Mathematics

In Figure, if $$PQ \parallel ST, \angle PQR=110^o$$ and $$\angle RST=130^o$$, find $$\angle QRS$$.


ANSWER

$$60^o$$


SOLUTION
Given :$$PQ || ST$$, $$\angle PQR = 110^o$$ and $$\angle RST = 130^o$$

Construction: A line $$XY$$ parallel to $$PQ$$ and $$ST$$ is drawn.

$$\angle PQR + \angle QRX = 180^o$$ (Angles on the same side of transversal.) 

$$110^o + \angle QRX = 180^o$$

$$\angle QRX = 180^o - 110^o$$

$$\angle QRX = 70^o$$

Also,$$\angle RST + \angle SRY = 180^o$$ (Angles on the same side of transversal.)

$$130^o + \angle SRY = 180^o$$

$$\angle SRY = 50^o$$

Now,$$\angle QRX +\angle SRY + \angle QRS = 180^o$$

$$70^o + 50^o + \angle QRS = 180^o$$

$$\angle QRS = 60^o$$

Hence, $$\angle QRS = 60^o$$
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