Law of Sines and Cosines Examples :
Here we are going to see some example problems based on law of sines and cosines.
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"
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