Atheater production group is making frames to support wall-like props. three-foot beams form right triangles with 10-foot beams to allow them to stand, as shown in the image.
what is x, the angle at which the diagonal beam meets the 10-foot beam at the top of the frame?
16.7°
17.5°
72.5°
73.3°

For this case, we can use the following trigonometric relationship:

Where,

C.O: opposite leg

AC; adjacent leg

x: angle

Substituting values we have:

From here, we clear the value of x.

We have then:

the angle at which the diagonal beam meets the 10-foot beam at the top of the frame is:

16.7°

x = ±sqrt(10)

step-by-step explanation:

5 - 2x² = -15

subtract 5 from each side

5-5 - 2x² = -15-5

-2x^2   = -20

divide by -2 on each side

-2x^2   /-2= -20/-2

x^2 =10

take the square root on each side

sqrt(x^2 )=±sqrt(10)

x = ±sqrt(10)

check:

5 - 2x² = -15

+sqrt(10)

5 - 2(sqrt(10))² = -15

5 -2(10) = -15

5 -20 = -15

-15 = -15

-sqrt(10)

5 - 2(-sqrt(10))² = -15

5 -2(10) = -15

5 -20 = -15

-15 = -15