Question

The area of a triangle is 1848. Two of the side lengths are 53 and 88 and the included angle is acute. Find the measure of the included angle, to the nearest tenth of a degree.

Answer 1

Area of triangle = 1848

Given the sides of the triangle

a =88

b=53

Using the formula of the area of a triangle

$\begin{gathered} \text{Area = }\frac{1}{2}ab\text{ sin}\theta \\ \text{where }\theta\text{ is the included angle} \end{gathered}$

Substitute the values of the sides and area, then simplify

$1848=\frac{1}{2}\times88\times53\times\sin \theta$$\begin{gathered} 3696=4664\sin \theta \\ \sin \theta=\frac{3696}{4664} \\ \sin \theta=0.7925 \\ \theta=\sin ^{-1}0.7925=52.42^0 \\ \theta\approx52.4^{0\text{ }}(nearest\text{ tenth)} \end{gathered}$

Hence,

The value of the included angle to the nearest tenth degree is 52.4