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Question

In a acute angle triangle ABC, if sin(A+BC)=12 and cos(B+CA)=12.
Find A,B and C.

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Solution

Given,
A+B+C=1800(1)
sin(A+BC)=12(2)
cos(B+CA)=12(3)
Now from (1)
A+B=ΠC(4)
Using (4) we can rewrite (2) as
sin(Π2C)=12
sin2C=12(sin(Πθ)=sinθ)
2C=300 or 1500 ( sin300=1/2
C=150 or 750 sin1500=1/2)
Again from (1)
B+C=ΠA
cos(B+CA)=12
cos(Π2A)=12
cos2A=12 ( cos(Πθ)=cosθ)
cos2A=12
2A=1350 ( cos1350=12)
A=67.50
Case I : For C=150 Case II : For C=750
A=67.50 A=67.50
B=1800(150+67.50) B=1800(67.50+750)
=180082.5=97.50 =37.50
Since ΔABC is acute angled, thereby n none of the angles can be more than 900.
Hence A=67.50, C=750, B=37.50

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