Question

If Δ ABC is right angled at C , then the value of cos{A +B } is (a) 0 (b) 1 (c) ½(d) √3/2.​

Answer 1

Answer:

$\tt (a) \: 0$

Step-by-step explanation:

$\angle A + \angle B + \angle C = 180^{\circ}$

$\tt \: \angle A + \angle B + 90^{\circ} = 180^{\circ}$

$\tt \angle A + \angle B = 90^{\circ}$

$\tt \cos (A+B) = \cos (90^{\circ})$

$\tt =0$