Question

an isosceles triangle has angle measures of 55, 55, and 70. the side across the 70 degree angle is 10 inches long. how long are the other sides?

Answer 1

Answer:

The other sides of triangles are 8.72 in and 8.72 in

Step-by-step explanation:

 In a triangle ABC. Please find the attachment for figure.

$\angle A=70^{\circ}$

$\angle B=\angle C=55^{\circ}$

Side BC=a = 10 in

Using sine law of trigonometry,

$\dfrac{a}{\sin A}=\dfrac{b}{\sin B}=\dfrac{c}{\sin C}$

Substitute the given value into formula.

$\dfrac{10}{\sin 70}=\dfrac{b}{\sin 55}=\dfrac{c}{\sin 55}$

$\dfrac{10}{\sin 70}=\dfrac{b}{\sin 55}$

Cross multiply and we get

$b=\sin 55\times \dfrac{10}{\sin 70}$

$b=8.72$ in

It is a isosceles triangle. Therefore, b=c=8.72 in

Hence, The other sides of triangles are 8.72 in and 8.72 in