How do you find the measure of angle B for the triangle with the following dimensions. A = 71° , a = 9.3, b = 8.5?

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How do you find the measure of angle B for the triangle with the following dimensions. A = 71° , a = 9.3, b = 8.5?

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Answer 1 · 2015-08-18T22:15:25

${B=59.8}^{o}$ $"B=59.8"^"o"$
Use the law of sines: $\frac{a}{sinA} = \frac{b}{sinB} = \frac{c}{sinC}$ $"a"/("sinA")="b"/"sinB"="c"/"sinC"$ .

Explanation:

Law of sines: $\frac{a}{sinA} = \frac{b}{sinB} = \frac{c}{sinC}$ $"a"/("sinA")="b"/"sinB"="c"/"sinC"$ .

${A=71}^{o}$ $"A=71"^"o"$
$a = 9.3$ $"a"=9.3$
$b = 8.5$ $"b"=8.5$
$B=?$ $"B=?"$

Since there are no values for side c and angle C, we can ignore this ratio. Since we have the sides a and b, but don't have angle B, we can invert the ratios for the law of sines.

$\frac{sinA}{a} = \frac{sinB}{b}$ $("sinA")/"a"=("sinB")/"b"$ =

$\frac{{sin71}^{o}}{8.5} = \frac{sinB}{9.3}$ $("sin71"^"o")/8.5=("sinB")/9.3$

Divide $sinA$ $"sinA"$ by side $a$ $"a"$ .

$0.101669 = \frac{sinB}{8.5}$ $0.101669=("sinB")/8.5$

Multiply both sides by $8.5$ $8.5$ .

$0.101669 \times 8.5 = sinB$ $0.101669xx8.5="sinB"$

$0.8641865 = sinB$ $0.8641865="sinB"$

Switch sides.

$sinB = 0.8641865$ $"sinB"=0.8641865$

Determine angle $B$ $"B"$ by taking the inverse sin of $0.85$ $0.85$ .

${B=sin}^{- 1} (0.8641865)$ $"B=sin"^(-1)(0.8641865)$

${B=59.8}^{o}$ $"B=59.8"^"o"$

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How do you find the measure of angle B for the triangle with the following dimensions. A = 71° , a = 9.3, b = 8.5?

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