Given: ∆ABC, AC = 12, AB = BC = 10 Find: m∠A, m∠C, m∠B
1 answer:
Answer:
- m∠A ≈ 53.13°
- m∠B ≈ 73.74°
- m∠C ≈ 53.13°
Step-by-step explanation:
An altitude to AC bisects it and creates two congruent right triangles. This lets you find ∠A = ∠C = arccos(6/10) ≈ 53.13°.
Since the sum of angles of a triangle is 180°, ∠B is the supplement of twice this angle, so is about 73.74°.
m∠A = m∠C ≈ 53.13°
m∠B ≈ 73.74°
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The mnemonic SOH CAH TOA reminds you of the relation between the adjacent side, hypotenuse, and trig function of an angle:
Cos = Adjacent/Hypotenuse
If the altitude from B bisects AC at X, triangle AXB is a right triangle with side AX adjacent to the angle A, and side AB as the hypotenuse. AX is half of AC, so has length 12/2 = 6, telling you the cosine of angle A is AX/AB = 6/10.
A diagram does not have to be sophisticated to be useful.
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