Hey everyone i was trying to setup this problem but i’m stuck can someone tell me setup for number 15 and 16 explain if you can : )

15. x = 10 3/11

16. ∠Q = 75°

Step-by-step explanation:

The two base angles of the isosceles triangle are congruent, so both are 180°-8x. The sum of those, together with the apex angle, will be 180°.

... 2(180 -8x) + (5x -67) = 180

... 360 -16x +5x -67 = 180 . . . . . eliminate parentheses

... -11x = -113 . . . . . subtract 293

... x = 113/11 = 10 3/11

Note on the solution

This value of x makes the interior base angles of the triangle be 97 9/11°. The apex angle is then -15 7/11°. The solution is correct—all the angles add up properly—but the geometry cannot exist.

The approach is the same. The sum of the two congruent base angles and the apex angle is 180°.

... 2(2x+15) + x = 180

... 5x + 30 = 180

... x = (180 -30)/5 = 30

... (2x +15)° = (2·30 +15)° = 75° = angle Q

it would rotate about 180 degrees.