Comments on: March 14 1.59pm http://www.exampaper.com.sg/questions/e-maths/march-14-159pm Sassy O Level Maths Tuition, Questions & Tips from Singapore's Favourite Private Tutor Sun, 05 Feb 2012 17:08:39 +0000 hourly 1 By: Miss Loi http://www.exampaper.com.sg/questions/e-maths/march-14-159pm#comment-13598 Miss Loi Thu, 18 Sep 2008 17:34:18 +0000 http://www.exampaper.com.sg/questions/e-maths/march-14-159pm#comment-13598 Thanks to <b>WS</b>, may our cutie π girl finally find her true love on March 14 1.59pm <strong class="highlight">2009</strong>. *Wonders where would <b>WS</b> be on that day* :P Thanks to WS, may our cutie π girl finally find her true love on March 14 1.59pm 2009.

*Wonders where would WS be on that day* :P

]]> By: WS http://www.exampaper.com.sg/questions/e-maths/march-14-159pm#comment-13501 WS Tue, 16 Sep 2008 17:01:21 +0000 http://www.exampaper.com.sg/questions/e-maths/march-14-159pm#comment-13501 <div class="highlight"> Yes to find the shaded area, your approach is summarized as follow: Area of Sector <var>OPQ</var> (= 3.6<var>π</var> cm<sup>2</sup> obtained in 1ii. above) - Area of Sector <var>USV</var> - Area of Δ <var>OSU</var> - Area of Δ <var>OSV</var> </div> 2. By symmetry, angle OSV = angle OSU = 0.8π rad Angle USV = 2π(angles at a point) - (2 x 0.8π) = 2π - 1.6π = 0.4π rad <span class="highlight">(which is obtained directly if one has used the 2nd method of calculating ∠<var>OSU</var> above - but follow whichever method you're comfortable with)</span> Area of sector USV = ½ (SU<sup>2</sup>)(0.4π) = ½X9X0.4π =1.8π cm<sup>2</sup> Again by symmetry, area of triangle OSU = area of triangle OSV = ½(OSxSU)xsin0.8π <span class="highlight">(using the ½<var>AB</var>sin<var>θ</var> area of Δ formula you've learned in your Sine Rule chapter)</span> = ½x9xsin0.8π = 2.645cm<sup>2</sup> (4 sig fig, from calculator) So shaded area PRQVU = sectorOPQ - sectorUSV - trianglesOSU&OSV = 3.6π - 1.8π - (2.645x2) = 1.8π - 5.29 = 0.365cm<sup>2</sup> <span class="highlight">(YEAH!)</span> (using calculator π, rounding off ans to 3 sig fig) It's long past 2pm... <span class="highlight">The cutie π girl is very touched by you burning the midnight oil for her! <3<3<3</span>

Yes to find the shaded area, your approach is summarized as follow:

Area of Sector OPQ (= 3.6π cm2 obtained in 1ii. above)
- Area of Sector USV
- Area of Δ OSU
- Area of Δ OSV

2. By symmetry, angle OSV = angle OSU = 0.8π rad
Angle USV = 2π(angles at a point) - (2 x 0.8π)
= 2π - 1.6π = 0.4π rad (which is obtained directly if one has used the 2nd method of calculating ∠OSU above - but follow whichever method you're comfortable with)

Area of sector USV = ½ (SU2)(0.4π)
= ½X9X0.4π =1.8π cm2

Again by symmetry, area of triangle OSU
= area of triangle OSV
= ½(OSxSU)xsin0.8π
(using the ½ABsinθ area of Δ formula you've learned in your Sine Rule chapter)

= ½x9xsin0.8π = 2.645cm2
(4 sig fig, from calculator)

So shaded area PRQVU = sectorOPQ - sectorUSV
- trianglesOSU&OSV
= 3.6π - 1.8π - (2.645x2)
= 1.8π - 5.29 = 0.365cm2 (YEAH!)
(using calculator π, rounding off ans to 3 sig fig)

It's long past 2pm... The cutie π girl is very touched by you burning the midnight oil for her! <3<3<3

]]> By: WS http://www.exampaper.com.sg/questions/e-maths/march-14-159pm#comment-13496 WS Tue, 16 Sep 2008 16:11:46 +0000 http://www.exampaper.com.sg/questions/e-maths/march-14-159pm#comment-13496 Full correction to 1iii. SU = SR (radii, centre S) = OS (S midpt of radius OR) = 6/2 = 3cm Since SU = OS, triangle OSU is isosceles and thus angle SUO = angle SOU = angle POQ/2 = 0.2π/2 = 0.1π rad Angle OSU = π(sum of angles in triangle) - (0.1πx2) = π - 0.2π = 0.8π rad So tedious... <div class="highlight"> <i>Yalor</i> very tedious <i>hor</i>? If one looks hard enough at the diagram, since <var>US</var> = <var>OS</var> = <var>VS</var>, a circle with center <var>S</var> will magically materialize: <img src="http://www.exampaper.com.sg/blog/wp-content/uploads/2008/09/circular-measure-pi-day-ans-2.gif" alt="Circular Measure Diagram Answer 2" /> And using one of the <em>Angle Properties of Circles</em> (which surfaced in the <a href="/questions/e-maths/locus-cave-maths" rel="nofollow">little link</a> Miss Loi included in the main article) where: <div class="attention"> <strong class="highlight">∠ at center = 2 × ∠ at circumference subtended by the same arc</strong> </div> So we get ∠<var>USV</var> = 2 × ∠<var>UOV</var> = 0.4<var>π</var> rad ∴ ∠<var>OSU</var> = <var>π</var> - (∠<var>USV</var>)/2 <span class="fineprint">(since it's given <var>OR</var> bisects ∠<var>POQ</var>)</span> = 0.8<var>π</var> rad ;) </div> Full correction to 1iii.

SU = SR (radii, centre S)
= OS (S midpt of radius OR) = 6/2 = 3cm
Since SU = OS, triangle OSU is isosceles and thus
angle SUO = angle SOU = angle POQ/2
= 0.2π/2 = 0.1π rad
Angle OSU = π(sum of angles in triangle) - (0.1πx2)
= π - 0.2π = 0.8π rad

So tedious...

Yalor very tedious hor?

If one looks hard enough at the diagram, since US = OS = VS, a circle with center S will magically materialize:

Circular Measure Diagram Answer 2

And using one of the Angle Properties of Circles (which surfaced in the little link Miss Loi included in the main article) where:

∠ at center = 2 × ∠ at circumference subtended by the same arc

So we get ∠USV = 2 × ∠UOV = 0.4π rad
∴ ∠OSU = π - (∠USV)/2 (since it's given OR bisects ∠POQ) = 0.8π rad ;)

]]> By: WS http://www.exampaper.com.sg/questions/e-maths/march-14-159pm#comment-13495 WS Tue, 16 Sep 2008 15:10:12 +0000 http://www.exampaper.com.sg/questions/e-maths/march-14-159pm#comment-13495 Hm..left unsolved for half yr... <div class="highlight"> * Our cutie π schoolgirl bemoans half a year of wasted youth :'( * It's always a good first thing to include all values in the question in the diagram, as you may be able to see things that sometimes others can't see *hair stands on neck* <img src="http://www.exampaper.com.sg/blog/wp-content/uploads/2008/09/circular-measure-pi-day-ans-1.gif" alt="Circular Measure Answer Diagram 1" /> </div> 1.i. Angle POQ = 36π/180 = 0.2π rad <span class="highlight">(Yup - just wish to reiterate once more that 180° = <var>π</var> radians NOT 2<var>π</var>!)</span> Arc PRQ = PO x 0.2π = 6 x 0.2π = 1.2π cm <span class="highlight">(Yup - simple substitution into arc length formula <var>s</var> = <var>rθ</var>, with radius <var>r</var> = 6 cm and <var>θ</var> = 0.2π)</span> ii.Area of sector OPQ = ½(6^2)(0.2π) = 3.6π cm<sup>2</sup> <span class="highlight">(simple substitution into sector area formula formula <m>A = 1/2 r^2 theta</m>, with radius <var>r</var> = 6 cm and <var>θ</var> = 0.2π)</span> <div class="highlight"> No sweat so far right? Remember again that all the <var>θ</var> in your arc and sector area formulae must be in <em>radians</em> i.e. in terms of <var>π</var> and/or switch your calculators into <span class="calc-button">RAD</span> mode. </div> iii.Angle SOU = Angle SOV = 0.2π/2 = 0.1π rad Angle OSU = π (ie 180 °) - (2x0.1π) = π - 0.2π = 0.8π rad Part 2 coming up later... <div class="highlight"> <i>Yalor</i> as seen from the diagram and described in part 2 of your workings below, since: <ol> <li><var>OS</var> = 3cm (as <var>S</var> is the mid-point of <var>OR</var> which is 6cm</li> <li><var>US</var> = 3cm (since it's given <var>S</var> is the center of arc <var>URV</var> with radius 3cm)</li> </ol> So <var>OS</var> = <var>US</var> ⇒ Δ <var>USO</var> is isosceles ⇒ ∠ <var>SUO</var> = ∠ <var>SOU</var> = 18° ⇒ ∠ <var>OSU</var> = 180° - 18° - 18° = 144° = 0.8<var>π</var> rad See your <a href="/questions/e-maths/march-14-159pm#comment-13496" rel="nofollow">next comment below</a> for an alternate approach ... </div> Hm..left unsolved for half yr...

* Our cutie π schoolgirl bemoans half a year of wasted youth :'( *

It's always a good first thing to include all values in the question in the diagram, as you may be able to see things that sometimes others can't see *hair stands on neck*

Circular Measure Answer Diagram 1

1.i.
Angle POQ = 36π/180 = 0.2π rad
(Yup - just wish to reiterate once more that 180° = π radians NOT 2π!)
Arc PRQ = PO x 0.2π = 6 x 0.2π = 1.2π cm
(Yup - simple substitution into arc length formula s = rθ, with radius r = 6 cm and θ = 0.2π)

ii.Area of sector OPQ = ½(6^2)(0.2π) = 3.6π cm2
(simple substitution into sector area formula formula A = 1/2 r^2 theta, with radius r = 6 cm and θ = 0.2π)

No sweat so far right? Remember again that all the θ in your arc and sector area formulae must be in radians i.e. in terms of π and/or switch your calculators into RAD mode.

iii.Angle SOU = Angle SOV = 0.2π/2 = 0.1π rad
Angle OSU = π (ie 180 °) - (2x0.1π) = π - 0.2π
= 0.8π rad

Part 2 coming up later...

Yalor as seen from the diagram and described in part 2 of your workings below, since:

  1. OS = 3cm (as S is the mid-point of OR which is 6cm
  2. US = 3cm (since it's given S is the center of arc URV with radius 3cm)

So OS = US
⇒ Δ USO is isosceles
⇒ ∠ SUO = ∠ SOU = 18°
⇒ ∠ OSU = 180° - 18° - 18° = 144° = 0.8π rad

See your next comment below for an alternate approach ...

]]> By: Miss Loi http://www.exampaper.com.sg/questions/e-maths/march-14-159pm#comment-5402 Miss Loi Mon, 17 Mar 2008 08:19:52 +0000 http://www.exampaper.com.sg/questions/e-maths/march-14-159pm#comment-5402 Easy there <b>HAM</b> ... easy ... it's just an innocent <strong class="big">π</strong>! Easy there HAM ... easy ... it's just an innocent π!

]]> By: HORNY ANG MOH http://www.exampaper.com.sg/questions/e-maths/march-14-159pm#comment-5399 HORNY ANG MOH Mon, 17 Mar 2008 06:22:49 +0000 http://www.exampaper.com.sg/questions/e-maths/march-14-159pm#comment-5399 Ehhhhh!!!! What notty commrnt arrrr??? This Pie tattoo got notty spot meh??? Where arrrr??? Ehhhhh!!!! What notty commrnt arrrr??? This Pie tattoo got notty spot meh??? Where arrrr???

]]> By: Miss Loi http://www.exampaper.com.sg/questions/e-maths/march-14-159pm#comment-5387 Miss Loi Mon, 17 Mar 2008 01:36:06 +0000 http://www.exampaper.com.sg/questions/e-maths/march-14-159pm#comment-5387 <b>HighwayBlogger</b>, if only it was THIS easy ... tattoo shops' business will skyrocket the week before every March 14 (just like the florists on Feb 14), <del>and guys will be sitting around everywhere waiting for the ... ummm ... 'poke'</del> *sorry that doesn't sound right* And SDU would've gone extinct much earlier! HighwayBlogger, if only it was THIS easy ... tattoo shops' business will skyrocket the week before every March 14 (just like the florists on Feb 14), and guys will be sitting around everywhere waiting for the ... ummm ... 'poke' *sorry that doesn't sound right*

And SDU would've gone extinct much earlier!

]]> By: HighwayBlogger http://www.exampaper.com.sg/questions/e-maths/march-14-159pm#comment-5383 HighwayBlogger Sun, 16 Mar 2008 19:49:34 +0000 http://www.exampaper.com.sg/questions/e-maths/march-14-159pm#comment-5383 Liddat also can! !! So was this your love story? Liddat also can! !! So was this your love story?

]]> By: Miss Loi http://www.exampaper.com.sg/questions/e-maths/march-14-159pm#comment-5367 Miss Loi Sun, 16 Mar 2008 04:06:02 +0000 http://www.exampaper.com.sg/questions/e-maths/march-14-159pm#comment-5367 Aiyoh <b>Soup</b> what kind of indices expression is that? *blushes* Aiyoh Soup what kind of indices expression is that? *blushes*

]]> By: ignorantsoup http://www.exampaper.com.sg/questions/e-maths/march-14-159pm#comment-5366 ignorantsoup Sun, 16 Mar 2008 03:58:20 +0000 http://www.exampaper.com.sg/questions/e-maths/march-14-159pm#comment-5366 Whoa really got Pi day ah..lol..n nice use of the link to the leap day tradition... [pmath]M^(iss).L^(oi).S^(o).C^(lever)[/pmath] Whoa really got Pi day ah..lol..n nice use of the link to the leap day tradition...
M^(iss).L^(oi).S^(o).C^(lever)

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