Two angles of a triangle are equal and the third angle is greater than each one of them by 18°.

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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