Answer:
[tex]x = 3.7[/tex]
Step-by-step explanation:
We can find the side of the hypotenuse by applying pythagorean theorem.
[tex] {5}^{2} + {5}^{2} = {x}^{2} [/tex]
[tex]25 + 25 = {x}^{2} [/tex]
[tex]50 = {x}^{2} [/tex]
[tex]x = 5 \sqrt{2} [/tex]
Now we can apply pythagorean theorem on the other triangle
[tex](5 \sqrt{2} ) {}^{2} + {x}^{2} = {8}^{2} [/tex]
[tex]50 + {x}^{2} = 64[/tex]
[tex] {x}^{2} = 14[/tex]
[tex]x = \sqrt{14} [/tex]
[tex]x = 3.7[/tex]
Answer:
Step-by-step explanation:
use Pythagoras's theorem to find the hypontenuse of the isosceles triangle on the right side.
hyp = sqrt [ 5^2 + 5^2 ]
hyp = sqrt [ 50 ]
hyp = 7.0710678
now we know the adjacent side of the triangle on the left , since it's the same as the hypotenuse of the triangle on the right.
SOH CAH TOA ( to remember how the trig funct. fit on triangles)
CAH Cos(Ф) = adj / hyp
cos(Ф) = 7.0710678 / 8
Ф = arcCos( 7.0710678 / 8 )
Ф = 27.8855668°
TOA Tan(Ф) = x / 7.0710678
Tan( 27.8855668° ) = x / 7.0710678
7.0710678 * Tan( 27.8855668° ) = x
3.741657 = x
x = 3.7 ( to the nearest 10th )