A geometry problem by Arpit Sah

Geometry Level pending

AB is a diameter of a circle of radius 10 units. C is a point on the circle such that length of arc BC = \(\frac{20πœ‹}{3}\) units. The angle bisector of angle ACB cuts the circle at D. Find the length of CD.

5 3 \sqrt{3} ( 2 \sqrt{2} +1) 10 2 \sqrt{2} + 10 3 \sqrt{3} 5 2 \sqrt{2} ( 3 \sqrt{3} +1) 5 3 \sqrt{3} + 5

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1 solution

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Let ∠ B O C = θ \angle BOC = \theta

r ΞΈ = 20 Ο€ 3 ⟢ 10 ΞΈ = 20 Ο€ 3 ⟢ ΞΈ = 2 Ο€ 3 = 12 0 ∘ r\theta = \frac{20\pi}{3} \longrightarrow 10\theta = \frac{20\pi}{3} \longrightarrow \theta = \frac{2\pi}{3} = 120^\circ

∠ A O C = ∠ A C O = ∠ C A O = 6 0 ∘ \angle AOC = \angle ACO = \angle CAO = 60^\circ

∠ A C D = ∠ B C D = 4 5 ∘ \angle ACD = \angle BCD = 45^\circ

So, ∠ O C D = 1 5 ∘ \angle OCD = 15^\circ and ∠ C O D = 15 0 ∘ \angle COD = 150^\circ

cos ⁑ 2 1 5 ∘ = 1 + cos ⁑ 3 0 ∘ 2 \cos^2 15^\circ = \frac{1+\cos 30^\circ}{2}

cos ⁑ 2 1 5 ∘ = 1 + 3 2 2 \cos^2 15^\circ = \frac{1+\frac{\sqrt{3}}{2}}{2}

cos ⁑ 2 1 5 ∘ = 2 + 3 4 \cos^2 15^\circ = \frac{2+\sqrt{3}}{4}

c o s 2 1 5 ∘ = 8 + 4 3 16 cos^2 15^\circ = \frac{8+4\sqrt{3}}{16}

c o s 2 1 5 ∘ = 6 + 2 12 + 2 16 cos^2 15^\circ = \frac{6+2\sqrt{12} + 2}{16}

c o s 2 1 5 ∘ = ( 6 + 2 ) 2 16 cos^2 15^\circ = \frac{(\sqrt{6} + \sqrt{2})^2 }{16}

c o s 1 5 ∘ = 6 + 2 4 cos 15^\circ = \frac{\sqrt{6} + \sqrt{2} }{4}

Now Using Sine Law in β–³ C O D \triangle COD ,

C D sin ⁑ 15 0 ∘ = C O sin ⁑ 1 5 ∘ \frac{CD}{\sin 150^\circ} = \frac{CO}{\sin 15^\circ}

C D = sin ⁑ 3 0 ∘ Γ— 10 sin ⁑ 1 5 ∘ CD = \frac{\sin 30^\circ \times 10}{\sin 15^\circ}

C D = 2 sin ⁑ 1 5 ∘ cos ⁑ 1 5 ∘ Γ— 10 sin ⁑ 1 5 ∘ CD = \frac{2\sin 15^\circ \cos 15^\circ \times 10}{\sin 15^\circ}

C D = 2 cos ⁑ 1 5 ∘ Γ— 10 CD = 2 \cos 15^\circ \times 10

C D = 5 2 ( 3 + 1 ) CD = 5\sqrt{2} (\sqrt{3}+1)

Your question is contradictory. How can the length of the chord CD (20Ο€/3) be greater than the diameter AB(20)?? In the above solution, the length of chord CD is assumed to be equal to the length of arc CD without any reference in the question. Please don't try to waste other's time by posting such question.

Sanjeet Raria - 7Β years ago

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