In a triangle ABC if tanB satisfy the inequation -3x2-4x-3-0 then

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Internal Angle Bisector Theorem

In a triangle...

Question

$I n$ a triangle $A B C$ if $t a n B$ satisfy the inequation $\sqrt{3} x^{2} - 4 x + \sqrt{3} < 0$ then

0^{0} < B < 30^{0}

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30^{0} < B < 45^{0}

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30^{0} < B < 60^{0}

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45^{0} < B < 90^{0}

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Solution

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\frac{{tan}^{2} 60^{0} + 4 {sin}^{2} 45^{0} + 3 {sec}^{2} 30^{0} + 5 {cos}^{2} 90^{0}}{csc 30^{0} + sec 60^{0} - {cot}^{2} 30^{0}} =

A B C

a = (\sqrt{3} + 1)

∠ B = 30^{0}, ∠ C = 45^{0}

A B C

tan C

\sqrt{3} x^{2} - 4 x + \sqrt{3} < 0

A B C

t a n A

\sqrt{3} x^{2} - 4 x + \sqrt{3} < 0

A = 60^{0}

B = 30^{0}

tan (A - B) = \frac{tan A - tan B}{1 + tan A tan B}

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