Mathematics

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


ANSWER

$$94^o$$


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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