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Q5. In a $\Delta\textup{ABC}$ , $\angle C = 3 \angle B = 2( \angle A + \angle B)$ . Find the three angles.

Answers (1)

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Given,

$\angle C = 3 \angle B = 2(\anlge A + \angle B)$

$\Rightarrow 3 \angle B = 2(\anlge A + \angle B)$

$\Rightarrow \angle B = 2 \angle A$

$\Rightarrow 2 \angle A -\angle B = 0..........(1)$

Also, As we know that the sum of angles of a triangle is 180, so

$\angle A +\angle B+ \angle C=180$

$\angle A +\angle B+ 3\angle B=180^0$

$\angle A + 4\angle B=180^0..........(2)$

Now From (1) we have

$\angle B = 2 \angle A.......(3)$

Putting this value in (2) we have

$\angle A + 4(2\angle A)=180^0.$

$\Rightarrow 9\angle A=180^0.$

$\Rightarrow \angle A=20^0.$

Putting this in (3)

$\angle B = 2 (20)=40^0$

And

$\angle C = 3 \angle B =3(40)=120^0$

Hence three angles of triangles $20^0,40^0\:and\:120^0.$

Posted by

Pankaj Sanodiya

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