QuestionJuly 14, 2025

Solve the triangle. a=6,b=14,c=18 Aapprox 16.2^circ (Round to one decimal place as needed.) Bapprox square ^circ (Round to one decimal place as needed.)

Solve the triangle. a=6,b=14,c=18 Aapprox 16.2^circ (Round to one decimal place as needed.) Bapprox square ^circ (Round to one decimal place as needed.)
Solve the triangle. a=6,b=14,c=18 Aapprox 16.2^circ (Round to one decimal place as needed.) Bapprox square ^circ (Round to one decimal place as needed.)

Solution
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Answer

B \approx 40.0^{\circ} Explanation 1. Use the Law of Cosines Calculate angle C using c^2 = a^2 + b^2 - 2ab \cdot \cos(C). Rearrange to find \cos(C) = \frac{a^2 + b^2 - c^2}{2ab}. Substitute values: \cos(C) = \frac{6^2 + 14^2 - 18^2}{2 \cdot 6 \cdot 14} = \frac{36 + 196 - 324}{168} = \frac{-92}{168}. Calculate C: C = \cos^{-1}\left(\frac{-92}{168}\right). 2. Calculate Angle C C \approx 123.8^{\circ} (rounded to one decimal place). 3. Use the Law of Sines Use \frac{\sin A}{a} = \frac{\sin B}{b} = \frac{\sin C}{c}. Solve for B: \sin B = \frac{b \cdot \sin C}{c}. Substitute values: \sin B = \frac{14 \cdot \sin(123.8^\circ)}{18}. 4. Calculate Angle B B \approx 40.0^{\circ} (rounded to one decimal place).

Explanation

1. Use the Law of Cosines<br /> Calculate angle $C$ using $c^2 = a^2 + b^2 - 2ab \cdot \cos(C)$. <br /> Rearrange to find $\cos(C) = \frac{a^2 + b^2 - c^2}{2ab}$.<br /> Substitute values: $\cos(C) = \frac{6^2 + 14^2 - 18^2}{2 \cdot 6 \cdot 14} = \frac{36 + 196 - 324}{168} = \frac{-92}{168}$.<br /> Calculate $C$: $C = \cos^{-1}\left(\frac{-92}{168}\right)$.<br /><br />2. Calculate Angle C<br /> $C \approx 123.8^{\circ}$ (rounded to one decimal place).<br /><br />3. Use the Law of Sines<br /> Use $\frac{\sin A}{a} = \frac{\sin B}{b} = \frac{\sin C}{c}$.<br /> Solve for $B$: $\sin B = \frac{b \cdot \sin C}{c}$.<br /> Substitute values: $\sin B = \frac{14 \cdot \sin(123.8^\circ)}{18}$.<br /><br />4. Calculate Angle B<br /> $B \approx 40.0^{\circ}$ (rounded to one decimal place).
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