In triangle ABC AB = BC = 3√5, CH height is 3. Find tg B.

| February 24, 2021 | education

Knowing the length of the side AB and the length of the height CH lowered to this side, we find the area of ​​the triangle ABC:

S = | AB | * | CH | / 2 = 3√5 * 3/2 = 9√5 / 2.

Using the formula for the area of ​​a triangle on two sides and the angle between them, we find the sine of the angle B:

sin (B) = 2 * S / (| AB | * | BC |) = 2 * (9√5 / 2) / (3√5 * 3√5) = 9√5 / (9 * 5) = √ 5/5.

Since sin (B)> 0 and 0 <b <180 °, angle B can only lie in the first quarter. Therefore, the cosine of angle B must be positive.

Knowing the sine of the angle B, we find cos (B).

cos (B) = √ (1 – sin ^ 2 (B)) = √ (1 – (√5 / 5) ^ 2) = √ (1 – 5/25) = √ (1 – 1/5) = √ (4/5) = 2 / √5 = 2√5 / 5.

Knowing sin (B) and cos (B), we find tg (B):

tg (B) = sin (B) / cos (B) = (√5 / 5) / (2√5 / 5) = 1/2.

Answer: tg (B) = 1/2.



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