Question

Use the Law of Sines to solve (if possible) the triangle. If two solutions exist, find both. Round your answers to two decimal places. (If not possible, enter IMPOSSIBLE.) A = 72°, a = 34, b = 21

Answer 1

Answer:

Given: A = 72°, a = 34, b = 21

Calculated: B = 35.97°, C = 72.03°, c = 34.00

Step-by-step explanation:

$\dfrac{\sin A}{a} = \dfrac{\sin B}{b}$

$\dfrac{\sin 72^\circ}{34} = \dfrac{\sin B}{21}$

$\sin B = \dfrac{21\sin 72^\circ}{34}$

$\sin B = 0.5874$

$B = \sin^{-1} 0.5874$

$B = 35.97^\circ$

C = 180° - 72° - 35.97°

C = 72.03°

$\dfrac{\sin A}{a} = \dfrac{\sin C}{c}$

$\dfrac{\sin 72^\circ}{34} = \dfrac{\sin 72.03^\circ}{c}$

$c = \dfrac{\sin 72.03^\circ \times 34}{\sin 72^\circ}$

$c = 34.00$

Given: A = 72°, a = 34, b = 21

Calculated: B = 35.97°, C = 72.03°, c = 34.00

Answer 2

Answer:

$B\approx35.97^\circ\\C\approx72.03^\circ\\c\approx34$

Step-by-step explanation:

Law of Sines

$\frac{sinA}{a}=\frac{sinB}{b}=\frac{sinC}{c}$

Given information

$A=72^\circ\\a=34\\b=21$

Check if solutions exist

As $A=72^\circ < 90^\circ$ and that $a > b\rightarrow 34 > 21$, then there exists only one possible triangle by the Ambiguous Case

Solve the triangle

$\frac{sin(72^\circ)}{34}=\frac{sin(B)}{21}\\ \\ 21sin(72^\circ)=34sin(B)\\\\\frac{21sin(72^\circ)}{34}=sin(B)\\ \\B=sin^{-1}(\frac{21sin(72^\circ)}{34})\\ \\B=35.97394255^\circ\approx35.97^\circ$

$A+B+C=180^\circ\\\\72^\circ+35.97394255^\circ+C=180^\circ\\\\107.97394255^\circ+C=180^\circ\\\\C=72.02605745^\circ\approx72.03^\circ$

$\frac{sin(72^\circ)}{34}=\frac{sin(72.02605745^\circ)}{c}\\\\c*sin(72^\circ)=34sin(72.02605745^\circ)\\\\c=\frac{34sin(72.02605745^\circ)}{sin(72^\circ)}\\ \\c=34.00502065\approx34$