The question is an illustration of right triangles
Triangle 1
This triangle is an isosceles right triangle.
So, the values of x and y are:
[tex]x = 13[/tex]
[tex]y = 13\sqrt 2[/tex]
Triangle 2
This triangle is an isosceles right triangle.
So, the values of x and y are:
[tex]x = 13[/tex]
[tex]y = 13[/tex]
Triangle 3
To calculate x, we make use of the following cosine ratio
[tex]\cos(60) = \frac 3x[/tex]
Make x the subject
[tex]x = 3 \div \cos(60)[/tex]
Solve for x
[tex]x = 3 \div 0.5[/tex]
[tex]x = 6[/tex]
The value of y is then calculated using:
[tex]\sin(60) = \frac y3[/tex]
Make y the subject
[tex]y = 3 * \sin(60)[/tex]
[tex]y = 3 * \frac{\sqrt 3}{2}[/tex]
[tex]y =\frac{3\sqrt 3}{2}[/tex]
Read more about right triangles at:
https://brainly.com/question/2217700
