Question

In a triangle abc,find the value of cos (a+b)+cosc

Answer 1

Answer:

A+B+C=180 so C=180-(A+B) SO COS(A+B)+COSC= cos(180-C)+cosC= -cosc+cosc=0

Answer 2

→the value of A° + B° + C° =18°

→ A+B = 180-C

Solving the values,

→ cos(a+b) + cosc

= cos(180°-c) + cosc

= -cos c + cosc (cos(180°-c) is in II quadrant so cosc is negative.)

= 0