Question

Find the length of the altitude of an isosceles triangle with a 64° vertex angle and a base 48 cm long. Round your answer to the nearest tenth.

Answer 1

Therefore,

$\begin{gathered} \tan (32^0)=\frac{24}{h} \\ \text{thus} \\ h=\frac{24}{\tan(32^0)}=38.4\operatorname{cm} \end{gathered}$

Therefore the length of the altitude is 38.4cm