Q8 of 120 Page 89

If ABC is right angled at C, then the value of cos (A+B) is

Given: ∠C = 90°


By the property of triangle, the sum of the three angles is equal to


∠A+∠B+∠C = 180°


⇒ ∠A+∠B+90° = 180° (∵ ∠C = 90° )


⇒ ∠A+∠B = 90°


Thus, cos(A+B) = cos 90° = 0

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