Law of Sines - Math

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

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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 }$c}{\sin(c)}$ .

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)}$

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

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

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

Cross multiply:

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

Multiply both sides by :

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

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

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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 }$c}{\sin(c)}$ .

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)}$

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

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

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

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

In this figure, angle $a=30^\circ$ . If side and , what is the value of angle ?

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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 }$c}{\sin(c)}$ .

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)}$

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

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

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

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

$b=56.44^\circ$

In this figure, if angle $a=18.5^\circ$ , side , and side , what is the value of angle ?

(NOTE: Figure not necessarily drawn to scale.)

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

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

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)}$

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

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

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

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

$b=10.41^\circ$

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

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Since we are given and want to find , we apply the Law of Sines, which states, in part,

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

$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 }$$

Substitute in the above equation:

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

Cross-multiply and solve for :

$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$

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

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

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

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

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

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

$\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 }$$

There are two angles between and that have any given positive sine other than 1 - we get the other by subtracting the previous result from :

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

This, however, is impossible, since this would result in the sum of the triangle measures being greater than . This leaves as the only possible answer.

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

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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 }$c}{\sin(c)}$ .

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)}$

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

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

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

Cross multiply:

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

Multiply both sides by :

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

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

Tap to see back →

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 }$c}{\sin(c)}$ .

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)}$

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

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

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

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

In this figure, angle $a=30^\circ$ . If side and , what is the value of angle ?

Tap to see back →

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 }$c}{\sin(c)}$ .

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)}$

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

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

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

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

$b=56.44^\circ$

In this figure, if angle $a=18.5^\circ$ , side , and side , what is the value of angle ?

(NOTE: Figure not necessarily drawn to scale.)

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

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

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)}$

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

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

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

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

$b=10.41^\circ$

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

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Since we are given and want to find , we apply the Law of Sines, which states, in part,

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

$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 }$$

Substitute in the above equation:

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

Cross-multiply and solve for :

$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$

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

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

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

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

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

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

$\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 }$$

There are two angles between and that have any given positive sine other than 1 - we get the other by subtracting the previous result from :

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

This, however, is impossible, since this would result in the sum of the triangle measures being greater than . This leaves as the only possible answer.

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

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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 }$c}{\sin(c)}$ .

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)}$

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

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

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

Cross multiply:

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

Multiply both sides by :

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

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

Tap to see back →

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 }$c}{\sin(c)}$ .

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)}$

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

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

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

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

In this figure, angle $a=30^\circ$ . If side and , what is the value of angle ?

Tap to see back →

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 }$c}{\sin(c)}$ .

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)}$

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

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

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

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

$b=56.44^\circ$

In this figure, if angle $a=18.5^\circ$ , side , and side , what is the value of angle ?

(NOTE: Figure not necessarily drawn to scale.)

Tap to see back →

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

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

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)}$

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

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

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

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

$b=10.41^\circ$

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

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Since we are given and want to find , we apply the Law of Sines, which states, in part,

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

$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 }$$

Substitute in the above equation:

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

Cross-multiply and solve for :

$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$

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

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

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

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

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

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

$\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 }$$

There are two angles between and that have any given positive sine other than 1 - we get the other by subtracting the previous result from :

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

This, however, is impossible, since this would result in the sum of the triangle measures being greater than . This leaves as the only possible answer.

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

Tap to see back →

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 }$c}{\sin(c)}$ .

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)}$

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

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

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

Cross multiply:

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

Multiply both sides by :

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

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

Tap to see back →

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 }$c}{\sin(c)}$ .

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)}$

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

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

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

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