Question-96785

Question Number 96785 by M±th+et+s last updated on 04/Jun/20

Answered by mr W last updated on 04/Jun/20

Commented by mr W last updated on 04/Jun/20

r + r \cos α = a

\Rightarrow \cos α = \frac{a}{r} - 1 = \frac{144}{80} - 1 = \frac{4}{5}

\tan β = \frac{r}{a - r \sin α} = \frac{1}{\frac{a}{r} - \sin α} = \frac{1}{\frac{144}{80} - \frac{3}{5}} = \frac{5}{6}

γ = \frac{π}{2} - 2 β

m = \frac{a}{\cos γ} = \frac{a}{\sin 2 β} = \frac{a}{2 \sin β \cos β}

= \frac{144}{2 \times \frac{5 \times 6}{5^{2} + 6^{2}}} = \frac{732}{5} = 146.4

Commented by M±th+et+s last updated on 04/Jun/20

g r e a t w o r k m r W . t h a n x