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Question Number 168921 by Shrinava last updated on 21/Apr/22

	Answered by mr W last updated on 22/Apr/22

	Commented by Tawa11 last updated on 22/Apr/22

	$Great sir .$
	Commented by mr W last updated on 22/Apr/22

	$s a y B C = a, D E = b$ $l e t B D = c$ $c^{2} = a^{2} + b^{2} - 2 a b \cos (π - θ - \frac{π}{3})$ $c^{2} = a^{2} + b^{2} + a b (\cos θ - \sqrt{3} \sin θ)$ $\frac{\sin ∠ A B D}{A D} = \frac{\sin (θ + \frac{π}{3})}{B D}$ $\Rightarrow \sin ∠ A B D = \frac{b \sin (θ + \frac{π}{3})}{c}$ $\cos ∠ A B D = \frac{a^{2} + c^{2} - b^{2}}{2 a c}$ ${A F}^{2} = a^{2} + c^{2} - 2 a c \cos (∠ A B D + \frac{π}{3})$ ${A F}^{2} = a^{2} + c^{2} - a c (\cos ∠ A B D - \sqrt{3} \sin ∠ A B D)$ ${A F}^{2} = a^{2} + c^{2} - a c \cos ∠ A B D + \sqrt{3} a b \sin (θ + \frac{π}{3})]$ ${A F}^{2} = a^{2} + c^{2} - a c \times \frac{a^{2} + c^{2} - b^{2}}{2 a c} + \sqrt{3} a b \sin (θ + \frac{π}{3})]$ ${A F}^{2} = \frac{a^{2} + b^{2} + c^{2}}{2} + \sqrt{3} a b \sin (θ + \frac{π}{3})$ ${A F}^{2} = \frac{a^{2} + b^{2} + a^{2} + b^{2} + a b (\cos θ - \sqrt{3} \sin θ)}{2} + \frac{\sqrt{3}}{2} a b (\sin θ + \sqrt{3} \cos θ)$ ${A F}^{2} = a^{2} + b^{2} + 2 a b \cos θ$ $A F = \sqrt{a^{2} + b^{2} + 2 a b \cos θ}$ $r e p l a c e θ w i t h - θ w e g e t$ $A G = \sqrt{a^{2} + b^{2} + 2 a b \cos (- θ)}$ $A G = \sqrt{a^{2} + b^{2} + 2 a b \cos θ}$ $i . e . A F = A G$ $i t c a n a l s o b e s h o w n t h a t F, A, G$ $a r e c o l l i n e a r .$ $F G = A F + A G = 2 \sqrt{a^{2} + b^{2} + 2 a b \cos θ}$ $⩽ 2 \sqrt{a^{2} + b^{2} + 2 a b} = 2 (a + b)$ $i . e . 2 (B C + D E) ⩾ F G$
	Commented by Shrinava last updated on 23/Apr/22

	$Cool dear sir$
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