Please Help Me, It's Due Tomorrow. (7th Grade Math)
1 answer:
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Answer:
∠A1 = 27.4°, ∠A2 = 56.6°, ∠C1 =104.6°, ∠C2=75.4°, a1 = 79.9 and a2 = 144.9
Step-by-step explanation:
From Sine rule
∴ b / sinB = c / sinC
From the question,
b = 129, c = 168 and ∠B = 48°
∴ 129 / sin48° = 168 / sinC
Then, sinC = (168×sin48)/129
sinC = 0.9678
C = sin⁻¹(0.9678)
C = 75.42
∠C2=75.4°
and
∴∠C1 = 180° - 75.4°
∠C1 =104.6°
For ∠A
∠A1 = 180° - (104.6°+48°) [sum of angles in a triangle]
∠A1 = 27.4°
and
∠A2 = 180° - (75.4° + 48°)
∠A2 = 180° - (123.4°)
∠A2 = 56.6°
For side a
a1/sinA1 = b/sinB
a1/ sin27.4° = 129/sin48
a1 = (129×sin27.4°)/sin48
a1 = 79.8845
a1 = 79.9
and
a2/sinA2 = b / sinB
a2/ sin56.6° = 129/sin48
a2 = (129×sin56.6°)/sin48
a2 = 144.9184
a2 = 144.9
Hence,
∠A1 = 27.4°, ∠A2 = 56.6°, ∠C1 =104.6°, ∠C2=75.4°, a1 = 79.9 and a2 = 144.9