Mathematics

In a $\triangle ABC$, $\angle A - \angle B = 30^{\circ}$ and $\angle B -\angle C = 42^{\circ}$; find $\angle A$.

$94^o$

SOLUTION
In $\triangle ABC$,
$\angle A - \angle B = 30$    ....(I)
$\angle B - \angle C = 42$     .....(II)
Also, sum of angles of the triangle $= 180$
$\angle A + \angle B + \angle C = 180$     ....(III)
On subtracting (I) and (II), we get
$\angle A-2\angle B+\angle C=-12^o$     ....(IV)
On subtracting (III) and (IV), we have
$3\angle B=192$
$\angle B=64^o$
From (I), we get
$\angle A=94^o$

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Single Correct Medium Published on 09th 09, 2020
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