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Question

If $$\triangle ABC$$ is a right angled triangle, right angled at C, then $$\cos (A + B) = $$


A
0
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B
12
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C
1
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D
32
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Solution

The correct option is D $$0$$
Given that $$\triangle ABC$$ is a right angled triangle, with $$\angle C=90^o$$
To find out: The value of $$\cos(A+B)$$
In $$\triangle ABC$$,
$$\angle A+\angle B+\angle C=180^o\quad \quad [\text{Angle sum property}]$$
$$\Rightarrow \angle A+\angle B+90^o=180^o$$
$$\Rightarrow \angle A+\angle B=180^o-90^o$$
$$\therefore \  \angle A+\angle B=90^o$$

Now, $$\cos(A+B)=\cos90^o$$
We know that, $$\cos 90^o=0$$
$$\therefore \ \cos(A+B)=0$$

Hence, the required value of $$\cos(A+B)$$ for the given triangle is $$0$$.

Mathematics

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