Problem 30 · 2018 Math Kangaroo
Stretch
Geometry & Measurement
pythagorean-triple
Two chords AB and AC are drawn in a circle with diameter AD. \(\angle BAC = 60^\circ\), \(AB = 24\) cm, point E lies on AC with \(EC = 3\) cm, and BE is perpendicular to AC. How long is the chord BD?
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Answer: D — \(2\sqrt{3}\) cm
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Hint 1 of 2
Drop into right triangle ABE to get the two legs \(AE\) and \(BE\) from the \(60^\circ\) angle.
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Hint 2 of 2
AD is a diameter, so \(\angle ABD = 90^\circ\); also the inscribed angle \(\angle ADB\) equals \(\angle ACB\) because both stand on arc \(AB\).
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Approach: find angle ACB from the right triangle, then use the right angle at B in triangle ABD
- In right triangle BEC (right angle at E): \(BE = 24\sin 60^\circ = 12\sqrt3\) and \(EC = 3\), so \(\tan(\angle ACB) = \dfrac{BE}{EC} = \dfrac{12\sqrt3}{3} = 4\sqrt3\).
- Because AD is a diameter, \(\angle ABD = 90^\circ\); and \(\angle ADB = \angle ACB\) since both inscribed angles stand on the same arc \(AB\).
- In right triangle ABD then \(BD = \dfrac{AB}{\tan(\angle ADB)} = \dfrac{24}{4\sqrt3} = \dfrac{6}{\sqrt3} = 2\sqrt3\).
- So \(BD = \) \(2\sqrt{3}\) cm.
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