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February 12th, 2018, 09:51 AM   #1
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Circle theorems and trigonometry

I am just completely clueless on how to even start on this question, so any help is appreciated.
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Last edited by skipjack; February 12th, 2018 at 01:56 PM.

 February 12th, 2018, 10:11 AM #2 Newbie   Joined: Dec 2017 From: Spain Posts: 15 Thanks: 1 Could you send another image? I'm not able to read the exercise...
February 12th, 2018, 10:41 AM   #3
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sorry yes attached below
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 February 12th, 2018, 12:11 PM #4 Global Moderator     Joined: Oct 2008 From: London, Ontario, Canada - The Forest City Posts: 7,783 Thanks: 1025 Math Focus: Elementary mathematics and beyond Can you use the inscribed angle theorem? If so (for part (a)), $$2\angle{BDC}=\angle{BOC} \\ 2\angle{DBC}=\angle{DOC} \\ 2\angle{BDC}+2\angle{DBC}=\angle{BOC}+\angle{DOC}= 180^\circ\implies\angle{BDC}+\angle{DBC}=90^\circ \implies\angle{BCD}=90 ^\circ$$ For part (b), $\overline{OB}=7\sin35^\circ$. Now use the fact that $\triangle{BOC}$ is isosceles $\implies\overline{BC}=2\overline{OB}\cos70^\circ$ . Last edited by greg1313; February 14th, 2018 at 02:10 AM.
 February 12th, 2018, 01:54 PM #5 Global Moderator   Joined: Dec 2006 Posts: 18,841 Thanks: 1564 As $OB = OC = OD$, $\angle OCB = \angle OBC = 70^\circ$ and $\angle ODC = \angle OCD$. Hence $\angle DOC = 140^\circ$, and so $\angle ODC + \angle OCD = 40^\circ$. It follows that$\angle OCD = 20^\circ$, and so $\angle BCD = 70^\circ + 20^\circ = 90^\circ$. You could alternatively obtain $\angle BCD$ in one step by use of the alternate segment theorem, which states that the angle between the tangent and chord equals the angle in the alternate segment. As $\sin(35^\circ) = OB/(7\text{ cm})$ and $\cos(70^\circ) = (BC/2)/OB$, $BC = 2\sin(35^\circ)\cos(70^\circ)\times 7\text{ cm}$, which can be evaluated by use of a calculator. If you're required to specify every theorem that you've used, you would need to add a few details to the above.

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