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Chapter 10: Problem 6

In questions \(1-10\) solve \(\triangle \mathrm{ABC}\) given \(\mathrm{AB}=32 \mathrm{~cm}, \mathrm{BC}=30 \mathrm{~cm}, \mathrm{AC}=41 \mathrm{~cm}\)

Short Answer

Expert verified
Question: Given a triangle ABC with side lengths AB = 32 cm, BC = 30 cm, and AC = 41 cm, find the angles of the triangle. Answer: The angles of the triangle ABC are ∠A ≈ 75.87°, ∠B ≈ 62.05°, and ∠C ≈ 42.08°.

Step by step solution

01

Use the Cosine Law to find angle A

Using the Cosine Law, which states that \(c^2 = a^2 + b^2 - 2ab \cos C\), we can find angle A (∠A) by plugging in the provided side lengths: \(a = 30\mathrm{~cm}, b = 41\mathrm{~cm}, c = 32\mathrm{~cm}\) \(32^2 = 30^2 + 41^2 - 2(30)(41) \cos A\) Solve for angle A: \(\cos A = \frac{30^2 + 41^2 - 32^2}{2(30)(41)}\) Now calculate the value of angle A: \(A \approx \arccos{(0.2411)} \approx 75.87^\circ\)
02

Use the Cosine Law to find angle B

Similarly, we can find angle B (∠B) using the Cosine Law: \(a = 32\mathrm{~cm}, b = 41\mathrm{~cm}, c = 30\mathrm{~cm}\) \(30^2 = 32^2 + 41^2 - 2(32)(41) \cos B\) Solve for angle B: \(\cos B = \frac{32^2 + 41^2 - 30^2}{2(32)(41)}\) Now calculate the value of angle B: \(B \approx \arccos{(0.4695)} \approx 62.05^\circ\)
03

Find angle C using the Triangle Angle Sum theorem

The Triangle Angle Sum theorem states that the sum of the angles in a triangle is \(180^\circ\). So, to find angle C (∠C), we can subtract the other two angles from \(180^\circ\): \(C = 180^\circ - A - B \approx 180^\circ - 75.87^\circ - 62.05^\circ \approx 42.08^\circ\)
04

Compile the results

The solved triangle ABC has the following angles and side lengths: ∠A ≈ 75.87°, ∠B ≈ 62.05°, ∠C ≈ 42.08° AB = 32 cm, BC = 30 cm, AC = 41 cm

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Most popular questions from this chapter

For questions \(1-10\) solve \(\triangle \mathrm{ABC}\) given \(B=18^{\circ}, C=110^{\circ}, \mathrm{BC}=12.3 \mathrm{~cm}\)

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In questions 1-11 \(\Delta \mathrm{ABC}\) has a right angle at \(\mathrm{C}\). Calculate \(\mathrm{AC}\) given \(\mathrm{BC}=10 \mathrm{~cm}\) and \(A=40^{\circ}\).

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