Mathematics

In the adjoining figure, AB$$||$$CD and PQ, QR intersects AB and CD both at E, F and G, H respectively. Given that $$\angle$$PEB$$=80^o$$, $$\angle$$QHD$$=120^o$$ and $$\angle$$PQR$$=x^o$$, find the value of x.


ANSWER

$$20$$


SOLUTION
$$\angle PEB=\angle PGD \,\,(\because AB||CD)$$
$$\therefore \angle PGD=80^o$$
$$\angle QGD=180-\angle PGD=180-80=100^o$$
$$\angle QHC=180-\angle QHD=180-120=60^o$$
In $$\triangle QGH:$$
$$x+\angle QGD+\angle QHC=180^o\\ \Rightarrow x=180-100-60=20^o$$
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