# If either side of an isosceles triangle measures 24 inches and its base measures 16 inches, then what are its angles?

Nov 14, 2015

$A = {\cos}^{-} 1 \frac{7}{9} , B = C = \frac{\pi}{2} - \frac{A}{2}$

#### Explanation:

$24 \times 24 \times 16 = A B \times A C \times B C$

we use "cos rule": ${16}^{2} = {24}^{2} + {24}^{2} - 2 \cdot 24 \cdot 24 \cdot \cos A$

$\frac{{16}^{2} - {24}^{2} - {24}^{2}}{- 2 \cdot 24 \cdot 24} = \cos A = - \frac{{\left({2}^{4}\right)}^{2}}{2 \cdot {\left({2}^{3}\right)}^{2} \cdot {3}^{2}} + 1 = 1 - \frac{2}{9}$

A + B + C = pi ; B = C = x

$A + 2 x = \pi R i g h t a r r o w x = \frac{\pi - A}{2}$