lf $D$ is the middle point of the side $B C$ of a triangle $A B C$ and $A D$ is perpendicular to $A C$ , then

3 b^{2} = a^{2} - c^{2}

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3 a^{2} = b^{2} - 3 c^{2}

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b^{2} = a^{2} - c^{2}

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a^{2} + b^{2} = 5 c^{2}

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Solution

The correct option is A $3 b^{2} = a^{2} - c^{2}$
$tan (π - θ) = \frac{b}{l}$
$tan θ = \frac{- b}{l}, l^{2} + b^{2} = (\frac{a^{2}}{4})$
$\frac{sin (p i - θ)}{b} = \frac{sin π / 2}{a / 2}$
$sin θ = \frac{2 b}{a}$
$cos θ = \frac{l^{2} + \frac{a^{2}}{4} - c^{2}}{2 \times l \times a / 2} = \frac{4 l^{2} + a^{2} - 4 c^{2}}{l a}$
$tan θ = \frac{2 b l}{4 l^{2} + a^{2} - 4 c^{2}} = \frac{- b}{l}$
$- 2 l^{2} = 4 l^{2} + a^{2} - 4 c^{2}$
$6 l^{2} + a^{2} - 4 c^{2} = 0$
$6 (\frac{a^{2}}{4}) - 6 b^{2} + a^{2} - 4 c^{2} = 0$
$3 c^{2} = a^{2} - c^{2}$

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Similar questions

D

B C

A B C

A D C

a^{2} : b^{2} : c^{2}

In the adjoining figure, if BC = a, AC = b, AB = c and $∠ C A B = 120^{\circ}$ , then the correct relation is-

In the adjoining figure, if $B C = a, A C = b, A B = c$ and $∠ C A B = 120^{\circ}$ , then the correct relation is

In the adjoining figure, if $B C = a, A C = b, A B = c$ and $∠ C A B = 120^{\circ}$ ,
then which of the following is the correct relation?

△ A B C

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

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

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