My Math Forum Circle theorems and trigonometry

 Elementary Math Fractions, Percentages, Word Problems, Equations, Inequations, Factorization, Expansion

February 12th, 2018, 09:51 AM   #1
Newbie

Joined: Feb 2018
From: England

Posts: 2
Thanks: 0

Circle theorems and trigonometry

I am just completely clueless on how to even start on this question, so any help is appreciated.
Attached Images
 82A94FD6-09DD-4443-BA90-5DC28BCF28C5.jpg (11.1 KB, 12 views)

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: 18 Thanks: 1 Could you send another image? I'm not able to read the exercise...
February 12th, 2018, 10:41 AM   #3
Newbie

Joined: Feb 2018
From: England

Posts: 2
Thanks: 0

sorry yes attached below
Attached Images

 February 12th, 2018, 12:11 PM #4 Global Moderator     Joined: Oct 2008 From: London, Ontario, Canada - The Forest City Posts: 7,923 Thanks: 1123 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: 20,386 Thanks: 2012 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.

 Tags circle, theorems, trigonometry

 Thread Tools Display Modes Linear Mode

 Similar Threads Thread Thread Starter Forum Replies Last Post jamesbrown Trigonometry 5 September 15th, 2016 12:12 PM Tangeton Geometry 1 March 10th, 2016 07:40 PM amni1234 Geometry 3 February 23rd, 2016 04:42 AM peterle1 Algebra 2 February 11th, 2010 10:46 AM David_Lete Algebra 5 April 11th, 2009 04:20 AM

 Contact - Home - Forums - Cryptocurrency Forum - Top