Question

In a triangle ABC ,AP is perpendicular to BC. If BC = 112cm , cot B = 4/3 and cot C = 12/5 , calculate the length of AP. (by trigonometry​)

Answer 1

Answer:

The length of AP is 30 cm.

Explanation:

Given,

In triangle ABC,

AP ⊥ BC,

And, BC = 112cm ,

$\because \cot x = \frac{\text{Adjacent leg}}{\text{opposite leg}}$

Thus,

$\cot B = \frac{BP}{AP}\text{ and }\cot C = \frac{CP}{AP}$

We have, cot B = $\frac{4}{3}$ and cot C = $\frac{12}{5}$,

$\implies \frac{BP}{AP}=\frac{4}{3}\text{ and }\frac{ CP}{AP}=\frac{12}{5}$

$\implies BP = \frac{4}{3}AP\text{ and }CP=\frac{12}{5}AP$

Now, BP + PC = BC = 112

$\implies \frac{4}{3}AP+\frac{12}{5}AP=112$

$\frac{20+36}{15}AP=112$

$\frac{56}{15}AP=112$

$\implies AP=\frac{112\times 15}{56}=2\times 15 = 30\text{ cm}$