Question

question 8 of 10
a triangle has two angles measuring 31° and 68°. the side across from the third angle is 45 units long. what are the lengths of the other two sides?
a. 81.6
b. 23.5
c. 24.8
d. 42.2

Explanation:

Answer:

First, find the third - angle: \(180-(31 + 68)=81^{\circ}\).
Let the side opposite the \(31^{\circ}\) angle be \(a\), the side opposite the \(68^{\circ}\) angle be \(b\), and the side opposite the \(81^{\circ}\) angle \(c = 45\).
Using the Law of Sines \(\frac{a}{\sin A}=\frac{b}{\sin B}=\frac{c}{\sin C}\).
For the side opposite the \(31^{\circ}\) angle: \(\frac{a}{\sin31^{\circ}}=\frac{45}{\sin81^{\circ}}\), so \(a=\frac{45\sin31^{\circ}}{\sin81^{\circ}}\approx23.5\).
For the side opposite the \(68^{\circ}\) angle: \(\frac{b}{\sin68^{\circ}}=\frac{45}{\sin81^{\circ}}\), so \(b=\frac{45\sin68^{\circ}}{\sin81^{\circ}}\approx42.2\).
So the answers are B. 23.5 and D. 42.2