The vertical angle of an isoceles triangle is three time the sum of its base angles. Find all the angles of the triangle.

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Solution

In $Δ A B C$
AB = AC

$∠ A = 3 (∠ B + ∠ C)$
Let $∠ B = x$
$∴ ∠ C = x$
$∴ ∠ B + ∠ C = 2 x, H e n c e ∠ A = 3 x$
Thus by angle sum property
$2 x + 3 x = 180$
$5 x = 180$
$x = \frac{180}{5}$
Thus, $∠ C = ∠ B = \frac{180}{5}$ = $36^{\circ}$
Angle A = $108^{\circ}$

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