In the diagram, ASRTB represents a piece of string passing over a pulley of radius 10cm in a vertical plane. O is the centre of the pulley and AMB is a hori...
Question 1 Report
In the diagram, ASRTB represents a piece of string passing over a pulley of radius 10cm in a vertical plane. O is the centre of the pulley and AMB is a horizontal straight line touching the pulley at M. Angle SAB = 90° and angle TBA = 60°.
(a) Calculate (i) the obtuse angle SOT ; (ii) arc SRT ; (iii) |BT|
(b) Find, correct to the nearest cm, the length of the string. (Take \(\pi = \frac{22}{7}\)).
Answer Details
From the diagram, the pulley has centre O and radius \(r = 10\text{ cm}\). The string runs \(A \to S\) (tangent on the left), over the arc \(S \to R \to T\) at the top, then \(T \to B\) (tangent on the right). \(AMB\) is horizontal, tangent to the pulley at M, with \(\angle SAB = 90^\circ\) and \(\angle TBA = 60^\circ\).
(a)(i) Obtuse angle SOT
From A, the tangents \(AS\) and \(AM\) meet the radii at right angles, so in quadrilateral \(ASOM\):
\[\angle SOM = 360^\circ - 90^\circ - 90^\circ - \angle SAM = 360^\circ - 90^\circ - 90^\circ - 90^\circ = 90^\circ\]From B, the tangents \(BT\) and \(BM\) give, in quadrilateral \(BTOM\):
\[\angle TOM = 360^\circ - 90^\circ - 90^\circ - \angle TBM = 360^\circ - 90^\circ - 90^\circ - 60^\circ = 120^\circ\]Going from S round the bottom through M to T gives \(90^\circ + 120^\circ = 210^\circ\). The angle on the top (through R), the obtuse \(\angle SOT\), is
\[\angle SOT = 360^\circ - 210^\circ = 150^\circ\](a)(ii) Arc SRT
\[\text{arc }SRT = \frac{150}{360}\times 2\pi r = \frac{150}{360}\times 2\times\frac{22}{7}\times 10 = \frac{5}{12}\times\frac{440}{7}\]\[= \frac{2200}{84} = 26.19\text{ cm} \approx 26\text{ cm}\](a)(iii) |BT|
In right triangle \(OTB\), \(OT = 10\text{ cm}\), \(\angle OTB = 90^\circ\), and \(OB\) bisects \(\angle TBM\) so \(\angle TBO = 30^\circ\):
\[\tan 30^\circ = \frac{OT}{BT} \;\Rightarrow\; BT = \frac{10}{\tan 30^\circ} = 10\sqrt{3} = 17.32\text{ cm}\](b) Length of the string
First find \(AS\). In right triangle \(OSA\), \(OS = 10\text{ cm}\), \(\angle OSA = 90^\circ\), and \(OA\) bisects \(\angle SAM\) so \(\angle SAO = 45^\circ\):
\[\tan 45^\circ = \frac{OS}{AS} \;\Rightarrow\; AS = \frac{10}{\tan 45^\circ} = 10\text{ cm}\]The string length is
\[AS + \text{arc }SRT + BT = 10 + 26.19 + 17.32 = 53.51\text{ cm}\]Length of the string \(\approx 54\text{ cm}\).
Summary: \(\angle SOT = 150^\circ\), arc \(SRT \approx 26\text{ cm}\), \(|BT| = 10\sqrt{3}\approx 17.32\text{ cm}\), string \(\approx 54\text{ cm}\).
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