In an isosceles triangle ABC, the height CH is drawn to the base of AB. lateral side CA is 15, sin B = 0.9. Find CH

Given: isosceles triangle ABC;
AB – base of triangle ABC;
CH height;
CA = 15;
sin B = 0.9.
Find the length of CH -?
Decision:
1) Since the triangle ABC is isosceles, then CA = BC = 15;
2) Consider the ВНС triangle. It is rectangular in the same way as CH height. Then sin B = (the ratio of the opposite leg to the hypotenuse) = CH / BC. Substitute in this formula instead of sin B and BC = 15. We get:
0.9 = CH / 15;
CH = 0.9 * 15;
CH = 13.5.
Answer: the CH height, carried to the base of the AB wound, is 13.5.