Question

Unit 8: Right Triangles & Trigonometry Homework 2: Special Right Triangles

Answer 1

Can you please help me? I still don't understand.

Answer 2

Answer:

Part 1) $x=14\sqrt{2}\ units$

Part 2) $y=14\ units$

Part 3) $z=14\sqrt{3}\ units$

Step-by-step explanation:

see the attached figure with letters to better understand the problem

step 1

In the right triangle ABC

we know that

The triangle ABC is a $45^o-90^o-45^o$

so

Is an isosceles right triangle

The legs are equal

therefore

AC=AB

$x=14\sqrt{2}\ units$

step 2

Find the length side BC

Applying the Pythagorean Theorem]

$BC^2=AB^2+AC^2$

$BC^2=(14\sqrt{2})^2+(14\sqrt{2})^2$

$BC^2=784\\BC=28\ units$

step 3

Find the value of y

In the right triangle BCD

$sin(30^o)=\frac{BD}{BC}$ ----> by SOH (opposite side divided by the hypotenuse)

$sin(30^o)=\frac{y}{28}$

Remember that

$sin(30^o)=\frac{1}{2}$

so

$\frac{y}{28}=\frac{1}{2}\\\\y=14\ units$

step 4

Find the value of z

In the right triangle BCD

$cos(30^o)=\frac{DC}{BC}$ ----> by CAH (adjacent side divided by the hypotenuse)

$cos(30^o)=\frac{z}{28}$

Remember that

$cos(30^o)=\frac{\sqrt{3}}{2}$

so

$\frac{z}{28}=\frac{\sqrt{3}}{2}\\z=14\sqrt{3}\ units$