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# help

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1.] In ABC, sin A : sin B : sinC = 2 : 3 : 4. Find cos(A + C).

May 19, 2020

#1
+21959
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In triangle(ABC),  sin(A) : sin(B) : sin(C)  =  2 : 3 : 4                     Find  cos(A + C)

In triangle(ABC), let    side a = 2x    side b = 3x    side c = 4x

Using the Law of Cosines:  cos(C)  =  ( a2 + b2 - c2 ]) / ( 2·a·b )

cos(A)  =  [ (3x)2 + (4x)2 - (2x)2 ] / [ 2·(3x)·(4x) ]  =  [ 9x2 + 16x2 - 4x2 ] / [ 24x2 ]  = 7/8

Using the Pythagorean Theorem:  sin(A)  =  sqrt(15) / 8

cos(C)  =  [ (2x)2 + (3x)2 - (4x)2 ] / [ 2·(2x)·(3x) ]  =  [  -3x2 ] / [ 12x2 ]  =  -1/4

Using the Pythagorean Theorem:  sinC)  =  sqrt(15) / 4

Using the formula:  cos(A + C)  =  cos(A) ·cos(C) - sin(A)·sin(C)

=  (7/8) · (-1/4) - ( sqrt(15) / 8 ) · ( sqrt(15) / 4 )

=  -11/16

May 19, 2020

#1
+21959
+1

In triangle(ABC),  sin(A) : sin(B) : sin(C)  =  2 : 3 : 4                     Find  cos(A + C)

In triangle(ABC), let    side a = 2x    side b = 3x    side c = 4x

Using the Law of Cosines:  cos(C)  =  ( a2 + b2 - c2 ]) / ( 2·a·b )

cos(A)  =  [ (3x)2 + (4x)2 - (2x)2 ] / [ 2·(3x)·(4x) ]  =  [ 9x2 + 16x2 - 4x2 ] / [ 24x2 ]  = 7/8

Using the Pythagorean Theorem:  sin(A)  =  sqrt(15) / 8

cos(C)  =  [ (2x)2 + (3x)2 - (4x)2 ] / [ 2·(2x)·(3x) ]  =  [  -3x2 ] / [ 12x2 ]  =  -1/4

Using the Pythagorean Theorem:  sinC)  =  sqrt(15) / 4

Using the formula:  cos(A + C)  =  cos(A) ·cos(C) - sin(A)·sin(C)

=  (7/8) · (-1/4) - ( sqrt(15) / 8 ) · ( sqrt(15) / 4 )

=  -11/16

geno3141 May 19, 2020