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Question

If in a triangle ΔABC,a2b2c2=0, then

A
π4<A<π2
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B
π2<A<π
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C
A=π2
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D
A<π4
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Solution

The correct option is B A=π2
Given, in a triangle ΔABC,a2b2c2=0.....(1).
We have,
cosA=b2+c2a22bc
or, cosA=0 [ Using (1)]
or, A=π2.

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