Special right triangles. find the value of x?
hypotenuse = leg ×
[tex] \sqrt{2} [/tex]
![Special right triangles. find the value of x? hypotenuse = leg ×[tex] \sqrt{2} [/tex] class=](https://us-static.z-dn.net/files/d95/cfb9b1d6faff251c84b38cb0b3996b95.jpg)
Answer:
x = (9√2)/2
Step-by-step explanation:
Reference angle = 45°
Hypotenuse length = 9
Adjacent length = x
Apply the trigonometric ratio CAH:
Cos 45° = Adj/Hyp
Cos 45° = x/9
Multiply both sides by 9
9*Cos 45 = x
9*1/√2 = x (cos 45 = 1/√2)
9/√2 = x
Rationalize
(9*√2)/2 = x
x = (9√2)/2