Question-172144

Question Number 172144 by mnjuly1970 last updated on 23/Jun/22

Answered by infinityaction last updated on 23/Jun/22

Commented by infinityaction last updated on 23/Jun/22

\cos α = \frac{2}{r} a n d \sin α = \frac{r}{6}

\cos^{2} α + \sin^{2} α = \frac{4}{r^{2}} + \frac{r^{2}}{36}

l e t r^{2} = p

1 = \frac{4}{p} + \frac{p}{36}

p^{2} - 36 p + 144 = 0

p = r^{2} = 18 - 6 \sqrt{5}

Commented by mr W last updated on 23/Jun/22

n i c e s o l u t i o n!

Commented by mnjuly1970 last updated on 23/Jun/22

t h a n k y o u s o m u c h \dots

Answered by mr W last updated on 23/Jun/22

Commented by mr W last updated on 23/Jun/22

y e s, i f o u n d i t t o o . t h a n k s!

Commented by mr W last updated on 23/Jun/22

γ = 90 ° - α = 45 ° + β

\frac{\sin α}{d} = \frac{\sin 45 °}{6} \dots (i)

\frac{d}{\sin 45 °} = \frac{4}{\sin (45 ° + β)} = \frac{4}{\cos α} \dots (i i)

(i) \times (i i) :

\frac{\sin α}{\sin 45 °} = \frac{4 \sin 45 °}{6 \cos α}

\Rightarrow 6 \sin α \cos α = 4 \sin^{2} 45 °

\Rightarrow 3 \sin 2 α = 2 \Rightarrow \sin 2 α = \frac{2}{3}

\cos 2 α = 2 \cos^{2} α - 1 = \sqrt{1 - {(\frac{2}{3})}^{2}} = \frac{\sqrt{5}}{3}

\Rightarrow \cos α = \frac{\sqrt{3 + \sqrt{5}}}{\sqrt{6}}

d = \frac{4 \sin 45 °}{\cos α} = \frac{4 \sqrt{3}}{\sqrt{3 + \sqrt{5}}}

a r e a o f s q u a r e

= {(\frac{d}{\sqrt{2}})}^{2} = \frac{d^{2}}{2} = \frac{8 \times 3}{3 + \sqrt{5}} = 6 (3 - \sqrt{5})

Commented by mnjuly1970 last updated on 23/Jun/22

t h a n k s a l o t s i r

Commented by infinityaction last updated on 23/Jun/22

\frac{d^{2}}{2} = \frac{1}{2} \times \frac{16 \times 3}{3 + \sqrt{5}} = 6 (3 - \sqrt{5}) = 18 - 6 \sqrt{5}

Commented by Tawa11 last updated on 24/Jun/22

Great sir

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