Question

Please Help Brainliest to whoever can give a clear explanation for this problem!

In triangle ABC, $\angle A = 90$, AC = 1, and AB = 5. Point D lies on ray AC such that $\angle DBC = 2\angle CBA$. Compute AD.

Answer 1

See the attached sketch. Let $\beta$ denote angle ABC and $\beta'$ denote angle CBD. Then $\beta'=2\beta$.

From the sketch, we see that

$\tan\beta=\dfrac15$

$\tan(\beta+\beta')=\tan(3\beta)=\dfrac{AD}5$

Write $\tan(3\beta)$ in terms of $\tan\beta$. Recall the angle sum identity for tangent:

$\tan(x+y)=\dfrac{\tan x+\tan y}{1-\tan x\tan y}$

This gives us

$\tan(3\beta)=\dfrac{\tan\beta+\frac{2\tan\beta}{1-\tan^2\beta}}{1-\tan\beta\cdot\frac{2\tan\beta}{1-\tan^2\beta}}=\dfrac{\tan\beta(\tan^2\beta-3)}{3\tan^2\beta-1}$

and plugging in $\tan\beta=\frac15$ leaves us with

$\tan(3\beta)=\dfrac{37}{55}$

and so AD = 5 * 37/55 = 37/11.