Question

Can someone teach me on how to do these type of problems

Answers


A. 22√2

B. 11 √6/2

C. 11√6/4

D. 11 √2/4

Answer 1

11
`\sqrt{x6/4`

Given:

The triangle on the left ( triangle 1) has a 60º, a 90º, and a side that equals 11.

So we know the triangle on the right (triangle 2) has a 45º and a 90º angle.

Triangle 2

Since triangle angles always have a sum of 180º, we can solve for the third angle of triangle 2. 180 - (45 + 90) = 45. So the third angle of triangle 2 is 45º.

This is a special type of right triangle called a 45-45-90. An image of the leg/hypotenuse is uploaded below. Meaning, if we solve for the leg that joins the two triangles, we can solve for the hypotenuse.

Triangle 1

To solve for the middle leg, we work with the information we have. So first, find the third angle. 190 - (60 + 90) = 30. This brings us to a second type of special right triangle. An image of the leg/hypotenuse is uploaded below.

Given that we have a side angle of 11, we know that is 2x due to orientation. So 2x=11 simplifies to x=5.5. We then plug that back in to find the leg that we want: 5.5
`√(3)` .

Triangle 2

Now that we have a side length for the second triangle we can solve. x for this triangle is 5.5
`√(3)` so to find the hypotenuse we plug into x
`√(2)`. This turns into (5.5
`√(3)`
`√(2)`) which simplifies into 5.5
`√(6)` = 13.47

Answers

The answers are not in the correct form. By going through and finding the decimal form of each, you find out that 11
`√(6/4)` is equivalent to 13.47, therefore your answer.