Answer:
Step-by-step explanation:
<u>Triangle 1</u>:
Problem:
Find leg
of a right triangle if hypotenuse
and angle
.
Solution:
≈ 
Explanation:
To find leg
use formula:

After substituting
and
we have:


<u>Triangle 2</u>:
Problem:
Find leg
of a right triangle if leg
and angle
.
Solution:
≈ 
Explanation:
To find leg
use formula:

After substituting
and
we have:

Answer:
13
Step-by-step explanation:
Its not in order its 2.34,9/4,2 3/4
ANSWER

EXPLANATION
The given triangle is a right triangle.
It was given that,

and

Using the Pythagoras Theorem, we can determine the value of c.




The ratio is the adjacent over the hypotenuse.

We rationalize to get:
