The angle of the triangle ABC from smallest to largest are approximately 22°, 27° and 131°.
The sides of the triangle are as follows;
AB = 20
BC = 12
AC = 10
Let's find ∠A using cosine rule,
a² = b² + c² - 2bc cos A
cos A = b² + c² - a² / 2bc
Therefore,
cos A = 10² + 20² - 12² / 2 × 10 × 20
cos A = 100 + 400 - 144 / 400
cos A = 356 / 400
cos A = 0.89
A = cos⁻¹ 0.89
A = 27.1267531173
∠A ≈ 27°
Using sine rule let's find ∠B
12 / sin 27 = 10 / sin B
cross multiply
12 sin B = 10 sin 27
sin B = 10 sin 27 / 12
sin B = 10 × 0.45399049974 / 12
sin B = 4.5399049974 / 12
B = sin⁻¹ 0.37832541645
B = 22.2299936087
∠B = 22°
Using sum of triangle theorem let's find ∠C
∠C = 180 - 27 - 22
∠C = 131°
The angles from smallest to largest are approximately 22°, 27° and 131°.
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Answer is D. hope this helps
