Question

If A = 30°, B = 45°, a = 6 √2, solve the triangle ABC​

Answer 1

Answer:

I don't know what will you say

Answer 2

Given :- If A = 30°, B = 45°, a = 6 √2, solve the triangle ABC ?

Solution :-

given that,

→ ∠A = 30°

→ ∠B = 45°

so,

→ ∠A + ∠B + ∠C = 180° { By angle sum property. }

→ 30° + 45° + ∠C = 180°

→ 75° + ∠C = 180°

→ ∠C = 180° - 75°

→ ∠C = 105°

now sine rule says that,

• a/sin A = b/sin B = c/sin C

comparing first two,

→ a/sin A = b/sin B

→ 6√2/sin 30° = b/sin 45°

→ 6√2/b = sin 30° / sin 45°

→ 6√2/b = (1/2) / (1/√2)

→ 6√2/b = (√2/2)

→ 6/b = 1/2

→ b = 12

again,

→ a/sin A = c/sin C

→ a/c = sin A / sin C

→ 6√2/c = sin 30° / sin 105°

→ 6√2/c = (1/2) / (√2 + √6/4)

→ 6√2/c = 2/(√2 + √6)

→ 3√2/c = 1/(√2 + √6)

→ c = 3√2(√2 + √6)

→ c = (6 + 3√12)

→ c = 3(2 + √12)

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