Question-47775

Question Number 47775 by behi83417@gmail.com last updated on 14/Nov/18

Commented by behi83417@gmail.com last updated on 14/Nov/18

ABC : e q u i l a t e r a l .

d i s t a n c e o f D f r o m : F, E, C = 2, 3, 6

(r e s p e c t i v e l y)

\to \dots . . AB = ?

Answered by behi83417@gmail.com last updated on 14/Nov/18

s a y : d i s t a n c e o f D f r o m B C = x,

D F = y, D E = z, ∡ D C F = α, ∡ D C B = β

a . x + a . y + a . z = 2 S_{A B C}, A B = a

a (x + y + z) = 2 \times a^{2} . \frac{\sqrt{3}}{4} \Rightarrow x + y + z = a \frac{\sqrt{3}}{2}

s i n α = \frac{1}{3}, c o s α = \frac{2 \sqrt{2}}{3}, β = 60 - α

s i n β = \frac{x}{6} \Rightarrow x = 6 s i n β = 6 s i n (60 - α) =

= 6 (\frac{\sqrt{3}}{2} . \frac{2 \sqrt{2}}{3} - \frac{1}{2} . \frac{1}{3}) = 2 \sqrt{6} - 1

\Rightarrow 2 \sqrt{6} - 1 + 2 + 3 = a . \frac{\sqrt{3}}{2} \Rightarrow a = \frac{4 (\sqrt{6} + 2)}{3} .

Answered by mr W last updated on 14/Nov/18

α = ∠ D C F

\sin α = \frac{2}{6} = \frac{1}{3}

\cos α = \frac{2 \sqrt{2}}{3}

h = E D + C D \times \cos (30 ° - α)

= 3 + 6 (\cos 30 ° \cos α + \sin 30 ° \sin α)

= 3 + 6 (\frac{\sqrt{3}}{2} \times \frac{2 \sqrt{2}}{3} + \frac{1}{2} \times \frac{1}{3})

= 2 (2 + \sqrt{6})

A B = \frac{2 \times 2 (2 + \sqrt{6})}{\sqrt{3}} = \frac{4 (2 \sqrt{3} + 3 \sqrt{2})}{3} \approx 10.3

Commented by behi83417@gmail.com last updated on 14/Nov/18

c o r r e c t a n s w e r . t h a n k s i n a d v a n c e s i r .