Question

Question 9

Answer 1

Answer:

$PX=4\\PY=4\sqrt{3}$

Step-by-step explanation:

Let $\angle PYZ=Q$

$\Rightarrow \angle XYP=90-\angle PYZ\\\angle XYP=90-\theta$

In the $\Delta PYZ$

$\tan\theta =\frac{opposite}{adjacent}=\frac{PZ}{PY}......(1)$

In the $\Delta YPX$

$\tan(90-\theta)=\frac{opposite}{adjacent}=\frac{PX}{PY}$

$\cot=\frac{PX}{PY}....(2)$

eqn(1)$\times$ eqn(2)

$\tan\times\cot\theta=\frac{PZ}{PY}\times\frac{PX}{PY}\\\\1=\frac{PZ\times PX}{PY^2}\ \ (as\ tan\theta\ =\frac{1}{\cot\theta})\\\\PY^2=PZ\times PX\\PY^2=12PX.......(3)$

Now in $\Delta XYP$

use Pythagorean theorem

$XY^2=PY^2+PX^2\\8^2=PY^2+PX^2\ \ (as\ XY\ =8)\\PX^2+PY^2=64\\\\PY^2=12PX\ \ (eqn(3)\\\\\Rightarrow PX^2+12PX=64\\\\PX^2+12PX-64=0\\\\PX^2+16-4PX-64=0\\\\(PX+16)(PX-4)=0\\\\PX=4,\ -16$

Length can not be negative

Hence $PX=4$

$PY^2=12PX=12\times4=48\\\\PY=\sqrt{48}=4\sqrt{3}$

Hence $PX=4,\ PY=4\sqrt{3}$