Find the perimeter of Triangle ABC

Question

This practice engineering board exam which I've tried seems very tricky (question no.2). The inscribed circle of $\triangle ABC$ is tangent to $AB$ at $P$ . If the radius is $21$, $AP=23$ and $PB=27$, find the perimeter of $\triangle ABC$ .I've attempted to use the Pythagorean Theorem and Heron's Formula but the problem is that I've missed the values of the two sides $BC$ and $CA$ .Can you please provide solutions by two methods(geometry and trigonometry)?

Answer 1

Let the circle touch side $BC$ and $CA$ at points $Q$ and $R$ respectively. Let $CQ=CR=m$.

$AR=AP=23$ and $BQ=BP=27$.

$AB=50, BC=27+m, CA=23+m$

Now, apply Heron's Formula and then use $\Delta=rs$.

Answer 2

Another approach:

$\tan(\frac{\alpha}2)=\frac{21}{23}$

$\tan(\frac{\beta}2)=\frac{21}{27}$

$\frac{\alpha}2+\frac{\beta}2+\frac{\gamma}2=90^o$

$\tan(\frac{\alpha}2+\frac{\beta}2)=\tan(90-\frac{\gamma}2)=cotan(\frac{\gamma}2)$

Puting values we get $\tan(\frac{\gamma}2)=\frac6{35}$

Now we use this formula:

$\tan(\frac{\gamma}2)=\frac r{p-c}$

Where r=21, p is half perimeter, and $c=AB=23+27=50$

Plugging these values we get:

$p=\frac r{tan(\frac{\gamma}2)}+c=\frac {21}{\frac 6{35}}+50=\frac {7\cdot 35}{2}+50$

So perimeter P is:

$P=2\times \left(\frac {7\cdot 35}{2}+50\right)=7\cdot 35+100=\boxed {345}$