QUESTION IMAGE
Question
solve the triangle.
b = 68°51, c = 38 m, a = 73 m
what is the length of side b?
b = □ m
(round to the nearest whole number as needed.)
what is the measure of angle a?
a = □°□
(round to the nearest whole number as needed.)
what is the measure of angle c?
c = □°□
(round to the nearest whole number as needed.)
Step1: Convert angle B to decimal degrees
First, convert \( 51' \) to degrees. Since \( 1^\circ = 60' \), we have \( 51'=\frac{51}{60}=0.85^\circ \). So \( B = 68^\circ + 0.85^\circ=68.85^\circ \).
Step2: Use the Law of Cosines to find side b
The Law of Cosines states that \( b^{2}=a^{2}+c^{2}-2ac\cos B \).
Given \( a = 73 \), \( c = 38 \), and \( B = 68.85^\circ \).
First, calculate \( \cos(68.85^\circ)\approx\cos(68.85)\approx0.3617 \)
Then, \( a^{2}=73^{2} = 5329 \), \( c^{2}=38^{2}=1444 \), and \( 2ac=2\times73\times38 = 5548 \)
So \( b^{2}=5329 + 1444- 5548\times0.3617 \)
\( b^{2}=6773-5548\times0.3617 \)
\( 5548\times0.3617\approx5548\times0.3617\approx2006.7 \)
\( b^{2}=6773 - 2006.7=4766.3 \)
Then \( b=\sqrt{4766.3}\approx69.04\approx69 \)
Step3: Use the Law of Sines to find angle A
The Law of Sines states that \( \frac{\sin A}{a}=\frac{\sin B}{b} \)
We know \( a = 73 \), \( b\approx69 \), \( B = 68.85^\circ \)
So \( \sin A=\frac{a\sin B}{b}=\frac{73\times\sin(68.85^\circ)}{69} \)
\( \sin(68.85^\circ)\approx0.9316 \)
\( 73\times0.9316\approx68.0068 \)
\( \sin A=\frac{68.0068}{69}\approx0.9856 \)
Then \( A=\arcsin(0.9856)\approx80.1^\circ \)
Convert \( 0.1^\circ \) to minutes: \( 0.1\times60 = 6' \), so \( A\approx80^\circ6' \)
Step4: Find angle C
Since the sum of angles in a triangle is \( 180^\circ \), \( C=180^\circ - A - B \)
\( A\approx80.1^\circ \), \( B = 68.85^\circ \)
\( C=180-(80.1 + 68.85)=180 - 148.95 = 31.05^\circ \)
Convert \( 0.05^\circ \) to minutes: \( 0.05\times60 = 3' \), so \( C\approx31^\circ3' \)
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- Length of side \( b \): \( \boxed{69} \)
- Measure of angle \( A \): \( \boxed{80^\circ 6'} \)
- Measure of angle \( C \): \( \boxed{31^\circ 3'} \)