Question

Please Help! Unit 8: Right Triangles & Trigonometry Homework 7: Law of Sines

Answer 1

Answer:

9= 71.67°

10= 60.65°

11= 86.59°

12= 62.30°

13= 34.51°

14= 51.71°

15= 22.87°

16= 44.63°

Step-by-step explanation:

The law of sine requires that if we have three sides of length A, B and C with their respective angles as a, b and c then these can be related by

$\frac{a}{Sin \ A\textdegree}=\frac{b}{Sin \ B\textdegree}=\frac{c}{Sin \ C\textdegree}$

9.

$\frac{11}{Sin \ 27^{\circ}}=\frac{b}{Sin \ B\textdegree}=\frac{c}{Sin \ C \textdegree}$

Making x the subject of the formula then

$x=sin^{-1}(\frac {23\ sin\ 27^{\circ}}{11}\approx 71.67^{\circ}$

10.

$\frac{8}{Sin \ 85\textdegree}=\frac{7}{Sin \ x\textdegree}$

Making x the subject of the formula then

$x=sin^{-1}(\frac {7\ sin\ 85^{\circ}}{8}\approx 60.65^{\circ}$

11.

$\frac{26}{Sin \ 74\textdegree}=\frac{27}{Sin \ x\textdegree}$

Making x the subject of the formula then

$x=sin^{-1}(\frac {27\ sin\ 74^{\circ}}{26}\approx 86.59^{\circ}$

12.

$\frac{12}{Sin \ 34\textdegree}=\frac{19}{Sin \ x\textdegree}$

Making x the subject of the formula then

$x=sin^{-1}(\frac {19\ sin\ 34^{\circ}}{12}\approx 62.30^{\circ}$

13.

$\frac{30}{Sin \ 91\textdegree}=\frac{17}{Sin \ B\textdegree}$

Making x the subject of the formula then

$x=sin^{-1}(\frac {17\ sin\ 91^{\circ}}{30}\approx 34.51^{\circ}$

14.

$\frac{25}{Sin \ 49\textdegree}=\frac{26}{Sin \ x\textdegree}$

Making x the subject of the formula then

$x=sin^{-1}(\frac {26\ sin\ 49^{\circ}}{25}\approx 51.71^{\circ}$

15.

$\frac{45}{Sin \ 119\textdegree}=\frac{20}{Sin \ x\textdegree}$

Making x the subject of the formula then

$x=Sin^{-1}(\frac{20\ Sin \ 119\textdegree}{45})\approx22.87\textdegree$

16.

$\frac{11}{Sin \ 105\textdegree}=\frac{8}{Sin \ x\textdegree}$

Making x the subject of the formula then

$x=sin^{-1}(\frac {8\ sin\ 105^{\circ}}{11}\approx 44.63^{\circ}$