Question:
Two angles of a triangle are equal and the third angle is greater than each one of them by 18°. Find the angles.
Solution:
Let $\angle A=\angle B$ and $\angle C=\angle A+18^{\circ}$.
Then,
$\angle A+\angle B+\angle C=180^{\circ} \quad[$ Sum of the angles of a triangle $]$
$\angle A+\angle A+\angle A+18^{\circ}=180^{\circ}$
$\Rightarrow 3 \angle A=162^{\circ}$
$\Rightarrow \angle A=54^{\circ}$
Since,
$\angle A=\angle B$
$\Rightarrow \angle B=54^{\circ}$
$\therefore \angle C=\angle A+18^{\circ}$
$=(54+18)^{\circ}$
$=72^{\circ}$
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