Question

If triangleABC is a right angle triangle at C.then what is the value of cos(A+B)

Answer 1

$\sf{Since \: ABC \: is \: right \: angled \: and \: angle \: C \: is \: 90°}$

Therefore,

$\sf{A+B=180°-C}$

$\sf{A+B=180°-90°}$

$\sf{A+B=90°}$

$\sf{∴ cos(A+B)=cos90° = 0}$

Answer 2

Answer:

Hi there!

As we know that the sum of all three angles of a triangle is 180°.

If ∆ABC is right angled at C

C = 90°

=> $\:\mathsf\red{∠A\: +\:∠B\: +\: ∠C\:=\:180°}$

=> $\:\sf{∠A\: +\:∠B\: +\: 90°\:=\:180°}$

=> $\:\sf{∠A\: +\:∠B\: =\:180°\:-\:90°}$

=> $\:\sf{∠A\: +\:∠B\: =\:90°}$

Multiplying cos both sides. We get,

=> $\:\sf{cos\:(∠A\: +\:∠B)\: =\:cos\:90°}$

We know cos 90° = 0, by trigonometric ratio.

=> $\:\sf{cos\:90°\:=\:0}$

So,

$\:\mathsf{cos(∠A\: +\:∠B)=\:cos\:90°\:=\:0}$

=> $\:\sf\purple{cos\:(∠A\:+\:∠B)\:=\:0}$

$\:\mathsf\blue{Hope\:this\:helps\:you!\:}$