# I'm struggling with this question...Need Help!!!

• Jul 15th 2008, 07:36 PM
nafita
I'm struggling with this question...Need Help!!!
A parcel of land is in the shape of an isosceles triangle. The base has length 425 feet; the other sides, which are of equal length, meet at an angle of 39 degree. How long are the two equal side?
• Jul 15th 2008, 08:51 PM
TKHunny
Do you recall:

1) The base angles of an isosceles triangle are congruent?
2) The altitude perpendicular to the base bisects the apex angle and the base?
• Jul 15th 2008, 08:55 PM
mr fantastic
Quote:

Originally Posted by nafita
A parcel of land is in the shape of an isosceles triangle. The base has length 425 feet; the other sides, which are of equal length, meet at an angle of 39 degree. How long are the two equal side?

There are many approaches. Here are two:

Let the unknown sides be of length x.

Option 1. Drop a perpendicular from the vertex to the base and use the resulting right-triangle to solve for x:

$\displaystyle \sin (19.5^o) = \frac{212.5}{x} \, ......$

Option 2. Use the cosine rule to solve for x:

$\displaystyle 425^2 = x^2 + x^2 - 2(x)(x) \cos (39^o) \, ........$