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Chapter 6: Problem 11

Solve each triangle. Round lengths to the nearest tenth and angle measures to the nearest degree. $$B=85^{\circ}, C=15^{\circ}, b=40$$

Short Answer

Expert verified
Solving the triangle gives lengths \(a \approx 7.3\) and angles \(B \approx 47^{\circ}, C \approx 31^{\circ}\).

Step by step solution

01

Apply the Law of Cosines to find side a

Apply the Law of Cosines in the form \(a^2 = b^2 + c^2 - 2bc \cdot \cos(A)\) to find side \(a\). Plugging the given values, we have \(a^2 = 5^2 + 3^2 - 2 \cdot 5 \cdot 3 \cdot \cos(102^{\circ})\), which after solving gives \(a \approx 7.3\).
02

Apply the Law of Sines to find angle B

Now use the Law of Sines in the form \(\frac{b}{sin(B)} = \frac{a}{sin(A)}\) to find angle \(B (ABC)\). Rearrange to \(sin(B) = \frac{b \cdot sin(A)}{a}\), plug given and calculated values, \(sin(B) = \frac{5 \cdot sin(102^{\circ})}{7.3}\), and then find the angle by applying the inverse sine operation, which gives \(B \approx 47^{\circ}\).
03

Find angle C

Angles in a triangle always add up to 180 degrees. So, to find \(C (BCA)\), subtract the known angles from 180, i.e., \(C = 180^{\circ} - A - B = 180^{\circ} - 102^{\circ} - 47^{\circ} \approx 31^{\circ}\).

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