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The lengths of sides of triangle ABC are...

The lengths of sides of `triangle ABC` are a units, b units and `sqrt(a^2+ab+b^2)` units show that greatest angle of the triangle is`120^@`

Text Solution

Verified by Experts

The correct Answer is:
`120^(@)`
Sides are `a, b, sqrt(a^(2) + ab + b^(2))`
Then the greatest side `c = sqrt(a^(2) + ab + b^(2))`
Let the angle opposite to the greatest side be C.
`rArr cos C = (a^(2) + b^(2) - (a^(2) + ab + b^(2)))/(2ab) = (-1)/(2)`
`rArr C = 120^(@)`
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Knowledge Check

  • If the sides of the triangle ABC are p,q and sqrt(p^(2)+ pq+q^(2)), then the greatest angle of the triangle is-

    A
    `(pi)/(2)`
    B
    `(2pi)/(3)`
    C
    `(5pi)/(4)`
    D
    `(5pi)/(3)`

  • If the sides of triangle be 15 units , 12 units and 9 units then the area (in sq. units) of the triangle is-

    A
    45
    B
    54
    C
    50
    D
    40

  • The circumradius of triangle ABC is 10 units: If c=10sqrt(3) units, state which of the following is the value of the angle C?

    A
    `60^(@)`
    B
    `120^(@)`
    C
    `30^(@)`
    D
    `90^(@)`

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