In an isosceles triangle KML, the sides LM and KM are 25. The sine of the angle L is 4/5. Find the height LK

In an isosceles triangle KML, the sides LM and KM are 25. The sine of the angle L is 4/5. Find the height LK drawn to the side KM of this isosceles triangle.

By condition, △ KML is given: KM = ML = 25, sin∠L = 4/5, LH is the height drawn to the KM side.
1. From the top M we draw the height MC to the base KL. Consider △ MCL: ∠MCL = 90 °, ML = 25 – hypotenuse, MC and LC – legs.
The sine of an acute angle in a right triangle is the ratio of the opposite leg to the hypotenuse, then:
sin∠L = MC / ML;
4/5 = MC / 25;
MC = 4 * 25/5 (proportional);
MC = 4 * 5 = 20.
By the Pythagorean theorem:
LC = √ (ML² – MC²) = √ (25² – 20²) = √ (625 – 400) = √ 225 = 15.
2. Since MC is the height drawn to the base of an isosceles triangle, then:
KC = LC = KL / 2.
KL / 2 = 15;
KL = 2 * 15 (proportional);
KL = 30.
2. Consider △ KHL: sin∠K = 4/5 (since ∠K = ∠L), ∠KHL = 90 °, KL = 30 is the hypotenuse.
The sine of an acute angle in a right triangle is the ratio of the opposite leg to the hypotenuse, then:
sin∠K = LH / KL;
4/5 = LH / 30;
LH = 4 * 30/5 = 4 * 6 = 24.
Answer: LH = 24.