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High School Math : Applying the Law of Sines

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Example Questions

Example Question #1 : Triangles

Rt_triangle_letters
In this figure, angle $a=30^\circ$ $\displaystyle a=30^\circ$ and side Z=15 $\displaystyle Z=15$. If angle $b=45^\circ$ $\displaystyle b=45^\circ$, what is the length of side $\displaystyle Y$?

Possible Answers:

$15\sqrt{2}$ $\displaystyle 15\sqrt{2}$

7.5 $\displaystyle 7.5$

$\displaystyle 30$

$30\sqrt{2}$ $\displaystyle 30\sqrt{2}$

$\frac{15\sqrt{2}}{2}$ $\displaystyle \frac{15\sqrt{2}}{2}$

Correct answer:

$15\sqrt{2}$ $\displaystyle 15\sqrt{2}$

Explanation:

For this problem, use the law of sines:

$\frac{\text{side opposite }a}{\sin(a)}=\frac{\text{side opposite }b}{\sin(b)}=\frac{\text{side opposite }c}{\sin(c)}$ $\displaystyle \frac{\text{side opposite }a}{\sin(a)}=\frac{\text{side opposite }b}{\sin(b)}=\frac{\text{side opposite }c}{\sin(c)}$.

In this case, we have values that we can plug in:

$\frac{\text{side opposite }a}{\sin(a)}=\frac{\text{side opposite }b}{\sin(b)}$ $\displaystyle \frac{\text{side opposite }a}{\sin(a)}=\frac{\text{side opposite }b}{\sin(b)}$

$\frac{Z}{\sin(a)}=\frac{Y}{\sin{(b)}}$ $\displaystyle \frac{Z}{\sin(a)}=\frac{Y}{\sin{(b)}}$

$\frac{15}{\sin(30^\circ)}=\frac{Y}{\sin{(45^\circ)}}$ $\displaystyle \frac{15}{\sin(30^\circ)}=\frac{Y}{\sin{(45^\circ)}}$

$\frac{15}{\frac{1}{2}}=\frac{Y}{\frac{\sqrt{2}}{2}}$ $\displaystyle \frac{15}{\frac{1}{2}}=\frac{Y}{\frac{\sqrt{2}}{2}}$

Cross multiply:

$15*\frac{\sqrt{2}}{2}=\frac{1}{2}Y$ $\displaystyle 15*\frac{\sqrt{2}}{2}=\frac{1}{2}Y$

Multiply both sides by $\displaystyle 2$:

$15\sqrt{2}=Y$ $\displaystyle 15\sqrt{2}=Y$

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Example Question #2 : Triangles

Rt_triangle_letters

In this figure $a=22^\circ$ $\displaystyle a=22^\circ$ and $c=85^\circ$ $\displaystyle c=85^\circ$. If X=30 $\displaystyle X=30$, what is $\displaystyle Z$?

Possible Answers:

20.1 $\displaystyle 20.1$

11.28 $\displaystyle 11.28$

0.997 $\displaystyle 0.997$

0.374 $\displaystyle 0.374$

30.09 $\displaystyle 30.09$

Correct answer:

11.28 $\displaystyle 11.28$

Explanation:

For this problem, use the law of sines:

In this case, we have values that we can plug in:

$\frac{\text{side opposite }a}{\sin(a)}=\frac{\text{side opposite }c}{\sin(c)}$ $\displaystyle \frac{\text{side opposite }a}{\sin(a)}=\frac{\text{side opposite }c}{\sin(c)}$

$\frac{Z}{\sin(a)}=\frac{X}{\sin{(c)}}$ $\displaystyle \frac{Z}{\sin(a)}=\frac{X}{\sin{(c)}}$

$\frac{Z}{\sin(22^\circ)}=\frac{30}{\sin{85^\circ}}$ $\displaystyle \frac{Z}{\sin(22^\circ)}=\frac{30}{\sin{85^\circ}}$

$\frac{Z}{0.375}=\frac{30}{0.997}$ $\displaystyle \frac{Z}{0.375}=\frac{30}{0.997}$

$\frac{Z}{0.375}=30.09$ $\displaystyle \frac{Z}{0.375}=30.09$

$\displaystyle Z=30.09*0.375$

Z=11.28 $\displaystyle Z=11.28$

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Example Question #3 : Applying The Law Of Sines

In $\Delta ABC$ $\displaystyle \Delta ABC$, $m \angle A = 75^{\circ }$ $\displaystyle m \angle A = 75^{\circ }$ , $m \angle B = 62^{\circ }$ $\displaystyle m \angle B = 62^{\circ }$, and BC = 16 $\displaystyle BC = 16$. To the nearest tenth, what is $\displaystyle AB$?

Possible Answers:

22.7 $\displaystyle 22.7$

27.9 $\displaystyle 27.9$

11.3 $\displaystyle 11.3$

14.6 $\displaystyle 14.6$

17.5 $\displaystyle 17.5$

Correct answer:

11.3 $\displaystyle 11.3$

Explanation:

Since we are given $\displaystyle BC$ and want to find $\displaystyle AB$, we apply the Law of Sines, which states, in part,

$\frac{ \sin m \angle A}{B C} = \frac{ \sin m \angle C}{AB}$ $\displaystyle \frac{ \sin m \angle A}{B C} = \frac{ \sin m \angle C}{AB}$

$m \angle A = 75^{\circ }$ $\displaystyle m \angle A = 75^{\circ }$ and

$m \angle C = 180 - \left ( m \angle A + m \angle B \right )= 180 -(75+ 62) = 180 -137 = 43^{\circ }$ $\displaystyle m \angle C = 180 - \left ( m \angle A + m \angle B \right )= 180 -(75+ 62) = 180 -137 = 43^{\circ }$

Substitute in the above equation:

$\frac{ \sin75^{\circ }}{16} = \frac{ \sin 43^{\circ }}{AB}$ $\displaystyle \frac{ \sin75^{\circ }}{16} = \frac{ \sin 43^{\circ }}{AB}$

Cross-multiply and solve for $\displaystyle AB$:

$AB\cdot \sin75^{\circ } = 16 \cdot \sin 43^{\circ }$ $\displaystyle AB\cdot \sin75^{\circ } = 16 \cdot \sin 43^{\circ }$

$AB = \frac{16 \cdot \sin 43^{\circ }}{\sin75^{\circ }} = \frac{16 \cdot \0.6820}{0.9659} \approx 11.3$ $\displaystyle AB = \frac{16 \cdot \sin 43^{\circ }}{\sin75^{\circ }} = \frac{16 \cdot \0.6820}{0.9659} \approx 11.3$

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Example Question #4 : Applying The Law Of Sines

In $\Delta ABC$ $\displaystyle \Delta ABC$, $m \angle A = 66^{\circ }$ $\displaystyle m \angle A = 66^{\circ }$ , AB = 16 $\displaystyle AB = 16$, and BC = 23 $\displaystyle BC = 23$. To the nearest tenth, what is $m \angle C$ $\displaystyle m \angle C$?

Possible Answers:

$39.5^{\circ } \textrm{ or } 140.5^{\circ }$ $\displaystyle 39.5^{\circ } \textrm{ or } 140.5^{\circ }$

$73.6^{\circ }$ $\displaystyle 73.6^{\circ }$

No triangle can exist with these characteristics.

$39.5^{\circ }$ $\displaystyle 39.5^{\circ }$

$73.6^{\circ } \textrm{ or }106.4^{\circ }$ $\displaystyle 73.6^{\circ } \textrm{ or }106.4^{\circ }$

Correct answer:

$39.5^{\circ }$ $\displaystyle 39.5^{\circ }$

Explanation:

Since we are given $m \angle A$ $\displaystyle m \angle A$ , $\displaystyle AB$, and $\displaystyle BC$, and want to find $m \angle C$ $\displaystyle m \angle C$, we apply the Law of Sines, which states, in part,

$\frac{ \sin m \angle C}{AB} = \frac{ \sin m \angle A}{B C}$ $\displaystyle \frac{ \sin m \angle C}{AB} = \frac{ \sin m \angle A}{B C}$.

Substitute and solve for $\sin m \angle C$ $\displaystyle \sin m \angle C$:

$\frac{ \sin m \angle C}{16} = \frac{ \sin 66^{\circ }}{23}$ $\displaystyle \frac{ \sin m \angle C}{16} = \frac{ \sin 66^{\circ }}{23}$

$\frac{ \sin m \angle C}{16} \cdot 16= \frac{ \sin 66^{\circ }}{23}\cdot 16$ $\displaystyle \frac{ \sin m \angle C}{16} \cdot 16= \frac{ \sin 66^{\circ }}{23}\cdot 16$

$\sin m \angle C = \frac{ \sin 66^{\circ }}{23}\cdot 16 \approx \frac{ 0.9135}{23}\cdot 16\approx 0.6355$ $\displaystyle \sin m \angle C = \frac{ \sin 66^{\circ }}{23}\cdot 16 \approx \frac{ 0.9135}{23}\cdot 16\approx 0.6355$

Take the inverse sine of 0.6355:

$m \angle C = \sin^{-1} 0.6355 \approx 39.5^{\circ }$ $\displaystyle m \angle C = \sin^{-1} 0.6355 \approx 39.5^{\circ }$

There are two angles between $0^{\circ }$ $\displaystyle 0^{\circ }$ and $180^{\circ }$ $\displaystyle 180^{\circ }$ that have any given positive sine other than 1 - we get the other by subtracting the previous result from $180^{\circ }$ $\displaystyle 180^{\circ }$:

$180 - 39.5 = 140.5^{\circ }$ $\displaystyle 180 - 39.5 = 140.5^{\circ }$

This, however, is impossible, since this would result in the sum of the triangle measures being greater than $180^{\circ }$ $\displaystyle 180^{\circ }$. This leaves $39.5^{\circ }$ $\displaystyle 39.5^{\circ }$ as the only possible answer.

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Example Question #1 : Law Of Sines

Rt_triangle_letters

In this figure, angle $a=30^\circ$ $\displaystyle a=30^\circ$. If side Z =12 $\displaystyle Z =12$ and Y=20 $\displaystyle Y=20$, what is the value of angle $\displaystyle b$?

Possible Answers:

Undefined

$90^\circ$ $\displaystyle 90^\circ$

$24^\circ$ $\displaystyle 24^\circ$

$0.83^\circ$ $\displaystyle 0.83^\circ$

$56.44^\circ$ $\displaystyle 56.44^\circ$

Correct answer:

$56.44^\circ$ $\displaystyle 56.44^\circ$

Explanation:

For this problem, use the law of sines:

In this case, we have values that we can plug in:

$\frac{\text{side opposite }a}{\sin(a)}=\frac{\text{side opposite }b}{\sin(b)}$ $\displaystyle \frac{\text{side opposite }a}{\sin(a)}=\frac{\text{side opposite }b}{\sin(b)}$

$\frac{Z}{\sin(a)}=\frac{Y}{\sin{(b)}}$ $\displaystyle \frac{Z}{\sin(a)}=\frac{Y}{\sin{(b)}}$

$\frac{12}{\sin(30^\circ)}=\frac{20}{\sin{(b)}}$ $\displaystyle \frac{12}{\sin(30^\circ)}=\frac{20}{\sin{(b)}}$

$24=\frac{20}{\sin(b)}$ $\displaystyle 24=\frac{20}{\sin(b)}$

$\sin(b)=\frac{20}{24}$ $\displaystyle \sin(b)=\frac{20}{24}$

$b=\sin^{-1}(\frac{20}{24})$ $\displaystyle b=\sin^{-1}(\frac{20}{24})$

$b=56.44^\circ$ $\displaystyle b=56.44^\circ$

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Example Question #1 : Applying The Law Of Sines

Rt_triangle_letters In this figure, if angle $a=18.5^\circ$ $\displaystyle a=18.5^\circ$, side Z =30.2 $\displaystyle Z =30.2$, and side Y=17.2 $\displaystyle Y=17.2$, what is the value of angle $\displaystyle b$?

(NOTE: Figure not necessarily drawn to scale.)

Possible Answers:

$61.22^\circ$ $\displaystyle 61.22^\circ$

Undefined

$33.6^\circ$ $\displaystyle 33.6^\circ$

$71.5^\circ$ $\displaystyle 71.5^\circ$

$10.41^\circ$ $\displaystyle 10.41^\circ$

Correct answer:

$10.41^\circ$ $\displaystyle 10.41^\circ$

Explanation:

First, observe that this figure is clearly not drawn to scale. Now, we can solve using the law of sines:

In this case, we have values that we can plug in:

$\frac{\text{side opposite }a}{\sin(a)}=\frac{\text{side opposite }b}{\sin(b)}$ $\displaystyle \frac{\text{side opposite }a}{\sin(a)}=\frac{\text{side opposite }b}{\sin(b)}$

$\frac{Z}{\sin(a)}=\frac{Y}{\sin{(b)}}$ $\displaystyle \frac{Z}{\sin(a)}=\frac{Y}{\sin{(b)}}$

$\frac{30.2}{\sin(18.5^\circ)}=\frac{17.2}{\sin{(b)}}$ $\displaystyle \frac{30.2}{\sin(18.5^\circ)}=\frac{17.2}{\sin{(b)}}$

$95.18=\frac{17.2}{\sin(b)}$ $\displaystyle 95.18=\frac{17.2}{\sin(b)}$

$\sin(b)=\frac{17.2}{95.18}$ $\displaystyle \sin(b)=\frac{17.2}{95.18}$

$b=\sin^{-1}(\frac{17.2}{95.18})$ $\displaystyle b=\sin^{-1}(\frac{17.2}{95.18})$

$b=10.41^\circ$ $\displaystyle b=10.41^\circ$

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High School Math : Applying the Law of Sines

Study concepts, example questions & explanations for High School Math

All High School Math Resources

Example Questions

Example Question #1 : Triangles

Example Question #2 : Triangles

Example Question #3 : Applying The Law Of Sines

Example Question #4 : Applying The Law Of Sines

Example Question #1 : Law Of Sines

Example Question #1 : Applying The Law Of Sines

All High School Math Resources

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