In triangle ABC, angle C is 90 °, AB = 15 √21, sine a = 0.4. Find the height CH.

In a right-angled triangle, the sine of an acute angle is equal to the ratio of the opposite leg to the hypotenuse.

B △ ABC: ∠ A – acute, BC – opposite leg, AB – hypotenuse.

BC: AB = sin ∠ A.

BC: 15√21 = 0.4.

BC = 15√21 x 0.4.

BC = 6√21.

By the Pythagorean theorem, we find AC:

AC ^ 2 + BC ^ 2 = AB ^ 2.

AC ^ 2 + (6√21) ^ 2 = (15√21) ^ 2.

AC ^ 2 + 36 x 21 = 225 x 21.

AC ^ 2 + 756 = 4 725.

AC ^ 2 = 4 725 – 756.

AC ^ 2 = 3 969.

AC = √3 969.

AC = 63.

S △ ABC can be found in two ways:

a) on both sides and the angle between them: S △ ABC = 1/2 x AB x AC x sin ∠ A.

b) by base and height: S △ ABC = 1/2 x AB x CH.

We find S △ ABC by the formula a):

S △ ABC = 15√21 x 63 x 0.4 / 2.

S △ ABC = 945√21 x 0.4 / 2.

S △ ABC = 472.5√21 x 0.4.

S △ ABC = 189√21.

We substitute the known data into the formula b) and express the CH from it:

15√21 x CH / 2 = 189√21.

15√21 x CH = 189√21 x 2.

15√21 x CH = 378√21.

CH = 378√21: 15√21.

CH = 378: 15.

CH = 25.2.

Answer: CH height is 25.2.