrm{\angle }PQS=\mathrm{\angle }PRT$.

Solution. $QS$ and $RT$ are straight extensions of $QR$. So at $Q$,

$$\mathrm{\angle }PQS+\mathrm{\angle }PQR={180}^{\circ }<span\; style="font-size:\; revert;\; font-family:\; var(--wp--preset--font-family--manrope);\; letter-spacing:\; -0.1px;">,</span>$$

and at $R$,

$$\mathrm{\angle }PRT+\mathrm{\angle }PRQ={180}^{\circ }\mathrm{.}$$

Given $\mathrm{\angle }PQR=\mathrm{\angle }PRQ$, subtracting from 180° gives

$$\mathrm{\angle }PQS={180}^{\circ }-\mathrm{\angle }PQR={180}^{\circ }-\mathrm{\angle }PRQ=\mathrm{\angle }PRT\mathrm{.}$$

Thus $\mathrm{\angle }PQS=\mathrm{\angle }PRT$

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Q4

In Fig. 6.16, if $x+y=w+z$, prove that $A,O,B$ are collinear (i.e. AOB is a line).

Solution. The four angles around $O$ (in order) are $x,w,z,y$.

So

$$x+w+z+y={360}^{\circ }\mathrm{}$$

Given $x+y=w+z$. Put $x+y=w+z=t$

Then

$$t+t={360}^{\circ }\Rightarrow 2t={360}^{\circ }\Rightarrow t={180}^{\circ }\mathrm{.}$$

Thus $w+z={180}^{\circ }$. But $w+z$ is the angle from ray $OB$ to ray $OA$; since it equals ${180}^{\circ }$, $OB$ and $OA$ are opposite rays, so $A,O,B$ are collinear. Hence AOB is a line.

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Q5

In Fig. 6.17, $POQ$ is a line. Ray $OR$ is perpendicular to line $P\mathrm{Q }$Ray $OS$ lies between rays $OP$ and $OR$.

Prove$$\mathrm{\angle }ROS=\frac{1}{2}(\mathrm{\angle }QOS-\mathrm{\angle }POS)\mathrm{.}$$

Solution. Because $OR$ is perpendicular to line $PQ$,

$$\mathrm{\angle }QOR={90}^{\circ }\text{and}\mathrm{\angle }POR={90}^{\circ }\mathrm{.}$$

Now $\mathrm{\angle }QOS=\mathrm{\angle }QOR+\mathrm{\angle }ROS={90}^{\circ }+\mathrm{\angle }ROS$
Also $\mathrm{\angle }POS=\mathrm{\angle }POR-\mathrm{\angle }ROS={90}^{\circ }-\mathrm{\angle }ROS$ (since $OS$ lies between $OP$ and $OR$).
Subtract:

$$\mathrm{\angle }QOS-\mathrm{\angle }POS=({90}^{\circ }+\mathrm{\angle }ROS)-({90}^{\circ }-\mathrm{\angle }ROS)=2\mathrm{\angle }ROS\mathrm{.}$$

Therefore $\mathrm{\angle }ROS=\frac{1}{2}(\mathrm{\angle }QOS-\mathrm{\angle }POS)$

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Q6

It is given that $\mathrm{\angle }XYZ={64}^{\circ }$ and $XY$ is produced to point $P$. Ray $YQ$ bisects $\mathrm{\angle }ZYP$. Find $\mathrm{\angle }XYQ$ and reflex $\mathrm{\angle }QYP$.

Solution. Since $XY$ is produced to $P$, rays $YX$ and $YP$ are opposite, so

$$\mathrm{\angle }ZYP={180}^{\circ }-\mathrm{\angle }XYZ={180}^{\circ }-{64}^{\circ }={116}^{\circ }\mathrm{.}$$

Ray $YQ$ bisects $\mathrm{\angle }ZYP$, so each half is ${116}^{\circ }\mathrm{/}2={58}^{\circ }$.

Thus the small angle $\mathrm{\angle }QYP={58}^{\circ }$

Now $\mathrm{\angle }XYQ$ is the angle from $YX$ to $YQ$. Moving from $YX$ to $YQ$ goes through $YZ$ then to $YQ$, so

$$\mathrm{\angle }XYQ=\mathrm{\angle }XYZ+\mathrm{\angle }ZYQ={64}^{\circ }+{58}^{\circ }={\mathbf{122}}^{\circ }\mathrm{.}$$

Reflex $\mathrm{\angle }QYP$ is the larger reflex angle at $Y$ corresponding to the small angle ${58}^{\circ }$, so

$$\text{reflex }\mathrm{\angle }QYP={360}^{\circ }-{58}^{\circ }={\mathbf{302}}^{\circ }\mathrm{.}$$

Answers: $\mathrm{\angle }XYQ={122}^{\circ },\text{ reflex }\mathrm{\angle }QYP={302}^{\circ }\mathrm{}$