Apply Laws of Sines and Cosines: CCSS.Math.Content.HSG-SRT.D.11 - Geometry

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Question

In a triangle where the side opposite a $62 ^{\circ}$ has length 10 find the side opposite a $66 ^{\circ}$ angle. Round you answer to the nearest hundredth.

Answer

In order to solve this, we need to recall the law of sines.

$\frac{1}{A} \sin{\left (a \right )} = \frac{1}{B} \sin{\left (b \right )}$

Where , and are angles, and , and , are opposite side lengths.

Now let's plug in 62 for , 10 for and 66 for .
Now our equation becomes

$\left(\frac{\sin( 62 )}{ 10 }\right)= \left(\frac{\sin( 66 )}{ b }\right)$

Now we rearrange the equation to solve for

$\b = \left(\frac{ 10 \sin( 66 )}{\sin( 62 )}\right) \\b = -11.3557732263804$

Now we round our answer to the nearest tenth.

Remember if your answer is negative, multiply it by -1, because side lengths can't be negative.

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Question

In a triangle where the side opposite a $41 ^{\circ}$ has length 6 find the side opposite a $39 ^{\circ}$ angle. Round you answer to the nearest hundredth.

Answer

In order to solve this, we need to recall the law of sines.

$\frac{1}{A} \sin{\left (a \right )} = \frac{1}{B} \sin{\left (b \right )}$

Where , and are angles, and , and , are opposite side lengths.

Now let's plug in 41 for , 6 for and 39 for .
Now our equation becomes

$\left(\frac{\sin( 41 )}{ 6 }\right)= \left(\frac{\sin( 39 )}{ b }\right)$

Now we rearrange the equation to solve for

$\b = \left(\frac{ 6 \sin( 39 )}{\sin( 41 )}\right) \\b = 6.45402256283484$

Now we round our answer to the nearest tenth.

Remember if your answer is negative, multiply it by -1, because side lengths can't be negative.

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Question

In a triangle where the side opposite a $60 ^{\circ}$ has length 3 find the side opposite a $51 ^{\circ}$ angle. Round you answer to the nearest hundredth.

Answer

In order to solve this, we need to recall the law of sines.

$\frac{1}{A} \sin{\left (a \right )} = \frac{1}{B} \sin{\left (b \right )}$

Where , and are angles, and , and , are opposite side lengths.

Now let's plug in 60 for , 3 for and 51 for .
Now our equation becomes

$\left(\frac{\sin( 60 )}{ 3 }\right)= \left(\frac{\sin( 51 )}{ b }\right)$

Now we rearrange the equation to solve for b

$\b = \left(\frac{ 3 \sin( 51 )}{\sin( 60 )}\right) \\b = 1.57596947141406$

Now we round our answer to the nearest tenth.

Remember if your answer is negative, multiply it by -1, because side lengths can't be negative.

Compare your answer with the correct one above

Question

In a triangle where the side opposite a $45 ^{\circ}$ has length 13 find the side opposite a $65 ^{\circ}$ angle. Round you answer to the nearest hundredth.

Answer

In order to solve this, we need to recall the law of sines.

$\frac{1}{A} \sin{\left (a \right )} = \frac{1}{B} \sin{\left (b \right )}$

Where , and are angles, and , and , are opposite side lengths.

Now let's plug in 45 for , 13 for and 65 for .
Now our equation becomes

$\left(\frac{\sin( 45 )}{ 13 }\right)= \left(\frac{\sin( 65 )}{ b }\right)$

Now we rearrange the equation to solve for b

$\b = \left(\frac{ 13 \sin( 65 )}{\sin( 45 )}\right) \\b = -15.9863244465377$

Now we round our answer to the nearest tenth.

Remember if your answer is negative, multiply it by -1, because side lengths can't be negative.

Compare your answer with the correct one above

Question

In a triangle where the side opposite a $26 ^{\circ}$ has length 12 find the side opposite a $41 ^{\circ}$ angle. Round you answer to the nearest hundredth.

Answer

In order to solve this, we need to recall the law of sines.

$\frac{1}{A} \sin{\left (a \right )} = \frac{1}{B} \sin{\left (b \right )}$

Where , and are angles, and , and , are opposite side lengths.

Now let's plug in 26 for , 12 for and 41 for .
Now our equation becomes

$\left(\frac{\sin( 26 )}{ 12 }\right)= \left(\frac{\sin( 41 )}{ b }\right)$

Now we rearrange the equation to solve for b

$\b = \left(\frac{ 12 \sin( 41 )}{\sin( 26 )}\right) \\b = -15.5778376306642$

Now we round our answer to the nearest tenth.

Remember if your answer is negative, multiply it by -1, because side lengths can't be negative.

Compare your answer with the correct one above

Question

In a triangle where the side opposite a $36 ^{\circ}$ has length 6 find the side opposite a $58 ^{\circ}$ angle. Round you answer to the nearest hundredth.

Answer

In order to solve this, we need to recall the law of sines.

$\frac{1}{A} \sin{\left (a \right )} = \frac{1}{B} \sin{\left (b \right )}$

Where , and are angles, and , and , are opposite side lengths.

Now let's plug in 36 for , 6 for and 58 for .
Now our equation becomes

$\left(\frac{\sin( 36 )}{ 6 }\right)= \left(\frac{\sin( 58 )}{ b }\right)$

Now we rearrange the equation to solve for b

$\b = \left(\frac{ 6 \sin( 58 )}{\sin( 36 )}\right) \\b = -3.69854363983686$

Now we round our answer to the nearest tenth.

Remember if your answer is negative, multiply it by -1, because side lengths can't be negative.

Compare your answer with the correct one above

Question

In a triangle where the side opposite a $26 ^{\circ}$ has length 11 find the side opposite a $17 ^{\circ}$ angle. Round you answer to the nearest hundredth.

Answer

In order to solve this, we need to recall the law of sines.

$\frac{1}{A} \sin{\left (a \right )} = \frac{1}{B} \sin{\left (b \right )}$

Where , and are angles, and , and , are opposite side lengths.

Now let's plug in 26 for , 11 for and 17 for .
Now our equation becomes

$\left(\frac{\sin( 26 )}{ 11 }\right)= \left(\frac{\sin( 17 )}{ b }\right)$

Now we rearrange the equation to solve for b

$\b = \left(\frac{ 11 \sin( 17 )}{\sin( 26 )}\right) \\b = 2.71142149312156$

Now we round our answer to the nearest tenth.

Remember if your answer is negative, multiply it by -1, because side lengths can't be negative.

Compare your answer with the correct one above

Question

In a triangle where the side opposite a $81 ^{\circ}$ has length 11 find the side opposite a $66 ^{\circ}$ angle. Round you answer to the nearest hundredth.

Answer

In order to solve this, we need to recall the law of sines.

$\frac{1}{A} \sin{\left (a \right )} = \frac{1}{B} \sin{\left (b \right )}$

Where , and are angles, and , and , are opposite side lengths.

Now let's plug in 81 for , 11 for and 66 for .
Now our equation becomes

$\left(\frac{\sin( 81 )}{ 11 }\right)= \left(\frac{\sin( 66 )}{ b }\right)$

Now we rearrange the equation to solve for b

$\b = \left(\frac{ 11 \sin( 66 )}{\sin( 81 )}\right) \\b = 12.2250984628794$

Now we round our answer to the nearest tenth.

Remember if your answer is negative, multiply it by -1, because side lengths can't be negative.

Compare your answer with the correct one above

Question

In a triangle where the side opposite a $70 ^{\circ}$ has length 5 find the side opposite a $50 ^{\circ}$ angle. Round you answer to the nearest hundredth.

Answer

In order to solve this, we need to recall the law of sines.

$\frac{1}{A} \sin{\left (a \right )} = \frac{1}{B} \sin{\left (b \right )}$

Where , and are angles, and , and , are opposite side lengths.

Now let's plug in 70 for , 5 for and 50 for .
Now our equation becomes

$\left(\frac{\sin( 70 )}{ 5 }\right)= \left(\frac{\sin( 50 )}{ b }\right)$

Now we rearrange the equation to solve for b

$\b = \left(\frac{ 5 \sin( 50 )}{\sin( 70 )}\right) \\b = -1.88083767684263$

Now we round our answer to the nearest tenth.

Remember if your answer is negative, multiply it by -1, because side lengths can't be negative.

Compare your answer with the correct one above

Question

In a triangle where the side opposite a $11 ^{\circ}$ has length 5 find the side opposite a $64 ^{\circ}$ angle. Round you answer to the nearest hundredth.

Answer

In order to solve this, we need to recall the law of sines.

$\frac{1}{A} \sin{\left (a \right )} = \frac{1}{B} \sin{\left (b \right )}$

Where , and are angles, and , and , are opposite side lengths.

Now let's plug in 11 for , 5 for and 64 for .
Now our equation becomes

$\left(\frac{\sin( 11 )}{ 5 }\right)= \left(\frac{\sin( 64 )}{ b }\right)$

Now we rearrange the equation to solve for b

$\b = \left(\frac{ 5 \sin( 64 )}{\sin( 11 )}\right) \\b = -3.41136278039312$

Now we round our answer to the nearest tenth.

Remember if your answer is negative, multiply it by -1, because side lengths can't be negative.

Compare your answer with the correct one above

Question

In a triangle where the side opposite a $78 ^{\circ}$ has length 6 find the side opposite a $52 ^{\circ}$ angle. Round you answer to the nearest hundredth.

Answer

In order to solve this, we need to recall the law of sines.

$\frac{1}{A} \sin{\left (a \right )} = \frac{1}{B} \sin{\left (b \right )}$

Where , and are angles, and , and , are opposite side lengths.

Now let's plug in 78 for , 6 for and 52 for .
Now our equation becomes

$\left(\frac{\sin( 78 )}{ 6 }\right)= \left(\frac{\sin( 52 )}{ b }\right)$

Now we rearrange the equation to solve for b

$\b = \left(\frac{ 6 \sin( 52 )}{\sin( 78 )}\right) \\b = 5.54498774628774$

Now we round our answer to the nearest tenth.

Remember if your answer is negative, multiply it by -1, because side lengths can't be negative.

Compare your answer with the correct one above

Question

In a triangle where the side opposite a $44 ^{\circ}$ has length 7 find the side opposite a $53 ^{\circ}$ angle. Round you answer to the nearest hundredth.

Answer

In order to solve this, we need to recall the law of sines.

$\frac{1}{A} \sin{\left (a \right )} = \frac{1}{B} \sin{\left (b \right )}$

Where , and are angles, and , and , are opposite side lengths.

Now let's plug in 44 for , 7 for and 53 for .
Now our equation becomes

$\left(\frac{\sin( 44 )}{ 7 }\right)= \left(\frac{\sin( 53 )}{ b }\right)$

Now we rearrange the equation to solve for b

$\b = \left(\frac{ 7 \sin( 53 )}{\sin( 44 )}\right) \\b = 6.67220075275393$

Now we round our answer to the nearest tenth.

Remember if your answer is negative, multiply it by -1, because side lengths can't be negative.

Compare your answer with the correct one above

Question

To the nearest hundredth, solve $\bigtriangleup ABC$ where, , $\angle B= 11 ^{\circ}$ , and $\angle C= 71 ^{\circ}$ .

Answer

The first step is to create a picture

Since we have 2 angles, we can solve for the 3rd angle easily.

$\\angle A =180- 11 - 71 \\angle A = 98^\circ$

Now we can use the Law of Sines to figure out one side

Recall the Law of Sines

$\frac{1}{a} \sin{\left (A \right )} = \frac{1}{AC} \sin{\left (B \right )}$

Now plug in 11 for B , 71 for A and 179 for a , and solve for the missing side.

$\\left(\frac{\sin( 71 )}{ 179 }\right)= \left(\frac{\sin( 11 )}{ AC }\right) \\AC = \left(\frac{ 179 \sin( 11 )}{\sin( 71 )}\right) \\AC = 36.1228336003367$

Remember to round to the nearest hundredth.

Now we need to solve for the last side of the triangle.

To solve for this, we can use the Law of Cosines.

Lets recall the Law of Cosines.

$\BC^{2} = a^{2} - 2 a b \cos{\left (BC \right )} + b^{2} \BC^2 = 1304.6544 + 32041 - 2 \cdot 36.12 \cdot 179 \cdot \cos( 98 ) \BC^{2} = 35145.2962015906 \BC = 187.470787595269 \BC = 187.47$

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Question

To the nearest hundredth, solve $\bigtriangleup ABC$ where, , $\angle B= 18 ^{\circ}$ , and $\angle C= 66 ^{\circ}$ .

Answer

The first step is to create a picture

Since we have 2 angles, we can solve for the 3rd angle easily.

$\\angle A =180- 18 - 66 \\angle A = 96^\circ$

Now we can use the Law of Sines to figure out one side

Recall the Law of Sines

$\frac{1}{a} \sin{\left (A \right )} = \frac{1}{AC} \sin{\left (B \right )}$

Now plug in 18 for B , 66 for A and 78 for a , and solve for the missing side.

$\\left(\frac{\sin( 66 )}{ 78 }\right)= \left(\frac{\sin( 18 )}{ AC }\right) \\AC = \left(\frac{ 78 \sin( 18 )}{\sin( 66 )}\right) \\AC = 26.3843745919819$

Remember to round to the nearest hundredth.

Now we need to solve for the last side of the triangle.

To solve for this, we can use the Law of Cosines.

Lets recall the Law of Cosines.

$\BC^{2} = a^{2} - 2 a b \cos{\left (BC \right )} + b^{2} \BC^2 = 695.9043999999999 + 6084 - 2 \cdot 26.38 \cdot 78 \cdot \cos( 96 ) \BC^{2} = 7210.06829431611 \BC = 84.9121210094066 \BC = 84.91$

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Question

To the nearest hundredth, solve $\bigtriangleup ABC$ where, , $\angle B= 32 ^{\circ}$ , and $\angle C= 82 ^{\circ}$ .

Answer

The first step is to create a picture

Since we have 2 angles, we can solve for the 3rd angle easily.

$\\angle A =180- 32 - 82 \\angle A = 66^\circ$

Now we can use the Law of Sines to figure out one side

Recall the Law of Sines

$\frac{1}{a} \sin{\left (A \right )} = \frac{1}{AC} \sin{\left (B \right )}$

Now plug in 32 for B , 82 for A and 95 for a , and solve for the missing side.

$\\left(\frac{\sin( 82 )}{ 95 }\right)= \left(\frac{\sin( 32 )}{ AC }\right) \\AC = \left(\frac{ 95 \sin( 32 )}{\sin( 82 )}\right) \\AC = 50.8370730019896$

Remember to round to the nearest hundredth.

Now we need to solve for the last side of the triangle.

To solve for this, we can use the Law of Cosines.

Lets recall the Law of Cosines.

$\BC^{2} = a^{2} - 2 a b \cos{\left (BC \right )} + b^{2} \BC^2 = 2584.7056000000002 + 9025 - 2 \cdot 50.84 \cdot 95 \cdot \cos( 66 ) \BC^{2} = 7680.792322545 \BC = 87.6401296356013 \BC = 87.64$

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Question

To the nearest hundredth, solve $\bigtriangleup ABC$ where, , $\angle B= 3 ^{\circ}$ , and $\angle C= 81 ^{\circ}$ .

Answer

The first step is to create a picture

Since we have 2 angles, we can solve for the 3rd angle easily.

$\\angle A =180- 3 - 81 \\angle A = 96^\circ$

Now we can use the Law of Sines to figure out one side

Recall the Law of Sines

$\frac{1}{a} \sin{\left (A \right )} = \frac{1}{AC} \sin{\left (B \right )}$

Now plug in 3 for B , 81 for A and 182 for a , and solve for the missing side.

$\\left(\frac{\sin( 81 )}{ 182 }\right)= \left(\frac{\sin( 3 )}{ AC }\right) \\AC = \left(\frac{ 182 \sin( 3 )}{\sin( 81 )}\right) \\AC = 9.64387615477605$

Remember to round to the nearest hundredth.

Now we need to solve for the last side of the triangle.

To solve for this, we can use the Law of Cosines.

Lets recall the Law of Cosines.

$\BC^{2} = a^{2} - 2 a b \cos{\left (BC \right )} + b^{2} \BC^2 = 92.92960000000001 + 33124 - 2 \cdot 9.64 \cdot 182 \cdot \cos( 96 ) \BC^{2} = 33583.7157964677 \BC = 183.25860360831 \BC = 183.26$

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Question

To the nearest hundredth, solve $\bigtriangleup ABC$ where, , $\angle B= 29 ^{\circ}$ , and $\angle C= 108 ^{\circ}$ .

Answer

The first step is to create a picture

Since we have 2 angles, we can solve for the 3rd angle easily.

$\\angle A =180- 29 - 108 \\angle A = 43^\circ$

Now we can use the Law of Sines to figure out one side

Recall the Law of Sines

$\frac{1}{a} \sin{\left (A \right )} = \frac{1}{AC} \sin{\left (B \right )}$

Now plug in 29 for B , 108 for A and 131 for a , and solve for the missing side.

$\\left(\frac{\sin( 108 )}{ 131 }\right)= \left(\frac{\sin( 29 )}{ AC }\right) \\AC = \left(\frac{ 131 \sin( 29 )}{\sin( 108 )}\right) \\AC = 66.7784292143583$

Remember to round to the nearest hundredth.

Now we need to solve for the last side of the triangle.

To solve for this, we can use the Law of Cosines.

Lets recall the Law of Cosines.

$\BC^{2} = a^{2} - 2 a b \cos{\left (BC \right )} + b^{2} \BC^2 = 4459.5684 + 17161 - 2 \cdot 66.78 \cdot 131 \cdot \cos( 43 ) \BC^{2} = 8824.54074913841 \BC = 93.9390267627806 \BC = 93.94$

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Question

To the nearest hundredth, solve $\bigtriangleup ABC$ where, , $\angle B= 27 ^{\circ}$ , and $\angle C= 120 ^{\circ}$ .

Answer

The first step is to create a picture

Since we have 2 angles, we can solve for the 3rd angle easily.

$\\angle A =180- 27 - 120 \\angle A = 33^\circ$

Now we can use the Law of Sines to figure out one side

Recall the Law of Sines

$\frac{1}{a} \sin{\left (A \right )} = \frac{1}{AC} \sin{\left (B \right )}$

Now plug in 27 for B , 120 for A and 116 for a , and solve for the missing side.

$\\left(\frac{\sin( 120 )}{ 116 }\right)= \left(\frac{\sin( 27 )}{ AC }\right) \\AC = \left(\frac{ 116 \sin( 27 )}{\sin( 120 )}\right) \\AC = 60.8098766383251$

Remember to round to the nearest hundredth.

Now we need to solve for the last side of the triangle.

To solve for this, we can use the Law of Cosines.

Lets recall the Law of Cosines.

$\BC^{2} = a^{2} - 2 a b \cos{\left (BC \right )} + b^{2} \BC^2 = 3697.8561000000004 + 13456 - 2 \cdot 60.81 \cdot 116 \cdot \cos( 33 ) \BC^{2} = 5321.95882107139 \BC = 72.9517568059289 \BC = 72.95$

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Question

To the nearest hundredth, solve $\bigtriangleup ABC$ where, , $\angle B= 27 ^{\circ}$ , and $\angle C= 123 ^{\circ}$ .

Answer

The first step is to create a picture

Since we have 2 angles, we can solve for the 3rd angle easily.

$\\angle A =180- 27 - 123 \\angle A = 30^\circ$

Now we can use the Law of Sines to figure out one side

Recall the Law of Sines

$\frac{1}{a} \sin{\left (A \right )} = \frac{1}{AC} \sin{\left (B \right )}$

Now plug in 27 for B , 123 for A and 111 for a , and solve for the missing side.

$\\left(\frac{\sin( 123 )}{ 111 }\right)= \left(\frac{\sin( 27 )}{ AC }\right) \\AC = \left(\frac{ 111 \sin( 27 )}{\sin( 123 )}\right) \\AC = 60.0866983976108$

Remember to round to the nearest hundredth.

Now we need to solve for the last side of the triangle.

To solve for this, we can use the Law of Cosines.

Lets recall the Law of Cosines.

$\BC^{2} = a^{2} - 2 a b \cos{\left (BC \right )} + b^{2} \BC^2 = 3610.8081 + 12321 - 2 \cdot 60.09 \cdot 111 \cdot \cos( 30 ) \BC^{2} = 4379.04653402366 \BC = 66.174364628787 \BC = 66.17$

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Question

To the nearest hundredth, solve $\bigtriangleup ABC$ where, , $\angle B= 15 ^{\circ}$ , and $\angle C= 97 ^{\circ}$ .

Answer

The first step is to create a picture

Since we have 2 angles, we can solve for the 3rd angle easily.

$\\angle A =180- 15 - 97 \\angle A = 68^\circ$

Now we can use the Law of Sines to figure out one side

Recall the Law of Sines

$\frac{1}{a} \sin{\left (A \right )} = \frac{1}{AC} \sin{\left (B \right )}$

Now plug in 15 for B , 97 for A and 100 for a , and solve for the missing side.

$\\left(\frac{\sin( 97 )}{ 100 }\right)= \left(\frac{\sin( 15 )}{ AC }\right) \\AC = \left(\frac{ 100 \sin( 15 )}{\sin( 97 )}\right) \\AC = 26.0762730956666$

Remember to round to the nearest hundredth.

Now we need to solve for the last side of the triangle.

To solve for this, we can use the Law of Cosines.

Lets recall the Law of Cosines.

$\BC^{2} = a^{2} - 2 a b \cos{\left (BC \right )} + b^{2} \BC^2 = 680.1664 + 10000 - 2 \cdot 26.08 \cdot 100 \cdot \cos( 68 ) \BC^{2} = 8726.2184087426 \BC = 93.4142302261417 \BC = 93.41$

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