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Question

In ABC,sinA+sinB+sinC=1+2 and cosA+cosB+cosC=2 if the triangle is

A
equilateral
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B
isosceles
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C
right angled
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D
right-angled isosceles
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Solution

The correct option is D right-angled isosceles
Given, sinA+sinB+sinC=1+2
Consider B=900
Hence we get
sinA+cosA=2.
1+2sinA.cosA=2
sin2A=1
2A=π2
A=C=π4.
Hence an isosceles right angled triangle.

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