Cirangle!

Geometry Level 3

The radius of the circle is 20 20 and ∠ A O B = 3 10 π . \angle AOB=\frac{3}{10} \pi.

If C C is a point on B O ‾ \overline{BO} such that A C ‾ \overline{AC} is the bisector of ∠ B A O , \angle BAO, what is the length of B C ‾ ? \overline{BC}?

Round your answer to 3 decimal places.


The answer is 9.518.

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Figel Ilham by Figel Ilham , 22, Indonesia

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1 solution

It is given that ∠ A O B = 3 π 10 ⇒ ∠ O A B = π − 3 π 10 2 = 7 π 20 \angle AOB = \dfrac{3\pi}{10} \quad \Rightarrow \angle OAB = \dfrac{\pi-\frac{3\pi}{10}}{2} = \dfrac{7\pi}{20}

⇒ ∠ O A C = 7 π 40 ⇒ ∠ O C A = π − 3 π 10 − 7 π 40 = 21 π 40 \Rightarrow \angle OAC = \dfrac{7\pi}{40} \quad \Rightarrow \angle OCA = \pi - \dfrac{3\pi}{10} - \dfrac{7\pi}{40} = \dfrac{21\pi}{40} .

Using Sine rule, we have:

O C sin ⁡ ( 7 π 40 ) = 20 sin ⁡ ( 21 π 40 ) ⇒ O C = 20 ( sin ⁡ ( 7 π 40 ) ) sin ⁡ ( 21 π 40 ) = 10.482 \dfrac{OC}{\sin{\left(\frac{7\pi}{40}\right)}} = \dfrac{20}{\sin{\left( \frac {21\pi} {40}\right)}} \quad \Rightarrow OC = \dfrac{20\left( \sin{\left( \frac {7\pi} {40}\right)} \right)}{\sin{\left( \frac {21\pi} {40} \right)}} = 10.482

Therefore, B C = 20 − O C = 9.518 BC = 20 - OC = \boxed{9.518}

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upvoted! (y)

Figel Ilham - 6 years ago

@Chew-Seong Cheong - In the first line of the solution you have taken angle AOB to be 3pi/20 ,but the question says it is 3pi/10. Pls explain sir. I am confused

rajdeep das - 4 years, 10 months ago

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Thanks, sorry, it should be 3 π / 10 3 \pi /10 .

Chew-Seong Cheong - 4 years, 10 months ago

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