document.write( "<A HREF=/cgi-bin/jump-to-question.mpl?question=1167805>Question 1167805</A>: <I><SPAN STYLE=\"word-break: break-all;\"> The figure below shows two pulleys of radii 6cm and 4cm with centres A and B respectively. AB = 8cm. The pulleys are connected by a string PQXRSY<BR>\n" );
document.write( "https://www.easyelimu.com/images/topicalrevision/maths/form_3/circles_chords_and_tangents/circle_q5.PNG<BR>\n" );
document.write( "calculate<BR>\n" );
document.write( "(a) Length PQ<BR>\n" );
document.write( "(b) PAS reflex<BR>\n" );
document.write( "(c) Length of arc PYS and QXR<BR>\n" );
document.write( "(d) The total length of the string PQXRSY </SPAN>\n" );
document.write( "</I><HR><TABLE width=100%><TR><TD><B><A HREF=http://www.algebra.com/>Algebra.Com's</A> Answer #852088 by </B><A HREF=/tutors/aboutme.mpl?userid=CPhill><B>CPhill(1959)</B></A><img src=\"/images/stars/stars-12.gif\" alt=\"\" >&nbsp;<A HREF=/tutors/aboutme.mpl?userid=CPhill><IMG SRC=http://www.algebra.com/images/aboutme-small.gif BORDER=0 alt=\"About Me\" VALIGN=MIDDLE></A>&nbsp;<IMG SRC=http://www.algebra.com/cgi-bin/show-face.mpl?user=CPhill><BR>You can <A HREF=\"/cgi-bin/embed-solution.mpl?show=1&solution=852088\">put this solution on YOUR website!</A><BR><SPAN STYLE=\"word-break: break-all;\"> Here's a step-by-step calculation for each part of the problem based on the provided figure and common geometry principles for pulleys and tangents.\r<BR>\n" );
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document.write( "First, let's define the radii and distance between centers:<BR>\n" );
document.write( "* Radius of pulley A ($R_A$) = 6 cm<BR>\n" );
document.write( "* Radius of pulley B ($R_B$) = 4 cm<BR>\n" );
document.write( "* Distance between centers AB ($d$) = 8 cm\r<BR>\n" );
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document.write( "We will use a reference angle $\theta$ for calculations related to the tangent lines. This angle is formed by the line connecting the centers (AB) and the radius to the tangent point (e.g., AP).<BR>\n" );
document.write( "In the right-angled triangle formed by drawing a line from the center of the smaller pulley (B) perpendicular to the radius of the larger pulley (AP), let's call the intersection point L.<BR>\n" );
document.write( "The sides of this triangle are:<BR>\n" );
document.write( "* Hypotenuse = AB = 8 cm<BR>\n" );
document.write( "* Side adjacent to $\angle BAL$ = $AL = R_A - R_B = 6 - 4 = 2$ cm.<BR>\n" );
document.write( "Therefore, $\cos(\angle BAL) = \frac{AL}{AB} = \frac{2}{8} = \frac{1}{4}$.<BR>\n" );
document.write( "Let $\theta = \angle BAL = \arccos(1/4)$ radians.<BR>\n" );
document.write( "Numerically, $\theta \approx 1.3181$ radians (or $75.52^\circ$).\r<BR>\n" );
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document.write( "---\r<BR>\n" );
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document.write( "**(a) Calculate Length PQ**\r<BR>\n" );
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document.write( "The length of the common external tangent (PQ) can be found using the Pythagorean theorem:<BR>\n" );
document.write( "$PQ = \sqrt{d^2 - (R_A - R_B)^2}$<BR>\n" );
document.write( "$PQ = \sqrt{8^2 - (6 - 4)^2}$<BR>\n" );
document.write( "$PQ = \sqrt{64 - 2^2}$<BR>\n" );
document.write( "$PQ = \sqrt{64 - 4}$<BR>\n" );
document.write( "$PQ = \sqrt{60} = \sqrt{4 \times 15} = 2\sqrt{15}$ cm.\r<BR>\n" );
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document.write( "**Length PQ = $2\sqrt{15}$ cm**\r<BR>\n" );
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document.write( "---\r<BR>\n" );
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document.write( "**(b) PAS reflex**\r<BR>\n" );
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document.write( "The angle $\theta = \arccos(1/4)$ is the angle formed by the line of centers AB and the radius AP (or AS).<BR>\n" );
document.write( "Due to the symmetry of the tangents, the angle $\angle PAB = \theta$ and $\angle SAB = \theta$.<BR>\n" );
document.write( "So, the smaller angle $\angle PAS = 2\theta = 2\arccos(1/4)$ radians.<BR>\n" );
document.write( "The question asks for the **reflex** angle PAS. This is the larger angle around center A.<BR>\n" );
document.write( "Reflex $\angle PAS = 2\pi - 2\theta$ radians.<BR>\n" );
document.write( "In degrees: $360^\circ - (2 \times 75.52^\circ) = 360^\circ - 151.04^\circ = 208.96^\circ$.\r<BR>\n" );
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document.write( "**PAS reflex = $2\pi - 2\arccos(1/4)$ radians (or approximately $208.96^\circ$)**\r<BR>\n" );
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document.write( "---\r<BR>\n" );
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document.write( "**(c) Length of arc PYS and QXR**\r<BR>\n" );
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document.write( "* **Length of arc PYS (on pulley A):**<BR>\n" );
document.write( "    This arc is on the larger pulley with radius $R_A = 6$ cm. The central angle for this arc is the reflex angle PAS calculated in part (b).<BR>\n" );
document.write( "    Arc length $L_{PYS} = R_A \times (\text{Reflex } \angle PAS \text{ in radians})$<BR>\n" );
document.write( "    $L_{PYS} = 6 \times (2\pi - 2\arccos(1/4))$ cm.<BR>\n" );
document.write( "    $L_{PYS} = 12(\pi - \arccos(1/4))$ cm.\r<BR>\n" );
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document.write( "* **Length of arc QXR (on pulley B):**<BR>\n" );
document.write( "    This arc is on the smaller pulley with radius $R_B = 4$ cm. From the diagram, arc QXR is the minor arc.<BR>\n" );
document.write( "    The central angle for arc QXR is formed by radii BQ and BR. In the context of direct tangents, the angle for the minor arc is $\pi - 2\theta$, where $\theta = \arccos(1/4)$.<BR>\n" );
document.write( "    Arc length $L_{QXR} = R_B \times (\pi - 2\theta)$ radians.<BR>\n" );
document.write( "    $L_{QXR} = 4 \times (\pi - 2\arccos(1/4))$ cm.\r<BR>\n" );
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document.write( "**Length of arc PYS = $12(\pi - \arccos(1/4))$ cm**<BR>\n" );
document.write( "**Length of arc QXR = $4(\pi - 2\arccos(1/4))$ cm**\r<BR>\n" );
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document.write( "---\r<BR>\n" );
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document.write( "**(d) The total length of the string PQXRSY**\r<BR>\n" );
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document.write( "The total length of the string is the sum of the two straight tangent segments (PQ and RS) and the two arc lengths (PYS and QXR).<BR>\n" );
document.write( "Due to symmetry, $PQ = RS$.\r<BR>\n" );
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document.write( "Total Length = $2 \times PQ + L_{PYS} + L_{QXR}$<BR>\n" );
document.write( "Total Length = $2 \times (2\sqrt{15}) + 12(\pi - \arccos(1/4)) + 4(\pi - 2\arccos(1/4))$<BR>\n" );
document.write( "Total Length = $4\sqrt{15} + 12\pi - 12\arccos(1/4) + 4\pi - 8\arccos(1/4)$<BR>\n" );
document.write( "Total Length = $4\sqrt{15} + 16\pi - 20\arccos(1/4)$ cm.\r<BR>\n" );
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document.write( "**The total length of the string PQXRSY = $4\sqrt{15} + 16\pi - 20\arccos(1/4)$ cm**\r<BR>\n" );
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document.write( "---\r<BR>\n" );
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document.write( "**Numerical Approximations (for reference, if needed):**<BR>\n" );
document.write( "* $\sqrt{15} \approx 3.873$<BR>\n" );
document.write( "* $\pi \approx 3.1416$<BR>\n" );
document.write( "* $\arccos(1/4) \approx 1.3181$ radians\r<BR>\n" );
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document.write( "* Length PQ $\approx 2 \times 3.873 = 7.746$ cm.<BR>\n" );
document.write( "* Reflex PAS $\approx 2\pi - 2(1.3181) = 6.2832 - 2.6362 = 3.6470$ radians.<BR>\n" );
document.write( "* $L_{PYS} \approx 6 \times 3.6470 = 21.882$ cm.<BR>\n" );
document.write( "* $L_{QXR} \approx 4 \times (\pi - 2 \times 1.3181) = 4 \times (3.1416 - 2.6362) = 4 \times 0.5054 = 2.022$ cm.<BR>\n" );
document.write( "* Total Length $\approx (2 \times 7.746) + 21.882 + 2.022 = 15.492 + 21.882 + 2.022 = 39.396$ cm. </SPAN>\n" );
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