1. ## iscoeles triangle

I know that a isoceles triangle has 2 equal sides and 2 equal angles. The sides of the 2 sides are 5m and the angles are 72.64 degrees, making the top angle 34.72 degrees. But i do not know how to work out the base length? can you help me please

2. Originally Posted by kath
I know that a isoceles triangle has 2 equal sides and 2 equal angles. The sides of the 2 sides are 5m and the angles are 72.64 degrees, making the top angle 34.72 degrees. But i do not know how to work out the base length? can you help me please
use the sine rule:

$\frac{a}{\sin(A)} = \frac{b}{\sin(B)} = \frac{c}{sin(C)}$

where side a is opposite angle A, side b opposite angle B and side c opposite angle C.

I'm going to define the side you need to find as side a and so A = 34.72

$\frac{a}{\sin(A)} = \frac{b}{\sin(B)}$ thus $\frac{a}{sin(34.72)} = \frac{5}{sin(72.64)}$

this gives

$a = \frac{5sin(34.72)}{sin(72.64)} = 2.98m (3sf)$

3. law of cosines ... $a^2 = b^2 + c^2 - 2 \cdot b \cdot c \cdot \cos{A}$

let $y$ = base length

$x$ = length of the two equal sides

$Y$ = measure of the vertex angle

$y^2 = x^2 + x^2 - 2 \cdot x \cdot x \cdot \cos{Y}$

$y^2 = 2x^2(1 - \cos{Y})$

$y = \sqrt{2x^2(1 - \cos{Y})}$