Question-88214

Question Number 88214 by M±th+et£s last updated on 09/Apr/20

Answered by ajfour last updated on 09/Apr/20

A B = t, r a d i u s = r, A D = D C = b

u p p e r s e c t i o n o f O B = c

t^{2} = 2 b^{2}; \tan α = \frac{r}{t + r}

\frac{\sqrt{r^{2} + c^{2}}}{2 b} = \frac{c}{r} = \frac{r}{t + r}

\Rightarrow \frac{\sqrt{r^{2} + c^{2}}}{t \sqrt{2}} = \frac{c}{r} = \frac{r}{t + r}

\Rightarrow c = \frac{r^{2}}{t + r}

\Rightarrow \frac{r^{2} + {(\frac{r^{2}}{t + r})}^{2}}{2 t^{2}} = \frac{r^{2}}{{(t + r)}^{2}}

l e t t / r = s

\Rightarrow \frac{1 + \frac{1}{{(s + 1)}^{2}}}{2 s^{2}} = \frac{1}{{(s + 1)}^{2}}

\Rightarrow {(s + 1)}^{2} + 1 = 2 s^{2}

s^{2} - 2 s - 2 = 0

s = 1 + \sqrt{3}

\tan α = \frac{1}{s + 1} = \frac{1}{2 + \sqrt{3}} = 2 - \sqrt{3} .

\Rightarrow α = \tan^{- 1} (2 - \sqrt{3}) = 15 ° .

Commented by mr W last updated on 09/Apr/20

c o r r e c t!

Commented by M±th+et£s last updated on 09/Apr/20

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

Answered by mr W last updated on 09/Apr/20

Commented by mr W last updated on 09/Apr/20

r = r a d i u s

A D = D C

\Rightarrow C F = \frac{O B}{2} = \frac{r}{2}

C D = C E \times \cos α = 2 r \cos α

C F = C D \times \sin α = 2 r \cos α \sin α = r \sin 2 α

\Rightarrow r \sin 2 α = \frac{r}{2}

\Rightarrow \sin 2 α = \frac{1}{2} \Rightarrow 2 α = 30 ° \Rightarrow α = 15 °

Commented by M±th+et£s last updated on 09/Apr/20

t h a n k y o u s i r