Question

solve the triangle.
a = 14.0 ft, b = 18.0 ft, c = 21.0 ft
what is the measure of angle a?
a = □°
(round to two decimal places as needed.)
what is the measure of angle b?
b = □°
(round to two decimal places as needed.)
what is the measure of angle c?
c = □°
(round to two decimal places as needed.)

Explanation:

Step1: Use the Law of Cosines to find angle A

The Law of Cosines formula for angle \( A \) is \( \cos A=\frac{b^{2}+c^{2}-a^{2}}{2bc} \).
Substitute \( a = 14.0 \), \( b = 18.0 \), \( c = 21.0 \) into the formula:
\( \cos A=\frac{18.0^{2}+21.0^{2}-14.0^{2}}{2\times18.0\times21.0}=\frac{324 + 441-196}{756}=\frac{569}{756}\approx0.7526 \)
Then \( A=\arccos(0.7526)\approx40.47^{\circ} \)

Step2: Use the Law of Cosines to find angle B

The Law of Cosines formula for angle \( B \) is \( \cos B=\frac{a^{2}+c^{2}-b^{2}}{2ac} \).
Substitute \( a = 14.0 \), \( b = 18.0 \), \( c = 21.0 \) into the formula:
\( \cos B=\frac{14.0^{2}+21.0^{2}-18.0^{2}}{2\times14.0\times21.0}=\frac{196 + 441 - 324}{588}=\frac{313}{588}\approx0.5323 \)
Then \( B=\arccos(0.5323)\approx57.83^{\circ} \)

Step3: Use the fact that the sum of angles in a triangle is \( 180^{\circ} \) to find angle C

We know that \( A + B + C=180^{\circ} \), so \( C = 180^{\circ}-A - B \)
Substitute \( A\approx40.47^{\circ} \) and \( B\approx57.83^{\circ} \) into the formula:
\( C=180-(40.47 + 57.83)=180 - 98.3=81.70^{\circ} \) (We can also verify using Law of Cosines for angle C, the result will be consistent)

Answer:

\( A\approx40.47^{\circ} \)
\( B\approx57.83^{\circ} \)
\( C\approx81.70^{\circ} \)