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Nov 2002 p1 q6
565
In the diagram, triangle ABC is right-angled and D is the mid-point of BC. Angle DAC is \(30^\circ\) and angle BAD is \(x^\circ\). Denoting the length of AD by l,
express each of AC and BC exactly in terms of l, and show that \(AB = \tfrac{1}{2}l\sqrt{7}\),
show that \(x = \tan^{-1}\!\left(\frac{2}{\sqrt{3}}\right) - 30^\circ\).
Solution
(i) To find AC, use the cosine of angle DAC: \(AC = l\cos 30^\circ = \frac{l\sqrt{3}}{2}\).
To find BC, use the sine of angle DAC: \(BC = 2l\sin 30^\circ = l\).
To find AB, use the Pythagorean theorem in triangle ABD:
\(AB = \sqrt{AD^2 + BD^2} = \sqrt{l^2 + \left(\frac{l}{2}\right)^2} = \frac{1}{2}l\sqrt{7}\).
(ii) Use the tangent of angle \((x + 30^\circ)\) in triangle ABC:
\(\tan(x + 30^\circ) = \frac{BC}{AC} = \frac{l}{\frac{l\sqrt{3}}{2}} = \frac{2}{\sqrt{3}}\).