In an isosceles triangle ABC with a base AC, the lateral side AB is equal to 32√3
| September 9, 2021 | education
In an isosceles triangle ABC with a base AC, the lateral side AB is equal to 32√3, and the cos of the angle A = 1/2. Find the height drawn to the base.
Determine the sine of angle A.
Sin2A + Cos2A = 1.
Sin2A = 1 – Cos2A = 1 – (1/2) ^ 2 = 1 – 1/4 = 3/4.
SinA = √3 / 2.
Then in a right-angled triangle ABN, BN = AB * SinA = 32 * √3 * √3 / 2 = 48 cm.
Answer: The length of the BH height is 48 cm.
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