in-triangle-ABC-BC-1-B-2-A-find-the-maximum-area-of-ABC-

Question Number 137327 by mr W last updated on 01/Apr/21

i n t r i a n g l e Δ A B C : B C = 1, ∠ B = 2 ∠ A .

f i n d t h e m a x i m u m a r e a o f Δ A B C .

Answered by EDWIN88 last updated on 01/Apr/21

∠ A + ∠ B + ∠ C = π; 3 ∠ A + ∠ C = π

let ∠ A = α; ∠ B = 2 α; ∠ C = π - 3 α

\frac{BC}{\sin α} = \frac{AC}{\sin 2 α} \Rightarrow \frac{1}{\sin α} = \frac{AC}{2 \sin α \cos α}

AC = 2 \cos α

Area of Δ ABC = \frac{1}{2} . 1 . AC . \sin (π - 3 α)

A (α) = \frac{1}{2} (2 \cos α \sin 3 α)

A (α) = \frac{1}{2} (\sin 4 α + \sin 2 α)

A^{'} (α) = \frac{1}{2} (4 \cos 4 α + 2 \cos 2 α) = 0

\Rightarrow 2 \cos 4 α + \cos 2 α = 0

\Rightarrow 2 (2 \cos^{2} 2 α - 1) + \cos 2 α = 0

\Rightarrow 4 \cos^{2} 2 α + \cos 2 α - 2 = 0

\cos 2 α = \frac{- 1 + \sqrt{33}}{8} = 0.593

\sin 2 α = \sqrt{1 - {(0.593)}^{2}} = 0.805

Thus A {(α)}_{\max} = \frac{1}{2} (2 \sin 2 α \cos 2 α + \sin 2 α)

A {(α)}_{\max} = \frac{1}{2} \sin 2 α (2 \cos 2 α + 1)

A {(α)}_{\max} = \frac{1}{2} (0.805) (2.186) = 0.8798

Commented by mr W last updated on 01/Apr/21

y e s . t h e e x a c t v a l u e i s

\frac{(\sqrt{33} + 3) \sqrt{(\sqrt{33} + 3) (\sqrt{33} - 1)}}{64} \approx 0.8801

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