in-triangle-ABC-a-2-b-c-2-1-2-B-C-2-find-h-a-S-ABC-d-a-R-A-

Question Number 75058 by behi83417@gmail.com last updated on 07/Dec/19

in triangle : ABC :

a = \sqrt{2}, b - c = \frac{\sqrt{2} + 1}{2}, \overset{}{B} - \overset{}{C} = \frac{π}{2}

find : h_{a}, S_{ABC}, d_{a}, R, \overset{}{A} .

Commented by mr W last updated on 07/Dec/19

b - c = \frac{1}{2} i s n o t p o s s i b l e s i r!

i t m u s t b e 1 < b - c < \sqrt{2}!

Commented by behi83417@gmail.com last updated on 07/Dec/19

thank you very much dear master .

now it is fixed .

Answered by mr W last updated on 07/Dec/19

l e t ∠ C = θ

\Rightarrow ∠ B = \frac{π}{2} + θ

\Rightarrow ∠ A = π - θ - (\frac{π}{2} + θ) = \frac{π}{2} - 2 θ

\frac{b}{\sin (\frac{π}{2} + θ)} = \frac{c}{\sin θ} = \frac{\sqrt{2}}{\sin (\frac{π}{2} - 2 θ)}

\Rightarrow b = \frac{\sqrt{2} \cos θ}{\cos 2 θ}

\Rightarrow c = \frac{\sqrt{2} \sin θ}{\cos 2 θ}

b - c = \frac{\sqrt{2} (\cos θ - \sin θ)}{\cos 2 θ} = δ

\frac{2 (1 - \sin 2 θ)}{\cos^{2} 2 θ} = δ^{2}

\frac{1}{1 + \sin 2 θ} = \frac{δ^{2}}{2}

\sin 2 θ = \frac{2}{δ^{2}} - 1 \Rightarrow 1 < δ < \sqrt{2}

\Rightarrow θ = \frac{1}{2} \sin^{- 1} (\frac{2}{δ^{2}} - 1)

w i t h δ = \frac{1 + \sqrt{2}}{2} w e g e t

θ = \frac{1}{2} \sin^{- 1} (23 - 16 \sqrt{2}) \approx 10.9475 °

t h e r e s t i s c l e a r .

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