Question

The vertex angle of an isosceles triangle measures 26° . The base is 30 units long. What is the length of a leg of the isosceles triangle to the nearest tenth?

Answer 1

Answer:

65.46 units

Step-by-step explanation:

Given the information:

• The vertex angle of an isosceles triangle measures 26°

=> the leg angle is:

= (180°  - 26° )/2

= 77°

• The base is 30 units

Let x is the length of the leg side (x>0)

We use cosine law to find x, we have:

$30^{2}$ = $x^{2} +x^{2} -2xxCos(26)$ Because the  isosceles has two equal leg sides

<=> $30^{2}$ = 2$x^{2}$ - 1.79$x^{2}$

<=> 0.21$x^{2}$  = $30^{2}$

<=> $x^{2} =4285$

<=> x = 65.46 units

Hope it will find you well.