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Chapter 10: Problem 4

In questions \(1-10\) solve \(\triangle \mathrm{ABC}\) given \(\mathrm{BC}=36 \mathrm{~cm}, \mathrm{AC}=92 \mathrm{~cm}, C=51^{\circ}\)

Short Answer

Expert verified
Answer: The side lengths of triangle ABC are approximately AB = 120.5 cm, BC = 36 cm, and AC = 92 cm. The angles are approximately A = 38.3°, B = 90.7°, and C = 51°.

Step by step solution

01

Check for the existence of triangle using Law of Sines

Use the Law of Sines to determine if the given information results in a unique triangle: \(\frac{\mathrm{BC}}{\sin A} = \frac{\mathrm{AC}}{\sin C}\) Plug in the given side lengths and angle C: \(\frac{36}{\sin A} = \frac{92}{\sin 51^{\circ}}\) Now, solve for \(\sin A\): \(\sin A = \frac{36 \times \sin 51^{\circ}}{92} \approx 0.6237\) Since \(0 \leq \sin A \leq 1\), there exists a unique triangle with the given side lengths and angle.
02

Find angle A using the inverse sine function

To find angle A, take the inverse sine of the value we found for \(\sin A\): \(A = \sin^{-1}(0.6237) \approx 38.3^{\circ}\)
03

Find angle B using the angle sum property of triangles

Since the sum of the angles in a triangle is always \(180^{\circ}\), we can find angle B as follows: \(B = 180^{\circ} - A - C \approx 180^{\circ} - 38.3^{\circ} - 51^{\circ} \approx 90.7^{\circ}\)
04

Find side AB using the Law of Sines

Finally, we can find the remaining side length (AB) using the Law of Sines: \(\frac{\mathrm{AB}}{\sin B} = \frac{\mathrm{AC}}{\sin C}\) Plug in the values of AC, C, and B: \(\mathrm{AB} = \frac{\mathrm{AC} \times \sin B}{\sin C} = \frac{92 \times \sin 90.7^{\circ}}{\sin 51^{\circ}} \approx 120.5 \mathrm{~cm}\) Now we have completely solved triangle ABC with side lengths AB \(\approx 120.5 \mathrm{~cm}\), BC \(= 36 \mathrm{~cm}\), and AC \(= 92 \mathrm{~cm}\), and angles A \(\approx 38.3^{\circ}\), B \(\approx 90.7^{\circ}\), and C \(= 51^{\circ}\).

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