Common Core: High School - Geometry : Prove Laws of Sines and Cosines: CCSS.Math.Content.HSG-SRT.D.10

Example Questions

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Example Question #1 : Prove Laws Of Sines And Cosines: Ccss.Math.Content.Hsg Srt.D.10

In a triangle where the side opposite a  has length 10 find the side opposite a  angle. Round you answer to the nearest hundredth.

Explanation:

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

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

Now we rearrange the equation to solve for

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.

Example Question #2 : Prove Laws Of Sines And Cosines: Ccss.Math.Content.Hsg Srt.D.10

In a triangle where the side opposite a  has length 6 find the side opposite a  angle. Round you answer to the nearest hundredth.

Explanation:

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

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

Now we rearrange the equation to solve for

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.

Example Question #3 : Prove Laws Of Sines And Cosines: Ccss.Math.Content.Hsg Srt.D.10

In a triangle where the side opposite a  has length 3 find the side opposite a  angle. Round you answer to the nearest hundredth.

Explanation:

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

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

Now we rearrange the equation to solve for b

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.

Example Question #4 : Prove Laws Of Sines And Cosines: Ccss.Math.Content.Hsg Srt.D.10

In a triangle where the side opposite a  has length 13 find the side opposite a  angle. Round you answer to the nearest hundredth.

Explanation:

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

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

Now we rearrange the equation to solve for b

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.

Example Question #5 : Prove Laws Of Sines And Cosines: Ccss.Math.Content.Hsg Srt.D.10

In a triangle where the side opposite a  has length 12 find the side opposite a  angle. Round you answer to the nearest hundredth.

Explanation:

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

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

Now we rearrange the equation to solve for b

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.

Example Question #6 : Prove Laws Of Sines And Cosines: Ccss.Math.Content.Hsg Srt.D.10

In a triangle where the side opposite a  has length 6 find the side opposite a  angle. Round you answer to the nearest hundredth.

Explanation:

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

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

Now we rearrange the equation to solve for b

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.

Example Question #7 : Prove Laws Of Sines And Cosines: Ccss.Math.Content.Hsg Srt.D.10

In a triangle where the side opposite a  has length 11 find the side opposite a  angle. Round you answer to the nearest hundredth.

Explanation:

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

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

Now we rearrange the equation to solve for b

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.

Example Question #8 : Prove Laws Of Sines And Cosines: Ccss.Math.Content.Hsg Srt.D.10

In a triangle where the side opposite a  has length 11 find the side opposite a  angle. Round you answer to the nearest hundredth.

Explanation:

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

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

Now we rearrange the equation to solve for b

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.

Example Question #9 : Prove Laws Of Sines And Cosines: Ccss.Math.Content.Hsg Srt.D.10

In a triangle where the side opposite a  has length 5 find the side opposite a  angle. Round you answer to the nearest hundredth.

Explanation:

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

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

Now we rearrange the equation to solve for b

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.

Example Question #10 : Prove Laws Of Sines And Cosines: Ccss.Math.Content.Hsg Srt.D.10

In a triangle where the side opposite a  has length 5 find the side opposite a  angle. Round you answer to the nearest hundredth.

Explanation:

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

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

Now we rearrange the equation to solve for b

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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