Question

An isosceles triangle ABC has legs of length 24 and a vertex angle that measures 36º . Determine the length of its base, BC , to the nearest tenth.

Answer 1

Answer: $BC=14.8$

Step-by-step explanation:

By definition, an Isosceles triangle has two equal sides and its opposite angles are congruent.

Observe the figure attached, where the isosceles triangle is divided into two equal right triangles.

So, in this case you need to use the following Trigonometric Identity:

$sin\alpha =\frac{opposite}{hypotenuse}$

In this case, you can identify that:

$\alpha =\angle BAD=\angle CAD=18\°\\\\opposite=BD=CD=x\\\\hypotenuse=AB=AC=24$

Substituting values, and solving for "x", you get:

$sin(18\°)=\frac{x}{24}\\\\24*sin(18\°)=x\\\\x=7.416$

Therefore, the length of BC rounded to nearest tenth, is:

$BC=2(7.416)=14.8$