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

Sagot :

ItzAkshara

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°

=> [tex]\:\mathsf\red{∠A\: +\:∠B\: +\: ∠C\:=\:180°}[/tex]

=> [tex]\:\sf{∠A\: +\:∠B\: +\: 90°\:=\:180°}[/tex]

=> [tex]\:\sf{∠A\: +\:∠B\: =\:180°\:-\:90°}[/tex]

=> [tex]\:\sf{∠A\: +\:∠B\: =\:90°}[/tex]

Multiplying cos both sides. We get,

=> [tex]\:\sf{cos\:(∠A\: +\:∠B)\: =\:cos\:90°}[/tex]

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

=> [tex]\:\sf{cos\:90°\:=\:0}[/tex]

So,

[tex]\:\mathsf{cos(∠A\: +\:∠B)=\:cos\:90°\:=\:0}[/tex]

=> [tex]\:\sf\purple{cos\:(∠A\:+\:∠B)\:=\:0}[/tex]

[tex]\:\mathsf\blue{Hope\:this\:helps\:you!\:}[/tex]

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