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
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:
$\displaystyle \frac{a}{\sin(A)} = \frac{b}{\sin(B)} = \frac{c}{sin(C)}$
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
$\displaystyle \frac{a}{\sin(A)} = \frac{b}{\sin(B)}$ thus $\displaystyle \frac{a}{sin(34.72)} = \frac{5}{sin(72.64)}$
this gives
$\displaystyle a = \frac{5sin(34.72)}{sin(72.64)} = 2.98m (3sf)$
law of cosines ... $\displaystyle a^2 = b^2 + c^2 - 2 \cdot b \cdot c \cdot \cos{A}$
let $\displaystyle y$ = base length
$\displaystyle x$ = length of the two equal sides
$\displaystyle Y$ = measure of the vertex angle
$\displaystyle y^2 = x^2 + x^2 - 2 \cdot x \cdot x \cdot \cos{Y}$
$\displaystyle y^2 = 2x^2(1 - \cos{Y})$
$\displaystyle y = \sqrt{2x^2(1 - \cos{Y})}$