# LAW OF SINES AND COSINES EXAMPLES

## About "Law of Sines and Cosines Examples"

Law of Sines and Cosines Examples :

Here we are going to see some example problems based on law of sines and cosines.

## Law of Sines and Cosines Examples - Examples

Question 1 :

In a triangle ABC, if cos C = sin A / 2 sin B, show that the triangle is isosceles.

Solution :

cos C = sin A / 2 sin B

cos C (2 sin B)  =  sin A

2 sin B cos C  =  sin A

sin (B + C) + sin (B - C)  =  sin A

A + B + C  =  180

B + C  =  180 - A

sin (180 - A) + sin (B - C)  =  sin A

sin A + sin (B - C)  =  sin A

sin (B - C)  =  0

B - C  =  0

B  =  C

Hence the given triangle is isosceles triangle.

Question 2 :

In a triangle ABC, prove that sin B/sinC  =  (c − a cosB)/(b − a cosC)

Solution :

sin B/sinC  =  (c − a cosB)/(b − a cosC)

Laws of sine :

a/sin A  =  b/sin B  =  c/sin C  =  2R

a  =  2R sin A, b = 2R sin B and c = 2R sin C

R.H.S

=  (c − a cosB)/(b − a cosC)

=  (2R sin C - 2R sin A cos B) / (2R sin B - 2R sin A cos C)

=  (sin C - sin A cos B) / (sin B - sin A cos C)

=  (sin C - (1/2) [sin (A+B) + sin (A-B)] / (sin B - (1/2)[sin(A+C) sin (A-C)]

=  (sin C - (1/2) [sin C + sin (A-B)] / (sin B - (1/2)[sinB sin (A-C)]

=  (1/2) [sin C - sin (A-B)]/(1/2)[sin B - sin (A-C)]

=  [sin (A+B) - sin (A-B)]/[sin (A+C) - sin (A-C)]

=  (2 cos A sin B)/(2 cos A sin C)

=  sin B / sin C

Hence proved.

Question 3 :

In a triangle ABC, prove that a cosA + b cosB + c cosC = 2a sinB sinC.

Solution :

L.H.S :

a cosA + b cosB + c cosC

a/sin A  =  b/sin B  =  c/sin C  =  2R

a = 2R sin A, b = 2R sin B  and c = 2R sin C

=  2R sin A cos A + 2R sin B cos B + 2R sin C cos C

=  R[sin 2A + sin 2B + sin 2C]

=  R[2 sin (A + B) cos (A - B) + sin 2C]

=  R[2 sin (180 - C) cos (A - B) + sin 2C]

=  R[2 sin C cos (A - B) + 2 sin C cos C]

=  2R sin C [cos (A - B) + cos C]

=  2R sin C [cos (A-B) + cos (180-(A+B))]

=  2R sin C [cos (A-B) - cos (A+B)]

=  2R sin C [2 sin A sin B]

Instead of 2R sin A, we may apply "a".

=  2a sin A sin B sin C

Hence proved.

After having gone through the stuff given above, we hope that the students would have understood, "Law of Sines and Cosines Examples"

Apart from the stuff given in "Law of Sines and Cosines Examples", if you need any other stuff in math, please use our google custom search here.

Kindly mail your feedback to v4formath@gmail.com

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The text passage contains multiple mathematical questions and their solutions, all related to the Law of Sines and Cosines in triangles.
multiple_questions_with_answers
Given a text passage, determine:
    1. If it contains a mathematical question (including word problems or calculations)
    2. If an answer to that mathematical question is provided in the text

First, describe the content structure you found in 1-2 sentences.

Then provide your analysis in the following format:
TYPE: <one of the following values>
- NO_QUESTION
- SINGLE_QUESTION
- SINGLE_QUESTION_WITH_ANSWER
- MULTIPLE_QUESTIONS
- MULTIPLE_QUESTIONS_WITH_ANSWERS
system
# LAW OF SINES AND COSINES EXAMPLES

## About "Law of Sines and Cosines Examples"

Law of Sines and Cosines Examples :

Here we are going to see some example problems based on law of sines and cosines.

## Law of Sines and Cosines Examples - Examples

Question 1 :

In a triangle ABC, if cos C = sin A / 2 sin B, show that the triangle is isosceles.

Solution :

cos C = sin A / 2 sin B

cos C (2 sin B)  =  sin A

2 sin B cos C  =  sin A

sin (B + C) + sin (B - C)  =  sin A

A + B + C  =  180

B + C  =  180 - A

sin (180 - A) + sin (B - C)  =  sin A

sin A + sin (B - C)  =  sin A

sin (B - C)  =  0

B - C  =  0

B  =  C

Hence the given triangle is isosceles triangle.

Question 2 :

In a triangle ABC, prove that sin B/sinC  =  (c − a cosB)/(b − a cosC)

Solution :

sin B/sinC  =  (c − a cosB)/(b − a cosC)

Laws of sine :

a/sin A  =  b/sin B  =  c/sin C  =  2R

a  =  2R sin A, b = 2R sin B and c = 2R sin C

R.H.S

=  (c − a cosB)/(b − a cosC)

=  (2R sin C - 2R sin A cos B) / (2R sin B - 2R sin A cos C)

=  (sin C - sin A cos B) / (sin B - sin A cos C)

=  (sin C - (1/2) [sin (A+B) + sin (A-B)] / (sin B - (1/2)[sin(A+C) sin (A-C)]

=  (sin C - (1/2) [sin C + sin (A-B)] / (sin B - (1/2)[sinB sin (A-C)]

=  (1/2) [sin C - sin (A-B)]/(1/2)[sin B - sin (A-C)]

=  [sin (A+B) - sin (A-B)]/[sin (A+C) - sin (A-C)]

=  (2 cos A sin B)/(2 cos A sin C)

=  sin B / sin C

Hence proved.

Question 3 :

In a triangle ABC, prove that a cosA + b cosB + c cosC = 2a sinB sinC.

Solution :

L.H.S :

a cosA + b cosB + c cosC

a/sin A  =  b/sin B  =  c/sin C  =  2R

a = 2R sin A, b = 2R sin B  and c = 2R sin C

=  2R sin A cos A + 2R sin B cos B + 2R sin C cos C

=  R[sin 2A + sin 2B + sin 2C]

=  R[2 sin (A + B) cos (A - B) + sin 2C]

=  R[2 sin (180 - C) cos (A - B) + sin 2C]

=  R[2 sin C cos (A - B) + 2 sin C cos C]

=  2R sin C [cos (A - B) + cos C]

=  2R sin C [cos (A-B) + cos (180-(A+B))]

=  2R sin C [cos (A-B) - cos (A+B)]

=  2R sin C [2 sin A sin B]

Instead of 2R sin A, we may apply "a".

=  2a sin A sin B sin C

Hence proved.

After having gone through the stuff given above, we hope that the students would have understood, "Law of Sines and Cosines Examples"

Apart from the stuff given in "Law of Sines and Cosines Examples", if you need any other stuff in math, please use our google custom search here.

Kindly mail your feedback to v4formath@gmail.com

## Recent Articles

1. ### SAT Math Videos

May 22, 24 06:32 AM

SAT Math Videos (Part 1 - No Calculator)

2. ### Simplifying Algebraic Expressions with Fractional Coefficients

May 17, 24 08:12 AM

Simplifying Algebraic Expressions with Fractional Coefficients
user
The text passage contains multiple mathematical questions and their solutions, all related to the Law of Sines and Cosines in triangles.

TYPE: MULTIPLE_QUESTIONS_WITH_ANSWERS