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In DeltaABC,2angleA=3angleB=6angleC, the...

In `DeltaABC,2angleA=3angleB=6angleC`, then find `angleA`.

A

`90^@`

B

`30^@`

C

`60^@`

D

`45^@`

Text Solution

Verified by Experts

The correct Answer is:
A
Let `2angleA=3angleB=6angleC=k` (say).
Then, `angleA=((k)/(2))^(@),angleB=((k)/(3))^(@)andangleC=((k)/(6))^(@)`
We know that the sum of the angles of a triangle is `180^(@)`.
`:.angleA+angleB+angleC=180^(@)`
`implies(k)/(2)+(k)/(3)+(k)/(6)=180`
`implies(3k+2k+k)=(180xx6)`
`implies6k=(180xx6)impliesk=180`.
`:.angleA=((k)/(2))^(@)=((180)/6)^(@)=90^(@),angleB=((k)/(3))^(@)=((180)/(3))^(@)=60^(@)`
and `angleC=((k)/(6))^(@)=((180)/(6))^(@)=30^(@)`
Hence, `angleA=90^(@),angleB=60^(@)andangleC=30^(@)`.
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