SOLUTION: If angles A, B, and C are the sides of a triangle such that sin(A+B)=1/cosC and cos(A+B)=cosC, then show that the triangle is a right triangle.

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Question 35670: If angles A, B, and C are the sides of a triangle such that sin(A+B)=1/cosC and cos(A+B)=cosC, then show that the triangle is a right triangle.
Answer by venugopalramana(3286) About Me  (Show Source):
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> If angles A, B, and C are the sides of a triangle such that sin(A+B)=1/cosC
> and cos(A+B)=cosC, then show that the triangle is a right triangle.
> SINCE IN A TRIANGLE A+B+C=180
SIN(A+B)=SIN(180-C)=SIN(C)=1/COS(C)
HENCE SIN(C)*COS(C)=1.......SIN(2C)/2=1......SIN2C=2....WRONG...IT MUST BE
SIN(A+B)=1-COSC...OR...=1
THEN SIN(C)=1-COS(C)
SIN(C)+COS(C)=1...DIVIDE THROUGHOUT BY SQRT(2)
SIN(C)/SQRT(2)+COS(C)/SQRT(2)=1/SQRT(2).......
PUT SIN(45)=COS(45)=1/SQRT(2)
SIN(C)COS(45)+COS(C)SIN(45)=SIN(45)
SIN(45+C)=SIN(45)=SIN(180-45)=SIN(135)
45+C=135
C=90...RIGHT ANGLED TRIANGLE.
COS(A+B)=COS(180-C)=-COS(C)=COS(C)...GIVEN
COS(C)=0...OR....C=90..RIGHT ANGLED TRIANGLE
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