Question

Given the triangle below, what is m/_A, rounded to the nearest tenth?

Answer 1

A

Explanation:

using the Cosine Rule in Δ ABC

cosA =
`(b^2+c^2-a^2)/(2bc)`

where a, b, c are the sides opposite A, B , C

here a = 10 , b = 12 , c = 8

cosA =
`(12^2+8^2-10^2)/(2(12)(8))` =
`(144+64-100)/(192)` =
`(108)/(192)` , then

∠ A =
`cos^(-1)` (
`(108)/(192)` ) ≈ 55.8° ( to the nearest tenth )