List the angles of ∆ abc from smallest to largest. Ab = 20, bc = 12, ca = 10.

Answers

Answer 1

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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