Question

question 5 of 25
an isosceles triangle has angle measures 55°, 55°, and 70°. the side across from the 70° angle is 10 inches long. how long are the other sides?
a. 11.47 inches
b. 10 inches
c. 8.19 inches
d. 8.72 inches

Explanation:

Step1: Apply the Law of Sines

The Law of Sines states that \(\frac{a}{\sin A}=\frac{b}{\sin B}=\frac{c}{\sin C}\). Let the side opposite the \(70^{\circ}\) angle be \(a = 10\) inches, and the sides opposite the \(55^{\circ}\) angles be \(b\) (we want to find \(b\)). So, \(\frac{a}{\sin A}=\frac{b}{\sin B}\), which becomes \(\frac{10}{\sin70^{\circ}}=\frac{b}{\sin55^{\circ}}\).

Step2: Solve for \(b\)

First, find \(\sin70^{\circ}\approx0.9397\) and \(\sin55^{\circ}\approx0.8192\). Then, from \(\frac{10}{0.9397}=\frac{b}{0.8192}\), we can cross - multiply: \(b=\frac{10\times0.8192}{0.9397}\).
Calculate \(b=\frac{8.192}{0.9397}\approx8.72\).

Answer:

D. 8.72 inches