Law of Sines - Trigonometry

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

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

Since the sum of angles in a triangle equals ,

$a + b + c = $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}$$

$Y \approx 38.7$

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

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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 .

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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 :

$\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:

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

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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.

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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}$$}{802}\right) = $6.81^{\circ}$ = $7^{\circ}$$

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

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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 $\small a$ .

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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.

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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.

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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 $\theta$ :

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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.

$\frac{ \sin \theta }{ 19 }$ = $\frac{ \sin ( $20^o$ )}{ 7 }$ cross-multiply

$7 \sin \theta = 19 \cdot \sin $(20^o$ )$ evaluate the right side

$7 \sin \theta = 6.4984$ divide by 7

$\sin \theta = 0.9283$ take the inverse sine

$\theta = \sin ^ {-1 } (0.9283 ) \approx 68.18 ^o$

Evaluate using law of sines:

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

In this case, we have a 90-degree angle across from x, but we don't currently know the angle across from the side length 7. We can figure out this angle by subtracting from :

Now we can set up and solve using law of sines:

$$\frac{ \sin $(65^o$ )}{ 7 }$ = $\frac{ \sin (90 ^o )}$ { x }$ cross-multiply

$\sin (90 ^o ) \cdot 7 = \sin $(65^o$ ) \cdot x$ evaluate the sines

$1 \cdot 7 = (0.9063) x$ divide by 0.9063

If , find the remaining angles and sides.

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The Law of Sines is a set of ratios that allows one to compute missing angles and side lengths of oblique triangles (non right angle triangles).

The Law of Sines:

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

To find the missing angle A we subtract the sum of the two known angles from 180° as the interior angles of all triangles equal 180°.

The LOS can be rearranged to solve for the missing side.

$$\frac{a}{\sin A}$=\frac{b}{\sin B}$\rightarrow a=\frac{b}{\sin B}$\cdot\sin A\rightarrow $\frac{25}{\sin 30}$\cdot\sin 45=35.4ft$

$$\frac{b}{\sin B}$=\frac{c}{\sin C}$\rightarrow c=\frac{b}{\sin B}$\cdot\sin C\rightarrow c=\frac{25}{\sin 30}$\cdot\sin 105=48.3ft$

Find the length of the line segment $\overline{AB}$ in the triangle below.

Round to the nearest hundredth of a centimeter.

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The law of sines states that

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

In this triangle, we are looking for the side length c, and we are given angle A, angle B, and side b. The sum of the interior angles of a triangle is ; using subtraction we find that angle C = .

We can now form a proportion that includes only one unknown, c:

Solving for c, we find that

.

What is the measure of $\angle F$ in $\triangle DEF$ below? Round to the nearest tenth of a degree.

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The law of sines tells us that $\frac{sinA}{a}$=\frac{sinB}{b}$=\frac{sinC}{c}$ , where a, b, and c are the sides opposite of angles A, B, and C. In $\triangle DEF$ , these ratios can be used to find $\angle F$ :

$\frac{sinD}{d}$=\frac{sinF}{f}$

$\frac{sin114^\circ}{16}$=\frac{sin F}{5}$

$sinF=\frac{5sin114^\circ}{16}$$

$F\approx16.6^\circ$

In the triangle below, $m\angle A=78^\circ$ , $m\angle B=40^\circ$ , and . What is the length of side to the nearest tenth?

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First, find $m\angle C$ . The sum of the interior angles of a triangle is $180^\circ$ , so $m\angle C = 180^\circ - 78^\circ - 40^\circ$ , or $62^\circ$ .

Using this information, you can set up a proportion to find side b:

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

$\frac{sin62^\circ}{14}$=\frac{sin40^\circ}{b}$

$b=\frac{14sin40^\circ}{sin62^\circ}$$

$b\approx10.2$

In the triangle below, $m\angle A=75^\circ$ , $m\angle B=21^\circ$ , and .

What is the length of side a to the nearest tenth?

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To use the law of sines, first you must find the measure of $\angle C$ . Since the sum of the interior angles of a triangle is $180^\circ$ , $m\angle C=84^\circ$ .

Law of sines:

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

$\frac{\sin 75^\circ}{a}$=\frac{\sin 84^\circ}{19.3}$

$a=\frac{19.3 \sin 75^\circ}{\sin 84^\circ}$$

$a\approx 18.7$

The triangle above has side lengths 3, 4, and 6. The angle opposite the side of length 6 measures 117.28 degrees, rounded to the nearest hundredth. Angle $\theta$ is opposite the side of length 3. What is the measure of $\theta$ , rounded to the nearest hundredth of a degree?

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Because the angle with measure 117.28 degrees is opposite the side of length 6, and angle $\theta$ is opposite the side of length 3, we can use the Law of Sines to solve for the measure of $\theta$ .

$\theta=\arcsin($\frac{3\sin(117.28^\circ)}{6}$)$

$\theta=26.38^\circ$

Find the length of side a using the law of sines. All angles are in degrees.

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The law of sines states that, given a triangle with sides a, b, and c and angles A, B, and C opposite to the corresponding sides,

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

Since the sides and angles given are directly opposite, we can use the law of sines.

$$\frac{a}{\sin $55^{\circ}$$}=\frac{10}{\sin $60^{\circ}$$}$

Solving for a, we get

$a=\frac{10\sin $55^{\circ}$$}{\sin 60}$

Evaluating the expression, we find that

Find the measure of angle A.

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The law of sines states that, given a triangle, the following relationship is always true:

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

where a, b, and c are sides and A, B, and C are the angles opposite to the sides.

This problem does not give us the length of the side opposite to the angle we want to find, so we have to find it indirectly.

We start by finding the measure of the unmarked angle, which we'll represent as B:

$$\frac{10}{\sin B}$=\frac{12}{\sin $65^{\circ}$$}$

Solving for $\sin B$ , we get

$$\frac{10\sin $65^{\circ}$$}{12}=\sin B$

$B=\arcsin(0.755)=49^\circ$

Now that we have the measures of two angles, we can find the measure of the third by the theorem:

$65^\circ+49^\circ+A=180^\circ$

Find the length of side .

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Recall the Law of Sines:

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

Start by finding the value of angle :

Plug in the given values into the Law of Sines:

$\frac{12}{\sin 81}$=\frac{c}{\sin 50}$

Rearrange the equation to solve for :

$c=(\sin 50)$\frac{12}{\sin 81}$=9.31$

Make sure to round to two places after the decimal.