In triangle ABC: angle C = 90 °, AC = 10; sin of angle A = 12/13. Find side BC

The following values are known in triangle ABC:

Angle C = 90 °;
AC = 10;
sin A = 12/13.
Find side BC.

Decision:

1) cos A = √ (1 – sin ^ 2 A) = √ (1 – (12/13) ^ 2) = √ (1 – 144/169) = √ (169/169 – 144/169) = √ ( 169 – 144) / √169 = √25 / √169 = 5/13;

2) cos A = AC / AB;

AB = AC / cos A;

Substitute the known values into the formula and calculate the value of the hypotenuse.

AB = 10 / (5/13) = 10 * 13/5 = 10/5 * 13 = 2 * 13 = 26;

2) Find the second leg BC of triangle ABC.

ВС = √ (AB ^ 2 – AC ^ 2) = √ (26 ^ 2 – 10 ^ 2) = √ (676 – 100) = √576 = 24;

Answer: BC = 24.