Question

Solve the problem below

Answer 1

Answer:

$\angle T=60^{\circ}$

Step-by-step explanation:

In all 30-60-90 triangles, the sides are in ratio $x:x\sqrt{3}:2x$, where $x$ is the side opposite to the 30 degree angle and $2x$ is the hypotenuse of the triangle. We know that two right triangles are created on both sides of the rectangle in the center. Notice that $8\sqrt{2}\cdot 2=16\sqrt{2}$ and since $16\sqrt{2}$ is the hypotenuse of the right triangle on the left, $8\sqrt{2}$ must be facing the 30 degree angle. Therefore, angle T must be 60 degrees.

Alternatively, the cosine of any angle in a right triangle is equal to its adjacent side divided by the hypotenuse.

Therefore, we have:

$\cos \angle T=\frac{8\sqrt{2}}{16\sqrt{2}},\\\cos \angle T=\frac{1}{2},\\\angle T=\arccos(\frac{1}{2}),\\\angle T=\boxed{60^{\circ}}$

Answer 2

Answer:

T = 60 degrees

Step-by-step explanation:

The dotted line is the height so it is a right angle

We are able to use trig functions since this is a right triangle

cos T = adj side / hyp

cos T = a/b

cos T = 8 sqrt(2) / 16 sqrt(2)

cos T = 1/2

Taking the inverse of each side

cos^-1 ( cosT) = cos^-1 ( 1/2)

T = 60 degrees