Question

Find the value of X round to the nearest tenth. I need help.

Answer 1

x ≈ 22.1°

Explanation:

Using the law of cosines, then

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

with a = 19, b = 14, c = 8 and C = x, thus

cos x =
`(19^2+14^2-8^2)/(2(19)(14))`

=
`(361+196-64)/(532)`

=
`(493)/(532)`, thus

x =
`cos^(-1)` (
`(493)/(532)` ) ≈ 22.1° ( to the nearest tenth )