Question

Question 9

In the diagram, XY = 8 and PZ = 12. What is the length of
PX and PY?
Round answers to the nearest tenth if necessary

Answer 1

`PX=4\\PY=4√(3)`

Explanation:

Let
`\angle PYZ=Q`

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

In the
`\Delta PYZ`

`\tan\theta =(opposite)/(adjacent)=(PZ)/(PY)......(1)`

In the
`\Delta YPX`

`\tan(90-\theta)=(opposite)/(adjacent)=(PX)/(PY)`

`\cot=(PX)/(PY)....(2)`

eqn(1)
`*` eqn(2)

`\tan*\cot\theta=(PZ)/(PY)*(PX)/(PY)\\\\1=(PZ* PX)/(PY^2)\ \ (as\ tan\theta\ =(1)/(\cot\theta))\\\\PY^2=PZ* 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*4=48\\\\PY=√(48)=4√(3)`

Hence
`PX=4,\ PY=4√(3)`