Question

question 10 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. 10 inches
b. 8.72 inches
c. 8.19 inches
d. 11.47 inches

Explanation:

Step1: Apply the Law of Sines

Let the side opposite the $70^{\circ}$ angle be $a = 10$ inches, and the sides opposite the $55^{\circ}$ angles be $b$ and $c$ (they are equal since it's an isosceles triangle). By the Law of Sines, $\frac{a}{\sin A}=\frac{b}{\sin B}$. Here, $A = 70^{\circ}$ and $B=55^{\circ}$, so $\frac{10}{\sin70^{\circ}}=\frac{b}{\sin55^{\circ}}$.

Step2: Solve for $b$

We can re - arrange the formula from Step 1 to get $b=\frac{10\times\sin55^{\circ}}{\sin70^{\circ}}$. We know that $\sin55^{\circ}\approx0.819$ and $\sin70^{\circ}\approx0.9397$. Then $b=\frac{10\times0.819}{0.9397}\approx8.19$ inches.

Answer:

C. 8.19 inches