I am having trouble on a Pre-Calculus question, can anyone help?

Question

An isosceles triangle has two equal sides of length 15 inches. The angle between the two equal sides is 32 degrees. Find the area A of the triangle in square inches. Sorry I don't have a picture but I'm hoping you can visual or draw out the picture yourself.

Answer 1

We know that when we cut the triangle into half by a vertical line, we form two identical right triangles with the angle half as given (16 deg).

The base of the right triangle, say x is,

x = 15 sin16

The height h is,

h = 15 cos16

Hence, the area of the right triangle (Ar) is

Ar = (1/2)*x*h

And finally the total area A is,

A = 2*Ar

Answer 2

Hello,

We have to form a triangle by dividing the top angle in half givin us a right triangle with a 16 degree angle and an opposite we will call x and an adjacent side of 15. So sine(16) = x/15 or x = 15sin(16) = 4.135. Then the other side (altitude) is tan(16) = 4.135 / y so y = 4.135/tan(16) = 14.20. Now the area of the big triangle is 14.2 * 4.135 = 59.63 Square inches.

Hope This Helps!!

Answer 3

Draw a diagram. The angle's bisector splits this into two right triangles, where the hypotenuse is 15 and the degree up top is 16. So if h is the height, then cos(16) = h/15. So h = 15cos(16). Likewise, the base of the right triangle is 15sin(16).

So the area is 2 * [ (1/2)15sin(16) * 15cos(16) ]