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June 2016 p12 q5
561
In the diagram, triangle ABC is right-angled at C and M is the mid-point of BC. It is given that angle ABC = \(\frac{1}{3} \pi\) radians and angle BAM = \(\theta\) radians. Denoting the lengths of BM and MC by x,
find AM in terms of x,
show that \(\theta = \frac{1}{6} \pi - \tan^{-1} \left( \frac{1}{2 \sqrt{3}} \right)\).
Solution
(i) Using the tangent ratio in triangle ABC, \(\tan \left( \frac{\pi}{3} \right) = \frac{AC}{2x}\). Therefore, \(AC = 2x \sqrt{3}\). Using the cosine ratio, \(\cos \left( \frac{\pi}{3} \right) = \frac{2x}{AB}\), we find \(AB = 4x\). Using the Pythagorean theorem in triangle ABM, \(AM = \sqrt{AB^2 - BM^2} = \sqrt{(4x)^2 - x^2} = \sqrt{16x^2 - x^2} = \sqrt{15x^2} = \sqrt{15}x\).
(ii) In triangle AMC, \(\tan(\angle MAC) = \frac{x}{AC} = \frac{x}{2x \sqrt{3}} = \frac{1}{2 \sqrt{3}}\). Therefore, \(\angle MAC = \tan^{-1} \left( \frac{1}{2 \sqrt{3}} \right)\). Since \(\angle BAM = \theta\) and \(\angle BAC = \frac{\pi}{6}\), we have \(\theta = \angle BAC - \angle MAC = \frac{\pi}{6} - \tan^{-1} \left( \frac{1}{2 \sqrt{3}} \right)\).
