Law of Cosines and Law of Sines - Trigonometry

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Question

If $\small B$ = $\small 90^o$ , $\small a=7$ , and $\small b=8$ find $\small \small \angle A$ to the nearest degree.

Answer

Notice that the given information is Angle-Side-Side, which is the ambiguous case. Therefore, we should test to see if there are no triangles that satisfy, one triangle that satisfies, or two triangles that satisfy this. From $\frac{\sin A}{a}=\frac{\sin B}{b}$ , we get $b\cdot \frac{\sin A}{a}=\sin B$ . In this equation, if $\sin B > 1$ , no angle A that satisfies the triangle can be found. If $\sin B = 1$ , $B=90^\circ$ and there is a right triangle determined. Finally, if $\sin B < 1$ , two measures of angle B can be calculated: an acute angle B and an obtuse angle $B'=180^\circ-B$ . In this case, there may be one or two triangles determined. If $B'+A\geq 180^\circ$ , then the angle B' is not a solution.

In this problem, $\sin B = 1$ , $B=90^\circ$ and there is one right triangle determined. Since we have the length of two sides of the triangle and the corresponding angle of one of the sides, we can use the Law of Sines to find the angle that we are looking for. This goes as follows:

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

Inputting the lengths of the triangle into this equation

$\small \frac{8}{\sin(90^o)} = \frac{7}{\sin\angle A}$

Isolating $\small \angle A$

$\small \angle A = \sin^{-1}(\frac{7\sin(90^o)}{8})$

$\small \angle A = 61^o$

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Question

If $\small c=10.3$ , $\small a=7.4$ , and $\small \angle A$ = $\small 34^o$ find $\small \angle C$ to the nearest degree.

Answer

Notice that the given information is Angle-Side-Side, which is the ambiguous case. Therefore, we should test to see if there are no triangles that satisfy, one triangle that satisfies, or two triangles that satisfy this. From $\frac{\sin A}{a}=\frac{\sin C}{c}$ , we get $c\cdot \frac{\sin A}{a}=\sin C$ . In this equation, if $\sin C >1$ , no $\angle A$ that satisfies the triangle can be found. If $\sin C=1, C=90^\circ$ and there is a right triangle determined. Finally, if $\sin C <1$ , two measures of $\angle C$ can be calculated: an acute $\angle C$ and an obtuse angle $C'=180^\circ-C$ . In this case, there may be one or two triangles determined. If $C'+A \geq 180^\circ$ , then the $\angle C'$ is not a solution.

In this problem, $\sin C =.559<1$ , so there may be one or two angles that satisfy this triangle. Since we have the length of two sides of the triangle and the corresponding angle of one of the sides, we can use the Law of Sines to find the angle that we are looking for. This goes as follows:

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

Inputting the values of the problem

$\small \frac{10.3}{\sin\angle C} = \frac{7.4}{\sin\angle 34^o}$

Rearranging the equation to isolate $\small \angle C$

$\small \angle C = \sin^{-1}(\frac{10.3\sin\angle34^o}{7.4})$

$\small \angle C =$ $\small 51^o$

When the original given angle ( $\angle A=34^\circ$ ) is acute, there will be:

One solution if the side opposite the given angle is equal to or greater than the other given side

No solution, one solution (right triangle), or two solutions if the side opposite the given angle is less than the other given side

In this problem, the side opposite the given angle is , which is less than the other given side . Therefore, we have a second solution. Find it by following the below steps:

$\angle C'=180^\circ-C=180^\circ-51^\circ=129^\circ$ .

$C'+A=129^\circ+34^\circ=163^\circ\leq 180^\circ$ , so $\angle C'$ is a solution.

Therefore there are two values for an angle, $\angle C=51^\circ$ and $\angle C'=129^\circ$

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Question

If $A=5^\circ$ , $\small b=10$ , $\small a=4$ , find $\small \angle B$ to the nearest tenth of a degree.

Answer

Notice that the given information is Angle-Side-Side, which is the ambiguous case. Therefore, we should test to see if there are no triangles that satisfy, one triangle that satisfies, or two triangles that satisfy this. From $\frac{\sin A}{a}=\frac{\sin B}{b}$ , we get $b\cdot \frac{\sin A}{a}=\sin B$ . In this equation, if $\sin B > 1$ , no $\angle A$ that satisfies the triangle can be found. If $\sin B = 1$ , $B=90^\circ$ and there is a right triangle determined. Finally, if $\sin B < 1$ , two measures of $\angle B$ can be calculated: an acute $\angle B$ and an obtuse $\angle$ $B'=180^\circ-B$ . In this case, there may be one or two triangles determined. If $B'+A\geq 180^\circ$ , then the $\angle B'$ is not a solution.

In this problem, $\sin\angle A=.087<1$ , so there may be one or two angles that satisfy this triangle. Since we have the length of two sides of the triangle and the corresponding angle of one of the sides, we can use the Law of Sines to find the angle that we are looking for. This goes as follows:

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

Inputting the values from the problem

$\small \frac{10}{\sin\angle B} = \frac{4}{\sin\angle 5^o}$

$\small \angle B = \sin^{-1}(\frac{10\sin\angle 5^o}{4})$

$\small \small \angle B = 12.6^o$

When the original given angle ( $\angle A=5^\circ$ ) is acute, there will be:

One solution if the side opposite the given angle is equal to or greater than the other given side

No solution, one solution (right triangle), or two solutions if the side opposite the given angle is less than the other given side

In this problem, the side opposite the given angle is , which is less than the other given side . Therefore, we have a second solution. Find it by following the below steps:

$\angle B'=180^\circ-B=180^\circ-12.6^\circ=167.4^\circ$

$\angle B'+\angle A=167.4^\circ+5^\circ=172.4^\circ$ $\leq180^\circ$ , so $\angle B'$ is a solution.

Therefore there are two values for an angle, $\angle B=12.6^\circ$ and $\angle B'=167.4^\circ$ .

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Question

Solve for . Image not drawn to scale. There may be more than one answer.

Answer

To solve, use Law of Sines, $\frac{a}{sinA} = \frac{b}{sinB}$ , where A is the angle across from side a, and B is the angle across from side b. In this case, our proportion is set up like this:

$\frac{7}{sin(40)}=\frac{8}{sin(x)}$ cross-multiply

evaluate the right side using a calculator

divide both sides by 7

solve for x by evaluating $sin^{-1}(0.734)$ in a calculator

There is another solution as well. If has a sine of 0.734, so will its supplementary angle, .

Since is still less than , is a possible value for x.

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Question

Solve for . Image not drawn to scale; there may be more than one solution.

Answer

To solve, use Law of Sines, $\frac{a}{sinA} = \frac{b}{sinB}$ , where A is the angle across from side a, and B is the angle across from side b. In this case, our proportion is set up like this:

$\frac{4}{sin(25)}=\frac{9}{sin(x)}$ Cross-multiply.

Evaluate the right side using a calculator.

Divide both sides by 4.

Solve for x by evaluating $sin^{-1}(0.951)$ in a calculator.

There is another solution as well. If has a sine of 0.951, so will its supplementary angle, .

Since is still less than , is a possible value for x.

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Question

If c=10.3, a=7.4, and $\angle A=122^\circ$ find $\angle C$ to the nearest degree.

Answer

Notice that the given information is Angle-Side-Side, which is the ambiguous case. Therefore, we should test to see if there are no triangles that satisfy, one triangle that satisfies, or two triangles that satisfy this. From $\frac{\sin A}{a}=\frac{\sin C}{c}$ , we get $c\cdot \frac{\sin A}{a}=\sin C$ . In this equation, if $\sin C >1$ , no $\angle A$ that satisfies the triangle can be found. If $\sin C=1, C=90^\circ$ and there is a right triangle determined. Finally, if $\sin C <1$ , two measures of $\angle C$ can be calculated: an acute $\angle C$ and an obtuse angle $C'=180^\circ-C$ . In this case, there may be one or two triangles determined. If $C'+A \geq 180^\circ$ , then the $\angle C'$ is not a solution.

In this problem, $\sin C = 1.18>1$ , which means that there are no solutions to $\angle C$ that satisfy this triangle. If you got answers for this triangle, check that you set up your Law of Sines equation properly at the start of the problem.

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Question

If $\small c=10.3$ , , and $\small \angle A$ = $\small 34^o$ find $\small \angle C$ to the nearest degree.

Answer

Notice that the given information is Angle-Side-Side, which is the ambiguous case. Therefore, we should test to see if there are no triangles that satisfy, one triangle that satisfies, or two triangles that satisfy this. From $\frac{\sin A}{a}=\frac{\sin C}{c}$ , we get $c\cdot \frac{\sin A}{a}=\sin C$ . In this equation, if $\sin C >1$ , no $\angle A$ that satisfies the triangle can be found. If $\sin C=1, C=90^\circ$ and there is a right triangle determined. Finally, if $\sin C <1$ , two measures of $\angle C$ can be calculated: an acute $\angle C$ and an obtuse angle $C'=180^\circ-C$ . In this case, there may be one or two triangles determined. If $C'+A \geq 180^\circ$ , then the $\angle C'$ is not a solution.

In this problem, $\sin C =.44<1$ , so there may be one or two angles that satisfy this triangle. Since we have the length of two sides of the triangle and the corresponding angle of one of the sides, we can use the Law of Sines to find the angle that we are looking for. This goes as follows:

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

Inputting the values of the problem

$\frac{10.3}{\sin C}=\frac{13}{\sin 34^\circ}$

Rearranging the equation to isolate $\small \angle C$

$\angle C=\sin^-^1\left (\frac{10.3\sin34^\circ}{13} \right )$

$\angle C = 26.3^\circ$

When the original given angle ( $\angle A=34^\circ$ ) is acute, there will be:

One solution if the side opposite the given angle is equal to or greater than the other given side

No solution, one solution (right triangle), or two solutions if the side opposite the given angle is less than the other given side

In this problem, the side opposite the given angle is , which is greater than the other given side . Therefore, we have only one solution, $\angle C = 26.3^\circ$ .

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Question

If $\small c=10.3$ , , and $\angle A =108^\circ$ find $\small \angle C$ to the nearest degree.

Answer

Notice that the given information is Angle-Side-Side, which is the ambiguous case. Therefore, we should test to see if there are no triangles that satisfy, one triangle that satisfies, or two triangles that satisfy this. From $\frac{\sin A}{a}=\frac{\sin C}{c}$ , we get $c\cdot \frac{\sin A}{a}=\sin C$ . In this equation, if $\sin C >1$ , no $\angle A$ that satisfies the triangle can be found. If $\sin C=1, C=90^\circ$ and there is a right triangle determined. Finally, if $\sin C <1$ , two measures of $\angle C$ can be calculated: an acute $\angle C$ and an obtuse angle $C'=180^\circ-C$ . In this case, there may be one or two triangles determined. If $C'+A \geq 180^\circ$ , then the $\angle C'$ is not a solution.

In this problem, $\sin C =.82<1$ , so there may be one or two angles that satisfy this triangle. Since we have the length of two sides of the triangle and the corresponding angle of one of the sides, we can use the Law of Sines to find the angle that we are looking for. This goes as follows:

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

Inputting the values of the problem

$\frac{12}{\sin 108^\circ}=\frac{10.3}{\sin C}$

Rearranging the equation to isolate $\small \angle C$

$\angle C=\sin^-^1\left (\frac{10.3\sin108^\circ}{12} \right )$

$\angle C=54.72^\circ$

When the original given angle ( $\angle A=108^\circ$ ) is obtuse, there will be:

No solution when the side opposite the given angle is less than or equal to the other given side

One solution if the side opposite the given angle is greater than the other given side

In this problem, the side opposite the given angle is , which is greater than the other given side . Therefore this problem has one and only one solution, $\angle C=54.72^\circ$

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Question

If , , and $\small \angle A$ = $\small 34^o$ find $\small \angle C$ to the nearest degree.

Answer

Notice that the given information is Angle-Side-Side, which is the ambiguous case. Therefore, we should test to see if there are no triangles that satisfy, one triangle that satisfies, or two triangles that satisfy this. From $\frac{\sin A}{a}=\frac{\sin C}{c}$ , we get $c\cdot \frac{\sin A}{a}=\sin C$ . In this equation, if $\sin C >1$ , no $\angle A$ that satisfies the triangle can be found. If $\sin C=1, C=90^\circ$ and there is a right triangle determined. Finally, if $\sin C <1$ , two measures of $\angle C$ can be calculated: an acute $\angle C$ and an obtuse angle $C'=180^\circ-C$ . In this case, there may be one or two triangles determined. If $C'+A \geq 180^\circ$ , then the $\angle C'$ is not a solution.

In this problem, $\sin C=2.01>1$ , which means that there are no solutions to $\angle C$ that satisfy this triangle. If you got answers for this triangle, check that you set up your Law of Sines equation properly at the start of the problem.

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Question

If c=70, a=50, and $\angle A=122^\circ$ find $\angle C$ to the nearest degree.

Answer

Notice that the given information is Angle-Side-Side, which is the ambiguous case. Therefore, we should test to see if there are no triangles that satisfy, one triangle that satisfies, or two triangles that satisfy this. From $\frac{\sin A}{a}=\frac{\sin C}{c}$ , we get $c\cdot \frac{\sin A}{a}=\sin C$ . In this equation, if $\sin C >1$ , no $\angle A$ that satisfies the triangle can be found. If $\sin C=1, C=90^\circ$ and there is a right triangle determined. Finally, if $\sin C <1$ , two measures of $\angle C$ can be calculated: an acute $\angle C$ and an obtuse angle $C'=180^\circ-C$ . In this case, there may be one or two triangles determined. If $C'+A \geq 180^\circ$ , then the $\angle C'$ is not a solution.

In this problem, $\sin C = 1.18>1$ , which means that there are no solutions to $\angle C$ that satisfy this triangle. If you got answers for this triangle, check that you set up your Law of Sines equation properly at the start of the problem.

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Question

In triangle $\triangle ABC$ , $\measuredangle A = 115^{\circ}$ , and . To the nearest tenth, what is ?

Answer

By the Law of Cosines,

$BC^2 = AB^2 +AC^2 - 2\left (AB\right )\left ( AC\right )\cos\angle A$

or, equivalently,

$BC = \sqrt[]{AB^2 +AC^2 - 2\left ( AB\right )\left ( AC\right )\cos \angle A}$

Substitute:

$BC = \sqrt[]{31^2 + 42^2 - 2\left ( 31\right )\left ( 42\right )\cos \left ( 115^{\circ}\right )}$

$\textup{BC}= \sqrt[]{961 + 1764 - 2604\cdot (-0.4226)} \approx \sqrt{2725 + 1100} \approx \sqrt{3825}\approx61.9$

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Question

A triangle has side lengths , and $\angle C=150^{\circ}$

Which of the following equations can be used to find the length of side ?

Answer

You are given the length of two sides of a triangle and the angle between them; therefore, you should use the Law of Cosines to find $\angle A$ , $\angle B$ , or, in this case, the length of .

$c^{2} = a^{2} + b^{2} - 2ab\cdot\cos C$

Substitute the given values for , , and $\angle C$ :

$c^{2}=(1)^{2}+(1.2)^{2}-2\cdot 1\cdot 1.2\cos 150^{\circ}$

At this point, if you are solving for , take the square root of both sides of the equation.

$\sqrt{c^{2}}=\sqrt{(1)^{2}+(1.2)^{2}-2\cdot 1\cdot 1.2\cos 150^{\circ}}$

$c=\sqrt{(1)^{2}+(1.2)^{2}-2\cdot 1\cdot 1.2\cos 150^{\circ}}$

This question merely asks for the equation, rather than the solution, so you need not simplify any further.

Two of the answer choices are equations derived from the Law of Sines. To use the Law of Sines, you must know at least one side and angle that correspond to one another, which is not the case here.

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Question

Given , and determine to the nearest degree the measure of $\angle B$ .

Answer

We are given three sides and our desire is to find an angle, this means we must utilize the Law of Cosines. Since the angle desired is $\angle B$ the equation must be rewritten as such:

$b^{2} = a^{2} + c^{2} - 2ac\cos B$

Substituting the given values:

$5.1^{2} = 6.2^{2} + 3.5^{2} - 2\left(6.2 \right )\left(3.5 \right )\cos B$

Rearranging:

$\cos B = \frac{5.1^{2} - 3.5^{2} - 6.2^{2}}{-2\left(3.5 \right )\left(6.2)}$

Solving the right hand side and taking the inverse cosine we obtain:

$B = 55^{\circ}$

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Question

If , and , determine the measure of $\angle C$ to the nearest degree.

Answer

This is a straightforward Law of Cosines problem since we are given three sides and desire one of the corresponding angles in the triangle. We write down the Law of Cosines to start:

$c^{2} = a^{2}+b^{2}-2ab\cos C$

Substituting the given values:

$12.08^{2} = 9.72^{2} + 13.91^{2} - 2\left(9.72 \right )\left(13.91 \right )\cos C$

Isolating the angle:

$\cos C = \frac{12.08^{2} - 9.72^{2} - 13.91^{2}}{-2\left(9.72 \right )\left(13.91 \right )}$

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

$C = \cos^{-1}\left(\frac{12.08^{2} - 9.72^{2} - 13.91^{2}}{-2\left(9.72 \right )\left(13.91 \right )}\right) = 58.31^{\circ} = 58^{\circ}$

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Question

Which famous theorem does the Law of Cosines boil down to for right triangles?

Answer

The Law of Cosines is as follows:

$a^2=b^2+c^2-2bc\cos(A)$

$b^2=a^2+c^2-2ac\cos(B)$

$c^2=a^2+b^2-2ab\cos(C)$

Notice these equations contains the Pythagorean Theorem, $a^{2}+b^{2}= c^{2}$ , within it.

The term at the end is the adjusting term for triangles which are not right triangles.

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Question

Using the Law of Cosines, determine the perimeter of the above triangle.

Answer

To apply the Law of Cosines, is the unknown, and are the respective given sides, and the given angle is .

Therefore, the equation becomes:

Which yields

Add to the other two given sides to get the perimeter,

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Question

Find the value of to the nearest tenth.

Answer

This is a prime example of a case that calls for using the Law of Cosines, which states

where , , and are the three sides of the triangle, and is the angle opposite side . Looking at our triangle, taking , then we have , , and . Plugging this into our formula, we get.

Using our calculator to approximate the cosine value gives

Simplifying further gives

$x^2\approx 586-512.31=73.69$

Solving by taking the square root gives

$x\approx8.6$

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Question

If $\small a=5$ , $\small b=12$ , and $\small c=13$ find $\small \angle C$ to the nearest degree.

Answer

The problem gives the lengths of three sides and asks to find an angle. We can use the Law of Cosines to solve for the angle. Because we are solving for $\small \angle C$ , we use the equation:

$\small \small \small c^2 = a^2 + b^2 - 2abcos\angle C$

Substituting the values from the problem gives

$\small 13^2 = 5^2 + 12^2 - 2(5)(12)cos\angle C$

Isolating $\small \angle C$ by itself gives

$\small \small \angle C = cos(\frac{13^2 - 5^2 - 12^2}{(-2)(5)(12)}) = 90^o$

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Question

If $\small a=4$ , $\small b=3$ , $\small c=5$ , find $\small \angle B$ to the nearest degree.

Answer

We are given the lengths of the three sides to a triangle. Therefore, we can use the Law of Cosines to find the angle being asked for. Since we are looking for $\small \angle B$ we use the equation,

$\small b^2 = a^2 + c^2 - 2ac\cos(\angle B)$

Inputting the values we are given,

$\small 3^2 = 4^2 + 5^2 - 2(4)(5)\cos\angle B$

Next we isolate $\small \angle B$ by itself to solve for it

$\small \angle B = \cos^{-1}(\frac{3^2-4^2-5^2}{(-2)(4)(5)})$

$\small \angle B = 37^o$

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Question

If $\small a=10$ , $\small b=6.2$ , and $\small c=4.8$ , find $\small \angle A$ to the nearest degree.

Answer

Because the problem provides all three sides of the triangle, we can use the Law of Cosines to solve this problem. Since we are solving for $\small \angle A$ , we use the equation

$\small a^2 = b^2 + c^2 - 2ac\cos\angle A$

Substitute in the given values

$\small 10^2 = 6.2^2 + 4.8^2 - 2(6.2)(4.8)\cos\angle A$

Isolate $\small \angle A$

$\small \angle A =\cos^{-1}( \frac{10^2-6.2^2-4.8^2}{(-2)(6.2)(4.8)})$

$\small \angle A = 130^o$

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