Exercise-6.2, Class 9th, Maths, Chapter 6, NCERT

Q1 (Fig. 6.23)

Given. $AB\parallel CD,CD\parallel EF$. Also $y:z=3:7$. Find $x$.

Reasoning (alternate angle-chase).
Because $CD\parallel EF$ and the labelled angle $z$ sits at the intersection of a transversal with line $EF$, the angle $z$ is equal to the angle made by the same transversal with line $CD$. Likewise, since $AB\parallel CD$, the corresponding angle on $AB$(labelled $x$) equals that angle too. So

$$x=z\mathrm{.}$$

The two adjacent interior angles $y$ and $z$ form a straight line, hence

$$y+z={180}^{\circ }\mathrm{.}$$

Write $y=3k,z=7k$. Then $3k+7k=10k={180}^{\circ }\Rightarrow k={18}^{\circ }$.

Thus

$$z=7k={126}^{\circ }\text{and}x=z={126}^{\circ }\mathrm{}$$

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

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Q2 (Fig. 6.24)

Given. $AB\parallel CD,EF\perp CD$. Also $\mathrm{\angle }GED={126}^{\circ }$. Find $\mathrm{\angle }AGE,\mathrm{\angle }GEF,\mathrm{\angle }FGE\mathrm{}$

Reasoning.

1. Line $GE$ meets the parallel pair $AB$ and $CD$. The angle $\mathrm{\angle }GED$ is the angle between $GE$ and $ED$ (where $ED$ lies on $CD$). The corresponding angle on the other parallel $AB$ (that is $\mathrm{\angle }AGE$) equals $\mathrm{\angle }GED$. So

$$\mathrm{\angle }AGE={126}^{\circ }\mathrm{.}$$

2. Because $EF\perp CD$ and $ED$ is along $CD$, the angle between $EF$ and $ED$ is ${90}^{\circ }$. The angle $\mathrm{\angle }GEF$ is the difference between $\mathrm{\angle }GED$ (GE vs ED) and the right angle (EF vs ED). So

$$\mathrm{\angle }GEF=\mathrm{\angle }GED-{90}^{\circ }={126}^{\circ }-{90}^{\circ }={36}^{\circ }\mathrm{.}$$

3. At point $G$, $\mathrm{\angle }AGE$ and $\mathrm{\angle }FGE$ are supplementary (they form a straight line along the points on the diagram), so

$$\mathrm{\angle }FGE={180}^{\circ }-\mathrm{\angle }AGE={180}^{\circ }-{126}^{\circ }={54}^{\circ }\mathrm{.}$$

Answers: $\boxed{{\displaystyle \mathrm{\angle }AGE={126}^{\circ },\mathrm{\angle }GEF={36}^{\circ },\mathrm{\angle }FGE={54}^{\circ }\mathrm{}}}$

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Q3 (Fig. 6.25)

Given. $PQ\parallel ST,\mathrm{\angle }PQR={110}^{\circ },\mathrm{\angle }RST={130}^{\circ }\mathrm{<span\; class="katex"\; style="font-size:\; revert;\; font-family:\; var(--wp--preset--font-family--manrope);\; letter-spacing:\; -0.1px;"><span\; class="katex-html"\; aria-hidden="true"><span\; class="base"><span\; class="mord">.</span></span></span></span><span\; style="font-size:\; revert;\; font-family:\; var(--wp--preset--font-family--manrope);\; letter-spacing:\; -0.1px;">\; Find </span>}$$\mathrm{\angle }QRS\mathrm{.}$

Reasoning (constructive).
Through point $R$ draw a line $r$ parallel to $ST$. With this auxiliary line:

• Because $PQ\parallel ST$ and $r\parallel ST$, we get $PQ\parallel r$. With transversal $QR$, the angle at $Q$ (given $\mathrm{\angle }PQR={110}^{\circ }$) corresponds to an angle at $R$ on the other side of the transversal; therefore the angle adjacent to $\mathrm{\angle }QRS$ on the left side equals ${180}^{\circ }-{110}^{\circ }={70}^{\circ }$

• Similarly, since $r\parallel ST$, the interior angle made by $RS$ with $ST$ (given $\mathrm{\angle }RST={130}^{\circ }$) implies the adjacent interior angle on the line through $R$ equals ${180}^{\circ }-{130}^{\circ }={50}^{\circ }$

Now the three adjacent angles at $R$ along the straight line through $R$ are ${70}^{\circ },\mathrm{\angle }QRS,{50}^{\circ }$; their sum is ${180}^{\circ }$. So

$${70}^{\circ }+\mathrm{\angle }QRS+{50}^{\circ }={180}^{\circ }\Rightarrow \mathrm{\angle }QRS={60}^{\circ }\mathrm{.}$$

Answer: $\boxed{{\displaystyle {60}^{\circ }}}$

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Q4 (Fig. 6.26)

Given. $AB\parallel CD,\mathrm{\angle }APQ={50}^{\circ },\mathrm{\angle }PRD={127}^{\circ }\mathrm{<span\; class="katex"\; style="font-size:\; revert;\; font-family:\; var(--wp--preset--font-family--manrope);\; letter-spacing:\; -0.1px;"><span\; class="katex-html"\; aria-hidden="true"><span\; class="base"><span\; class="mord">.</span></span></span></span><span\; style="font-size:\; revert;\; font-family:\; var(--wp--preset--font-family--manrope);\; letter-spacing:\; -0.1px;">\; Find </span>}$$x$ and $y\mathrm{}$

Reasoning.
Interpret the figure angles with respect to transversals and parallels:

• The angle labelled $x$ corresponds to $\mathrm{\angle }PQR$ (the angle made where the transversal meets the top parallel). Since $\mathrm{\angle }APQ={50}^{\circ }$ is an alternate interior/corresponding angle with $\mathrm{\angle }PQR$, we have

  $$x={50}^{\circ }\mathrm{.}$$

• The given $\mathrm{\angle }PRD={127}^{\circ }$ is the angle formed by the transversal on the right side with line $CD$. The angle $y$ is the angle between the same transversal and the other parallel direction but on the left; in the figure $y$and the ${50}^{\circ }$ angle at the top add to the external angle ${127}^{\circ }$. Thus

  $$y={127}^{\circ }-{50}^{\circ }={77}^{\circ }\mathrm{}$$

Answers: $\boxed{{\displaystyle x={50}^{\circ },y={77}^{\circ }\mathrm{}}}$

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Q5 (Fig. 6.27) — reflection between two parallel mirrors

Given. $PQ$ and $RS$ are parallel mirrors. Ray $AB$ strikes $PQ$ at $B$, reflects to $BC$ which hits $RS$ at $C$, and after reflection goes along $CD$. Prove $AB\parallel CD\mathrm{}$

Reasoning (angle-based, law of reflection).

1. At point $B$ the incoming ray $AB$ and reflected ray $BC$ make equal angles with the normal to mirror $PQ$. Let the normal at $B$ be $n_1$. So angle of incidence = angle of reflection relative to $n_1$.

2. At $C$ the incoming ray $BC$ and outgoing ray $CD$ make equal angles with the normal $n_2$ to mirror $RS$.

3. Because the mirrors $PQ$ and $RS$ are parallel, their normals $n_1$ and $n_2$ are also parallel. Now follow the oriented angle of the ray:

   • The change in direction from $AB$ to $BC$ is twice the angle between $AB$ and $n_1$ (symmetric reflection).

   • The change from $BC$ to $CD$ is twice the angle between $BC$ and $n_2$.

   • Adding these two directed changes gives the total turning from $AB$ to $CD$.

   But because $n_1\parallel n_2$, the angle $AB$ makes with $n_1$ is the same as the angle $CD$ makes with $n_2$ (by chasing corresponding/reflected angles). The two reflections therefore undo each other’s turning in such a way that the net orientation of $CD$ is parallel to $AB$.

4. Concretely: let the acute angle between the incident ray $AB$ and normal $n_1$ be $\alpha$. After first reflection the ray $BC$ is at angle $-\alpha$ relative to $n_1$. Relative to $n_2$ (parallel to $n_1$) this is also $-\alpha$. Reflection at $C$changes $-\alpha$ to $+\alpha$. Thus the final ray $CD$ makes the same directed angle $\alpha$ with $n_2$ as $AB$ made with $n_1$. So $AB$ and $CD$ are parallel.

Conclusion: $\boxed{{\displaystyle AB\parallel CD\mathrm{}}}$

(That is the standard clean reflection- + parallel-normals argument; it shows the two reflections produce no net change in direction other than translation, so the entering and exiting rays are parallel.)