If the sides of a triangle are sin- cos- -1-sin-cos- 0-2 the largest angle is

The correct option is C 120^∘ The largest side is c=√(1+sinαcosα) cosC=a^2+b^2 c^22ab =sin^2α+cos^2α (1+sinαcosα)2sinαcosα =1 1 sinαco ...

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Standard XII

Physics

Vector Addition

If the sides ...

Question

If the sides of a triangle are $sin α, cos α, \sqrt{1 + sin α cos α}, 0 < α < \frac{π}{2}$ the largest angle is

60^{\circ}

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90^{\circ}

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120^{\circ}

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150^{\circ}

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Solution

The correct option is C
$120^{\circ}$

The largest side is $c = \sqrt{1 + sin α cos α}$

$cos C = \frac{a^{2} + b^{2} - c^{2}}{2 a b}$

$= \frac{{sin}^{2} α + {cos}^{2} α - (1 + sin α cos α)}{2 sin α cos α}$

$= \frac{1 - 1 - sin α cos α}{2 sin α cos α}$

$= \frac{- sin α cos α}{2 sin α cos α}$

$= \frac{- 1}{2}$

$∴ C = 180^{\circ} - 60^{\circ} = 120^{\circ}$

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If the sides of a triangle are sinα,cosα,√1+sinαcosα,0<α<π2 the largest angle is

The correct option is C 120∘The largest side is c=√1+sinαcosαcosC=a2+b2−c22ab=sin2α+cos2α−(1+sinαcosα)2sinαcosα=1−1−sinαcosα2sinαcosα=−sinαcosα2sinαcosα=−12∴C=180∘−60∘=120∘

If the sides of a triangle are $sin α, cos α, \sqrt{1 + sin α cos α}, 0 < α < \frac{π}{2}$ the largest angle is