Question

Find the value of x and y. ( ANSWER NEEDS TO BE IN REDUCED RADICAL FORM )

Answer 1

Answer:

Part 1) $x=8\sqrt{2}\ units$

Part 2) $y=4\sqrt{6}\ units$

Step-by-step explanation:

see the attached figure with letters to better understand the problem

In the right triangle ABD

Find the length side BD

$sin(45\°)=BD/AB$

$BD=(AB)sin(45\°)$

we have

$AB=8\ units$

$sin(45\°)=\frac{\sqrt{2}}{2}$

substitute the given values

$BD=(8)\frac{\sqrt{2}}{2}$

$BD=4\sqrt{2}\ units$

In the right triangle DBC

Find the length side BC

$sin(30\°)=BD/BC$

$BC=BD/sin(30\°)$

we have

$sin(30\°)=\frac{1}{2}$

$BD=4\sqrt{2}\ units$

substitute the given values

$BC=4\sqrt{2}/\frac{1}{2}$

$BC=8\sqrt{2}\ units$

therefore

The value of x is

$x=8\sqrt{2}\ units$

In the right triangle DBC

Find the length side DC

$cos(30\°)=DC/BC$

$DC=(BC)cos(30\°)$

we have

$BC=8\sqrt{2}\ units$

$cos(30\°)=\frac{\sqrt{3}}{2}$

substitute the given values

$DC=(8\sqrt{2})\frac{\sqrt{3}}{2}$

$DC=4\sqrt{6}\ units$

therefore

the value of y is

$y=4\sqrt{6}\ units$