In a triangle ABC, the angle C is 90 °, AB = 2.6, tg A = 5 \ 12. Find BC.

In triangle ABC we find the leg BC.

It is known:

Angle C is 90 °;
AB = 2.6;
tg A = 5/12.
Solution:

1) sin a = BC / AB;

Hence, BC = AB * sin a;

sin a – unknown;

2) Find sin a by the formula:

1 + tg ^ 2 a = 1 / cos ^ 2 a;

1 + (5/12) ^ 2 = 1 / cos ^ 2 a;

1 + 25/144 = 1 / cos ^ 2 a;

169/144 = 1 / cos ^ 2 a;

cos ^ 2 a = 144/169;

cos a = 12/13;

3) Find sin a.

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

4) Find from the formula BC = AB * sin a leg BC;

BC = 2.6 * 5/13 = 13/13 = 1.

Answer: BC = 1.