Trigonometry - Law of Sines

Given sides , and angle $\angle A = 100^{\circ}$ determine the corresponding value for $\angle B$

$29.5^{\circ}$

$-29.5^{\circ}}$

Undefined

$5^{\circ}$

$60^{\circ}$

Explanation

The Law of Sines is used here since we have Side - Angle - Side. We setup our equation as follows:

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

Next, we substitute the known values:

$\frac{10}{\sin 100} = \frac{5}{\sin B}$

Now we cross multiply:

$10\sin B = 5\sin 100$

Divide by 10 on both sides:

$\sin B = 0.5\sin100$

Finally taking the inverse sine to obtain the desired angle:

$B = sin^{-1}\left(0.5\sin100\right ) = 29.5^{\circ}$

Let $\angle A = 17^{\circ}$ , $\angle C = 110^{\circ}$ and , determine the length of side .

Explanation

We have two angles and one side, however we do not have $\angle B$ . We can determine the angle using the property of angles in a triangle summing to $180^{\circ}$ :

$\angle B = 180^{\circ} - 17^{\circ} - 110^{\circ} = 53^{\circ}$

Now we can simply utilize the Law of Sines:

$\frac{5}{\sin 17} = \frac{b}{\sin 53}$

Cross multiply and divide:

$\frac{5\sin 53}{\sin 17} = b$

Reducing to obtain the final solution:

Triangle

In the above triangle, $a = 35^{\circ}$ and $c = 93^{\circ}$ . If , what is to the nearest tenth? (note: triangle not to scale)

Explanation

If we solve for , we can use the Law of Sines to find .

Since the sum of angles in a triangle equals $180^{\circ}$ ,

$a + b + c = 180^{\circ}$

$35^{\circ} + b + 93^{\circ} = 180^{\circ}$

$b = 52^{\circ}$

Now, using the Law of Sines:

$\frac{\textup{side opposite a}}{\sin\left ( a\right )} = \frac{\textup{side opposite b}}{\sin\left ( b\right )} = \frac{\textup{side opposite c}}{\sin\left ( c\right )}$

$\frac{{X}}{\sin\left ( c\right )} = \frac{{Y}}{\sin\left ( b\right )}$

$\frac{{49}}{\sin\left ( 93^{\circ}\right )} = \frac{{Y}}{\sin\left ( 52^{\circ}\right )}$

$\frac{49}{0.999} \approx \frac{Y}{0.788}$

$Y \approx \frac{49 \cdot 0.788 }{0.999}$

$Y \approx 38.7$

If , , and determine the length of side , round to the nearest whole number.

Explanation

This is a straightforward Law of Sines problem as we are given two angles and a corresponding side:

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

Substituting the known values:

$\frac{a}{\sin 39} = \frac{19}{\sin 100}$

Solving for the unknown side:

$a = \sin 39 \left(\frac{19}{\sin 100} \right ) = 12.14 = 12$

If $B = 72^{\circ}$ , , and determine the measure of $\angle A$ , round to the nearest degree.

$7^{\circ}$

$6^{\circ}$

$8^{\circ}$

$72^{\circ}$

$91^{\circ}$

Explanation

This is a straightforward Law of Sines problem since we are given one angle and two sides and are asked to determine the corresponding angle.

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

Substituting the given values:

$\frac{100}{\sin A} = \frac{802}{\sin 72^{\circ}}$

Now rearranging the equation:

$\sin A = \frac{100\sin 72^{\circ}}{802}$

The final step is to take the inverse sine of both sides:

$A = \sin^{-1}\left(\frac{100\sin 72^{\circ}}{802}\right) = 6.81^{\circ} = 7^{\circ}$

Screen_shot_2015-03-07_at_5.09.32_pm

By what factor is larger than in the triangle pictured above.

It isn't

Explanation

The Law of Sines states

$\frac{a}{\sin(A)}=\frac{b}{\sin(B)}=\frac{c}{\sin(C)}$

so for a and b, that sets up

$\frac{a}{\sin(35)}=\frac{b}{\sin(60)}, b=a\frac{\sin(60)}{\sin(35)}=1.51a$

If $\small b=4$ , $\small \angle B$ = $\small 2^o$ , and $\small \small \angle A$ = $\small 40^o$ , find the length of side $\small a$ .

$\small 74$

$\small 19$

$\small 14$

$\small 26$

Explanation

We are given two angles and the length of the corresponding side to one of those angles. Because the problem is asking for the corresponding length of the other angle we can use the Law of Sines to find the length of the side $\small a$ . The equation for the Law of Sines is

$\small \frac{a}{\sin\angle A} = \frac{b}{\sin\angle B}$

If we rearrange the equation to isolate we obtain

$\small a = \frac{b\sin\angle A}{\sin\angle B}$

Substituting on the values given in the problem

$\small a = \frac{4\sin40^o}{\sin2^o}$

$\small a = 74$

If $\small c=18$ , $\small \small \angle B$ = $\small 22^o$ , and $\small \angle C$ = $\small 7^o$ , find the length of side $\small b$ to the nearest whole number.

$\small 55$

$\small 30$

$\small 63$

$\small 71$

Explanation

Because we are given the two angles and the length of the corresponding side to one of those angles, we can use the Law of Sines to find the length of the side that we need. So we use the equation

$\small \frac{b}{\sin\angle B} = \frac{c}{\sin\angle C}$

Rearranging the equation to isolate $\small b$ gives

$\small b = \frac{c\sin\angle B}{\sin\angle C}$

Substituting in the values from the problem gives

$\small b = \frac{18\sin22^o}{\sin7^o}$

$\small b = 55$

If $\small \angle A = 40^o$ , $\small b = 3.5$ , and $\small \angle B = 35^o$ , find to the nearest whole number.

$\small 4$

$\small 7$

$\small 8$

$\small 11$

Explanation

We can use the Law of Sines to find the length of the missing side, because we have its corresponding angle and the length and angle of another side. The equation for the Law of Sines is

$\small \frac{a}{\sin\angle A} = \frac{b}{\sin\angle B}$

Isolating $\small a$ gives us

$\small a = \frac{b\sin\angle A}{\sin\angle B}$

Finally, substituting in the values of the of from the problem gives

$\small a= \frac{3.5\sin40^o}{\sin35^o}$

$\small a = 4$

Solve for $\theta$ :
Sines 1

Explanation

To solve, use the law of sines, $\frac{ \sin A }{a } = \frac{ \sin B }{b}$ where a is the side across from the angle A, and b is the side across from the angle B.