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Question

In figure, $$PQ \parallel RS$$.
If $$\angle 1 =(2a+b)^o$$ and $$\angle 6=(3a-b)^o$$ then the measure of $$\angle 2$$ in terms of $$b$$ is

A
$$(108-b)^o$$
B
$$(2a+b)^o$$
C
$$(3-b)^o$$
D
$$(180-b)^o$$
Solution
Verified by Toppr

Correct option is C. $$(108-b)^o$$
Here, $$PQ$$ parallel to $$RS$$ and $$I$$ is a transversal.
$$\Rightarrow \angle 2= \angle 6=(3a-b)^o$$ (1) [ corresponding angles ]
Now,
$$ \angle 1 +\angle 2=180^o$$ [linear angles]
$$\because \angle 1=(2a+b)^o$$
$$\therefore \angle 2=180^o -(2a+b)^o $$ ....(2)
From equation 1 and 2, we get
$$\Rightarrow (3a-b)^o =180^o -(2a+b)^o$$
$$\Rightarrow 3a-b+2a+b=180^o$$
$$\Rightarrow 5a=180^o$$
$$\Rightarrow a=\dfrac{180^o}{5}$$
$$\Rightarrow a=36^o$$
$$\therefore a=36^o$$
Now,
$$\Rightarrow \angle 2=(3a-b)^o$$ [ from equation 1]
$$\Rightarrow \angle 2=(3\times 36-b)^o$$
$$\Rightarrow \angle 2=(108-b)^0$$
hence, option $$C$$ is correct

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