Given: ∆abc, ac = 12,
ab = bc = 10
find: m∠a, m∠c, m∠b

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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