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

1 Answer
Vinícius Ferraz
Nov 14, 2015

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

Explanation:

24 times 24 times 16 = AB times AC times BC

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{(2^4)^2}{2 cdot (2^3)^2 cdot 3^2} + 1 = 1 - 2/9

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

A + 2x = pi Rightarrow x = frac{pi - A}{2}

Related questions
See all questions in Angles with Triangles and Polygons
Impact of this question
1954 views around the world
You can reuse this answer
Creative Commons License