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?

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- Jul 15th 2008, 07:36 PMnafitaI'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 PMTKHunny
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 PMmr fantastic
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:

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