Question-112076

Question Number 112076 by I want to learn more last updated on 06/Sep/20

Commented by Aina Samuel Temidayo last updated on 06/Sep/20

Is OD the angle bisector of D ?

Commented by som(math1967) last updated on 06/Sep/20

yes you are correct

Commented by Aina Samuel Temidayo last updated on 06/Sep/20

DC is not a chord . I think your

solution is incorrect .

Answered by mr W last updated on 06/Sep/20

s a y r a d i u s O C = O B = 1

B C = \frac{2}{\cos 20 °}

\frac{C D}{\sin 30 °} = \frac{B D}{\sin 40 °} = \frac{B C}{\sin 70 °} = \frac{2}{\cos^{2} 20 °}

\Rightarrow C D = \frac{4}{\cos^{2} 20 °}

{O D}^{2} = 1^{2} + \frac{4^{2}}{\cos^{4} 20 °} - 2 \times \frac{4}{\cos^{2} 20 °} \times \cos 40 °

= 1 + \frac{16}{\cos^{4} 20 °} - \frac{8 (2 \cos^{2} 20 - 1)}{\cos^{2} 20 °}

= \frac{16}{\cos^{4} 20 °} + \frac{8}{\cos^{2} 20 °} - 15

= \frac{1}{\cos^{4} 20 °} (16 + 8 \cos^{2} 20 ° - 15 \cos^{4} 20 °)

\frac{\sin x}{O C} = \frac{\sin 20 °}{O D}

\Rightarrow \sin x = \frac{\sin 20 ° \cos^{2} 20 °}{\sqrt{16 + 8 \cos^{2} 20 ° - 15 \cos^{4} 20 °}}

\Rightarrow x \approx 5.139 °

Commented by I want to learn more last updated on 06/Sep/20

Wow, thanks sir, i appreciate .

Sir, what condition can i say radii OC = OB = 1 ? ?

Commented by mr W last updated on 06/Sep/20

y o u c a n s a y O C = O B = r, i t {d o e s n}^{'} t

m a t t e r, s i n c e w e a r e c a l c u l a t i n g t h e

a n g l e s o n l y .

Commented by I want to learn more last updated on 06/Sep/20

Ohh, i understand sir . Thanks so much .