A geometry problem by Vivek Vijayan

Geometry Level 2

Two chords AB and CD of circle with center O, meet at the point P such that ∠ A O C = 5 0 o , ∠ B O D = 4 0 o . \angle AOC = 50^o, \angle BOD = 40^o. What is the acute angle between lines A B AB and C D CD ?

7 5 o 75^o 4 5 o 45^o 6 0 o 60^o 4 0 o 40^o

Vivek Vijayan by Vivek Vijayan , 34, India

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2 solutions

Vivek Vijayan
Jul 17, 2014

Draw the side BC, now, ∠ B O D = 4 0 o g i v e s ∠ B C D = 2 0 o \angle BOD =40^o gives \angle BCD = 20^o and ∠ A O C = 5 0 o g i v e s ∠ A B C = 2 5 o \angle AOC = 50^o gives \angle ABC = 25^o

This gives ∠ B P C = 180 − ( 20 + 25 ) = 13 5 o \angle BPC = 180-(20+25) = 135^o Hence the acute angle between the chords is ( 360 − ( 135 × 2 ) ) ÷ 2 = 4 5 o (360-(135 \times 2))÷2 = 45^o

3 Helpful 0 Interesting 0 Brilliant 0 Confused

You can just deduct 135 degrees from 180 degrees since it's a supplementary angle.

Astro Enthusiast - 6 years, 11 months ago

By the inscribed angle theorem, we have

∠ D C B = 1 2 ( 40 ) = 2 0 ∘ \angle DCB=\dfrac{1}{2}(40)=20^\circ

∠ A B C = 1 2 ( 50 ) = 2 5 ∘ \angle ABC=\dfrac{1}{2}(50)=25^\circ

By the exterior angle theorem, we have

x = 20 + 25 = x=20+25= 45 \boxed{45}

1 Helpful 0 Interesting 1 Brilliant 0 Confused

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