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Question Number 16682    Answers: 1   Comments: 0

Question Number 16681    Answers: 1   Comments: 0

If tan x+tan ((π/3)+x)+tan (((2π)/3)+x)=3 a) tan x=1 b) tan 2x=1 c) tan 3x=1 d) None of above

$$\mathrm{If} \\ $$$$\mathrm{tan}\:{x}+\mathrm{tan}\:\left(\frac{\pi}{\mathrm{3}}+{x}\right)+\mathrm{tan}\:\left(\frac{\mathrm{2}\pi}{\mathrm{3}}+{x}\right)=\mathrm{3} \\ $$$$\left.{a}\right)\:\mathrm{tan}\:{x}=\mathrm{1} \\ $$$$\left.{b}\right)\:\mathrm{tan}\:\mathrm{2}{x}=\mathrm{1} \\ $$$$\left.{c}\right)\:\mathrm{tan}\:\mathrm{3}{x}=\mathrm{1} \\ $$$$\left.{d}\right)\:\mathrm{None}\:\mathrm{of}\:\mathrm{above} \\ $$

Question Number 16675    Answers: 2   Comments: 0

The number of intersecting points on the graph for sin x = (x/(10)) for x ∈ [−π, π] is

$$\mathrm{The}\:\mathrm{number}\:\mathrm{of}\:\mathrm{intersecting}\:\mathrm{points}\:\mathrm{on} \\ $$$$\mathrm{the}\:\mathrm{graph}\:\mathrm{for}\:\mathrm{sin}\:{x}\:=\:\frac{{x}}{\mathrm{10}}\:\mathrm{for}\:{x}\:\in\:\left[−\pi,\:\pi\right] \\ $$$$\mathrm{is} \\ $$

Question Number 16699    Answers: 0   Comments: 8

Related to Q16675: Find the number of intersection points of graph sin x=(x/(10)). Let′s see sin x = (x/n) with n>1. For n≤1 there is one intersection point. Let x=2kπ+t with k∈N ∧ t∈[0,2π] sin x=sin t cos x=cos t we find the point on f(x)=sin x where its tangent is g(x)=(x/n). f′(x)=cos x=cos t g′(x)=(1/n) cos t=(1/n) t=cos^(−1) (1/n) sin t=(n/(√(n^2 +1))) so that f(x) intersects with g(x), ((sin x)/x)≥(1/n) ⇒n sin x≥x ⇒n sin t≥2kπ+t ⇒k≤((n sin t −t)/(2π))=(((n^2 /(√(n^2 +1)))−cos^(−1) (1/n))/(2π)) k_(max) =⌊(((n^2 /(√(n^2 +1)))−cos^(−1) (1/n))/(2π))⌋ number of intersecting points is m=2×2(k_(max) +1)−1=4k_(max) +3 for n=10 k_(max) =⌊(((n^2 /(√(n^2 +1)))−cos^(−1) (1/n))/(2π))⌋ =⌊((((10^2 )/(√(10^2 +1)))−cos^(−1) (1/(10)))/(2π))⌋=⌊1.35⌋=1 ⇒m=4×1+3=7 for n=20 k_(max) =⌊((((20^2 )/(√(20^2 +1)))−cos^(−1) (1/(20)))/(2π))⌋=⌊2.94⌋=2 ⇒m=4×2+3=11

$$\mathrm{Related}\:\mathrm{to}\:\mathrm{Q16675}: \\ $$$$ \\ $$$$\mathrm{Find}\:\mathrm{the}\:\mathrm{number}\:\mathrm{of}\:\mathrm{intersection}\:\mathrm{points} \\ $$$$\mathrm{of}\:\mathrm{graph}\:\mathrm{sin}\:\mathrm{x}=\frac{\mathrm{x}}{\mathrm{10}}. \\ $$$$ \\ $$$$\mathrm{Let}'\mathrm{s}\:\mathrm{see}\:\mathrm{sin}\:\mathrm{x}\:=\:\frac{\mathrm{x}}{\mathrm{n}}\:\mathrm{with}\:\mathrm{n}>\mathrm{1}. \\ $$$$\mathrm{For}\:\mathrm{n}\leqslant\mathrm{1}\:\mathrm{there}\:\mathrm{is}\:\mathrm{one}\:\mathrm{intersection}\:\mathrm{point}. \\ $$$$ \\ $$$$\mathrm{Let}\:\mathrm{x}=\mathrm{2k}\pi+\mathrm{t}\:\mathrm{with}\:\mathrm{k}\in\mathbb{N}\:\wedge\:\mathrm{t}\in\left[\mathrm{0},\mathrm{2}\pi\right] \\ $$$$\mathrm{sin}\:\mathrm{x}=\mathrm{sin}\:\mathrm{t} \\ $$$$\mathrm{cos}\:\mathrm{x}=\mathrm{cos}\:\mathrm{t} \\ $$$$ \\ $$$$\mathrm{we}\:\mathrm{find}\:\mathrm{the}\:\mathrm{point}\:\mathrm{on}\:\mathrm{f}\left(\mathrm{x}\right)=\mathrm{sin}\:\mathrm{x}\:\mathrm{where}\:\mathrm{its} \\ $$$$\mathrm{tangent}\:\mathrm{is}\:\mathrm{g}\left(\mathrm{x}\right)=\frac{\mathrm{x}}{\mathrm{n}}. \\ $$$$\mathrm{f}'\left(\mathrm{x}\right)=\mathrm{cos}\:\mathrm{x}=\mathrm{cos}\:\mathrm{t} \\ $$$$\mathrm{g}'\left(\mathrm{x}\right)=\frac{\mathrm{1}}{\mathrm{n}} \\ $$$$\mathrm{cos}\:\mathrm{t}=\frac{\mathrm{1}}{\mathrm{n}} \\ $$$$\mathrm{t}=\mathrm{cos}^{−\mathrm{1}} \frac{\mathrm{1}}{\mathrm{n}} \\ $$$$\mathrm{sin}\:\mathrm{t}=\frac{\mathrm{n}}{\sqrt{\mathrm{n}^{\mathrm{2}} +\mathrm{1}}} \\ $$$$ \\ $$$$\mathrm{so}\:\mathrm{that}\:\mathrm{f}\left(\mathrm{x}\right)\:\mathrm{intersects}\:\mathrm{with}\:\mathrm{g}\left(\mathrm{x}\right), \\ $$$$\frac{\mathrm{sin}\:\mathrm{x}}{\mathrm{x}}\geqslant\frac{\mathrm{1}}{\mathrm{n}} \\ $$$$\Rightarrow\mathrm{n}\:\mathrm{sin}\:\mathrm{x}\geqslant\mathrm{x} \\ $$$$\Rightarrow\mathrm{n}\:\mathrm{sin}\:\mathrm{t}\geqslant\mathrm{2k}\pi+\mathrm{t} \\ $$$$\Rightarrow\mathrm{k}\leqslant\frac{\mathrm{n}\:\mathrm{sin}\:\mathrm{t}\:−\mathrm{t}}{\mathrm{2}\pi}=\frac{\frac{\mathrm{n}^{\mathrm{2}} }{\sqrt{\mathrm{n}^{\mathrm{2}} +\mathrm{1}}}−\mathrm{cos}^{−\mathrm{1}} \frac{\mathrm{1}}{\mathrm{n}}}{\mathrm{2}\pi} \\ $$$$\mathrm{k}_{\mathrm{max}} =\lfloor\frac{\frac{\mathrm{n}^{\mathrm{2}} }{\sqrt{\mathrm{n}^{\mathrm{2}} +\mathrm{1}}}−\mathrm{cos}^{−\mathrm{1}} \frac{\mathrm{1}}{\mathrm{n}}}{\mathrm{2}\pi}\rfloor \\ $$$$ \\ $$$$\mathrm{number}\:\mathrm{of}\:\mathrm{intersecting}\:\mathrm{points}\:\mathrm{is} \\ $$$$\mathrm{m}=\mathrm{2}×\mathrm{2}\left(\mathrm{k}_{\mathrm{max}} +\mathrm{1}\right)−\mathrm{1}=\mathrm{4k}_{\mathrm{max}} +\mathrm{3} \\ $$$$ \\ $$$$\mathrm{for}\:\mathrm{n}=\mathrm{10} \\ $$$$\mathrm{k}_{\mathrm{max}} =\lfloor\frac{\frac{\mathrm{n}^{\mathrm{2}} }{\sqrt{\mathrm{n}^{\mathrm{2}} +\mathrm{1}}}−\mathrm{cos}^{−\mathrm{1}} \frac{\mathrm{1}}{\mathrm{n}}}{\mathrm{2}\pi}\rfloor \\ $$$$=\lfloor\frac{\frac{\mathrm{10}^{\mathrm{2}} }{\sqrt{\mathrm{10}^{\mathrm{2}} +\mathrm{1}}}−\mathrm{cos}^{−\mathrm{1}} \frac{\mathrm{1}}{\mathrm{10}}}{\mathrm{2}\pi}\rfloor=\lfloor\mathrm{1}.\mathrm{35}\rfloor=\mathrm{1} \\ $$$$\Rightarrow\mathrm{m}=\mathrm{4}×\mathrm{1}+\mathrm{3}=\mathrm{7} \\ $$$$ \\ $$$$\mathrm{for}\:\mathrm{n}=\mathrm{20} \\ $$$$\mathrm{k}_{\mathrm{max}} =\lfloor\frac{\frac{\mathrm{20}^{\mathrm{2}} }{\sqrt{\mathrm{20}^{\mathrm{2}} +\mathrm{1}}}−\mathrm{cos}^{−\mathrm{1}} \frac{\mathrm{1}}{\mathrm{20}}}{\mathrm{2}\pi}\rfloor=\lfloor\mathrm{2}.\mathrm{94}\rfloor=\mathrm{2} \\ $$$$\Rightarrow\mathrm{m}=\mathrm{4}×\mathrm{2}+\mathrm{3}=\mathrm{11} \\ $$

Question Number 16754    Answers: 2   Comments: 1

Question Number 16665    Answers: 0   Comments: 1

Question Number 16663    Answers: 0   Comments: 0

Question Number 16656    Answers: 1   Comments: 0

A particle moves with a speed of 10 ms^(−1) from the point (2, −2) in the direction 3i^∧ + 4j^∧ . The position vector after 3 s is

$$\mathrm{A}\:\mathrm{particle}\:\mathrm{moves}\:\mathrm{with}\:\mathrm{a}\:\mathrm{speed}\:\mathrm{of}\:\mathrm{10} \\ $$$$\mathrm{ms}^{−\mathrm{1}} \:\mathrm{from}\:\mathrm{the}\:\mathrm{point}\:\left(\mathrm{2},\:−\mathrm{2}\right)\:\mathrm{in}\:\mathrm{the} \\ $$$$\mathrm{direction}\:\mathrm{3}\overset{\wedge} {{i}}\:+\:\mathrm{4}\overset{\wedge} {{j}}.\:\mathrm{The}\:\mathrm{position}\:\mathrm{vector} \\ $$$$\mathrm{after}\:\mathrm{3}\:\mathrm{s}\:\mathrm{is} \\ $$

Question Number 16647    Answers: 0   Comments: 4

A ball is dropped vertically from a height d above the ground. It hits the ground and bounces up vertically to a height (d/2). Neglecting subsequent motion and air resistance its velocity V varies with height h above the ground is

$$\mathrm{A}\:\mathrm{ball}\:\mathrm{is}\:\mathrm{dropped}\:\mathrm{vertically}\:\mathrm{from}\:\mathrm{a} \\ $$$$\mathrm{height}\:{d}\:\mathrm{above}\:\mathrm{the}\:\mathrm{ground}.\:\mathrm{It}\:\mathrm{hits}\:\mathrm{the} \\ $$$$\mathrm{ground}\:\mathrm{and}\:\mathrm{bounces}\:\mathrm{up}\:\mathrm{vertically}\:\mathrm{to}\:\mathrm{a} \\ $$$$\mathrm{height}\:\frac{{d}}{\mathrm{2}}.\:\mathrm{Neglecting}\:\mathrm{subsequent} \\ $$$$\mathrm{motion}\:\mathrm{and}\:\mathrm{air}\:\mathrm{resistance}\:\mathrm{its}\:\mathrm{velocity} \\ $$$${V}\:\mathrm{varies}\:\mathrm{with}\:\mathrm{height}\:{h}\:\mathrm{above}\:\mathrm{the} \\ $$$$\mathrm{ground}\:\mathrm{is} \\ $$

Question Number 16645    Answers: 1   Comments: 1

A body is at rest at x = 0. At t = 0, it starts moving in the positive x-direction with a constant acceleration. At the same instant another body passes through x = 0 moving in the positive x-direction with a constant speed. The position of the first body is given by x_1 (t) after time ′t′ and that of the second body by x_2 (t) after the same time interval. Which of the following graphs correctly describes (x_1 − x_2 ) as a function of time ′t′?

$$\mathrm{A}\:\mathrm{body}\:\mathrm{is}\:\mathrm{at}\:\mathrm{rest}\:\mathrm{at}\:{x}\:=\:\mathrm{0}.\:\mathrm{At}\:{t}\:=\:\mathrm{0},\:\mathrm{it} \\ $$$$\mathrm{starts}\:\mathrm{moving}\:\mathrm{in}\:\mathrm{the}\:\mathrm{positive}\:{x}-\mathrm{direction} \\ $$$$\mathrm{with}\:\mathrm{a}\:\mathrm{constant}\:\mathrm{acceleration}.\:\mathrm{At}\:\mathrm{the} \\ $$$$\mathrm{same}\:\mathrm{instant}\:\mathrm{another}\:\mathrm{body}\:\mathrm{passes} \\ $$$$\mathrm{through}\:{x}\:=\:\mathrm{0}\:\mathrm{moving}\:\mathrm{in}\:\mathrm{the}\:\mathrm{positive} \\ $$$${x}-\mathrm{direction}\:\mathrm{with}\:\mathrm{a}\:\mathrm{constant}\:\mathrm{speed}.\:\mathrm{The} \\ $$$$\mathrm{position}\:\mathrm{of}\:\mathrm{the}\:\mathrm{first}\:\mathrm{body}\:\mathrm{is}\:\mathrm{given}\:\mathrm{by} \\ $$$${x}_{\mathrm{1}} \left({t}\right)\:\mathrm{after}\:\mathrm{time}\:'{t}'\:\mathrm{and}\:\mathrm{that}\:\mathrm{of}\:\mathrm{the} \\ $$$$\mathrm{second}\:\mathrm{body}\:\mathrm{by}\:{x}_{\mathrm{2}} \left({t}\right)\:\mathrm{after}\:\mathrm{the}\:\mathrm{same} \\ $$$$\mathrm{time}\:\mathrm{interval}.\:\mathrm{Which}\:\mathrm{of}\:\mathrm{the}\:\mathrm{following} \\ $$$$\mathrm{graphs}\:\mathrm{correctly}\:\mathrm{describes}\:\left({x}_{\mathrm{1}} \:−\:{x}_{\mathrm{2}} \right)\:\mathrm{as} \\ $$$$\mathrm{a}\:\mathrm{function}\:\mathrm{of}\:\mathrm{time}\:'{t}'? \\ $$

Question Number 16641    Answers: 0   Comments: 0

Prove that p is a prime number if and only if every equiangular polygon with p sides of rational lengths is regular.

$$\mathrm{Prove}\:\mathrm{that}\:{p}\:\mathrm{is}\:\mathrm{a}\:\mathrm{prime}\:\mathrm{number}\:\mathrm{if}\:\mathrm{and} \\ $$$$\mathrm{only}\:\mathrm{if}\:\mathrm{every}\:\mathrm{equiangular}\:\mathrm{polygon}\:\mathrm{with} \\ $$$${p}\:\mathrm{sides}\:\mathrm{of}\:\mathrm{rational}\:\mathrm{lengths}\:\mathrm{is}\:\mathrm{regular}. \\ $$

Question Number 16638    Answers: 1   Comments: 0

∫_( 0) ^( (π/4)) tan^2 (3x) dx

$$\int_{\:\mathrm{0}} ^{\:\frac{\pi}{\mathrm{4}}} \:\mathrm{tan}^{\mathrm{2}} \left(\mathrm{3x}\right)\:\mathrm{dx} \\ $$

Question Number 16635    Answers: 0   Comments: 0

Solve for x, y, z (√x) + (√y) + (√z) = 4 x^2 + y^2 + z^2 = 40.125 e^(xyz) = 4.771

$$\mathrm{Solve}\:\mathrm{for}\:\:\mathrm{x},\:\mathrm{y},\:\mathrm{z} \\ $$$$\sqrt{\mathrm{x}}\:+\:\sqrt{\mathrm{y}}\:+\:\sqrt{\mathrm{z}}\:=\:\mathrm{4} \\ $$$$\mathrm{x}^{\mathrm{2}} \:+\:\mathrm{y}^{\mathrm{2}} \:+\:\mathrm{z}^{\mathrm{2}} \:=\:\mathrm{40}.\mathrm{125} \\ $$$$\mathrm{e}^{\mathrm{xyz}} \:=\:\mathrm{4}.\mathrm{771} \\ $$

Question Number 16634    Answers: 1   Comments: 2

Consider a rubber ball freely falling from a height h = 4.9 m onto a horizontal elastic plate. Assume that the duration of collision is negligible and the collision with the plate is totally elastic. Then the velocity as a function of time and the height as function of time will be

$$\mathrm{Consider}\:\mathrm{a}\:\mathrm{rubber}\:\mathrm{ball}\:\mathrm{freely}\:\mathrm{falling} \\ $$$$\mathrm{from}\:\mathrm{a}\:\mathrm{height}\:{h}\:=\:\mathrm{4}.\mathrm{9}\:\mathrm{m}\:\mathrm{onto}\:\mathrm{a} \\ $$$$\mathrm{horizontal}\:\mathrm{elastic}\:\mathrm{plate}.\:\mathrm{Assume}\:\mathrm{that} \\ $$$$\mathrm{the}\:\mathrm{duration}\:\mathrm{of}\:\mathrm{collision}\:\mathrm{is}\:\mathrm{negligible} \\ $$$$\mathrm{and}\:\mathrm{the}\:\mathrm{collision}\:\mathrm{with}\:\mathrm{the}\:\mathrm{plate}\:\mathrm{is} \\ $$$$\mathrm{totally}\:\mathrm{elastic}.\:\mathrm{Then}\:\mathrm{the}\:\mathrm{velocity}\:\mathrm{as}\:\mathrm{a} \\ $$$$\mathrm{function}\:\mathrm{of}\:\mathrm{time}\:\mathrm{and}\:\mathrm{the}\:\mathrm{height}\:\mathrm{as} \\ $$$$\mathrm{function}\:\mathrm{of}\:\mathrm{time}\:\mathrm{will}\:\mathrm{be} \\ $$

Question Number 16633    Answers: 0   Comments: 1

(√1) . (√2) . (√3). ... (√n) = S_n what is S_n ?

$$\sqrt{\mathrm{1}}\:.\:\sqrt{\mathrm{2}}\:.\:\sqrt{\mathrm{3}}.\:...\:\sqrt{\mathrm{n}}\:\:=\:\mathrm{S}_{\mathrm{n}} \\ $$$$\mathrm{what}\:\mathrm{is}\:\:\mathrm{S}_{\mathrm{n}} \:? \\ $$

Question Number 16632    Answers: 0   Comments: 0

e^(xy) + y^2 = sin(x + y) Express the variable y interms of x only.

$$\mathrm{e}^{\mathrm{xy}} \:+\:\mathrm{y}^{\mathrm{2}} \:=\:\mathrm{sin}\left(\mathrm{x}\:+\:\mathrm{y}\right) \\ $$$$\mathrm{Express}\:\mathrm{the}\:\mathrm{variable}\:\mathrm{y}\:\mathrm{interms}\:\mathrm{of}\:\mathrm{x}\:\mathrm{only}. \\ $$

Question Number 16623    Answers: 1   Comments: 0

A ball is thrown up in a lift with a velocity u relative to the lift. If it returns to the lift in time t, then acceleration of the lift is [Answer: ((2u − gt)/t) upwards]

$$\mathrm{A}\:\mathrm{ball}\:\mathrm{is}\:\mathrm{thrown}\:\mathrm{up}\:\mathrm{in}\:\mathrm{a}\:\mathrm{lift}\:\mathrm{with}\:\mathrm{a} \\ $$$$\mathrm{velocity}\:{u}\:\mathrm{relative}\:\mathrm{to}\:\mathrm{the}\:\mathrm{lift}.\:\mathrm{If}\:\mathrm{it} \\ $$$$\mathrm{returns}\:\mathrm{to}\:\mathrm{the}\:\mathrm{lift}\:\mathrm{in}\:\mathrm{time}\:{t},\:\mathrm{then} \\ $$$$\mathrm{acceleration}\:\mathrm{of}\:\mathrm{the}\:\mathrm{lift}\:\mathrm{is}\:\left[\boldsymbol{\mathrm{Answer}}:\right. \\ $$$$\left.\frac{\mathrm{2}{u}\:−\:{gt}}{{t}}\:\mathrm{upwards}\right] \\ $$

Question Number 16621    Answers: 1   Comments: 1

A body moves in a straight line with a velocity whose square decreases linearly with displacement between two points A and B as shown. Determine the acceleration of the particle. [Answer: (8/3) ms^(−2) ]

$$\mathrm{A}\:\mathrm{body}\:\mathrm{moves}\:\mathrm{in}\:\mathrm{a}\:\mathrm{straight}\:\mathrm{line}\:\mathrm{with}\:\mathrm{a} \\ $$$$\mathrm{velocity}\:\mathrm{whose}\:\mathrm{square}\:\mathrm{decreases} \\ $$$$\mathrm{linearly}\:\mathrm{with}\:\mathrm{displacement}\:\mathrm{between} \\ $$$$\mathrm{two}\:\mathrm{points}\:{A}\:\mathrm{and}\:{B}\:\mathrm{as}\:\mathrm{shown}. \\ $$$$\mathrm{Determine}\:\mathrm{the}\:\mathrm{acceleration}\:\mathrm{of}\:\mathrm{the} \\ $$$$\mathrm{particle}.\:\left[\boldsymbol{\mathrm{Answer}}:\:\frac{\mathrm{8}}{\mathrm{3}}\:\mathrm{ms}^{−\mathrm{2}} \right] \\ $$

Question Number 16600    Answers: 1   Comments: 8

Solve simultaneously x^2 + y^2 = 61 .............. equation (i) x^3 − y^3 = 91 .............. equation (ii)

$$\mathrm{Solve}\:\mathrm{simultaneously} \\ $$$$\mathrm{x}^{\mathrm{2}} \:+\:\mathrm{y}^{\mathrm{2}} \:=\:\mathrm{61}\:\:\:\:\:\:\:\:\:..............\:\mathrm{equation}\:\left(\mathrm{i}\right) \\ $$$$\mathrm{x}^{\mathrm{3}} \:−\:\mathrm{y}^{\mathrm{3}} \:=\:\mathrm{91}\:\:\:\:\:\:\:\:\:..............\:\mathrm{equation}\:\left(\mathrm{ii}\right) \\ $$

Question Number 16598    Answers: 1   Comments: 0

Prove that cos^3 2θ + 3 cos 2θ = 4(cos^6 θ − sin^6 θ)

$$\mathrm{Prove}\:\mathrm{that} \\ $$$$\mathrm{cos}^{\mathrm{3}} \:\mathrm{2}\theta\:+\:\mathrm{3}\:\mathrm{cos}\:\mathrm{2}\theta\:=\:\mathrm{4}\left(\mathrm{cos}^{\mathrm{6}} \:\theta\:−\:\mathrm{sin}^{\mathrm{6}} \:\theta\right) \\ $$

Question Number 16595    Answers: 1   Comments: 4

Question Number 16592    Answers: 1   Comments: 1

please what does the question mean by the overlapping portion of A and B.

$$\mathrm{please}\:\mathrm{what}\:\mathrm{does}\:\mathrm{the}\:\mathrm{question}\:\mathrm{mean} \\ $$$$\mathrm{by}\:\mathrm{the}\:\mathrm{overlapping}\:\mathrm{portion}\:\mathrm{of}\:\mathrm{A}\: \\ $$$$\mathrm{and}\:\mathrm{B}. \\ $$

Question Number 16589    Answers: 1   Comments: 0

x^2 −∣3x+2∣+x ≥ 0 find range of values of x agreeing with above inequality.

$$\:\:\:\mathrm{x}^{\mathrm{2}} −\mid\mathrm{3x}+\mathrm{2}\mid+\mathrm{x}\:\geqslant\:\mathrm{0} \\ $$$$\:\mathrm{find}\:\mathrm{range}\:\mathrm{of}\:\mathrm{values}\:\mathrm{of}\:\mathrm{x}\:\mathrm{agreeing} \\ $$$$\mathrm{with}\:\mathrm{above}\:\mathrm{inequality}. \\ $$

Question Number 16579    Answers: 0   Comments: 3

Question Number 16570    Answers: 0   Comments: 1

A,B and E are three circles all with radius 1 unit.A and E touch at P whole B and E touch at Q. ∠POQ=x° where O is the centre of E.Find the area of the overlapping portion of A and B if 0≤x≤60°

$$\mathrm{A},\mathrm{B}\:\mathrm{and}\:\mathrm{E}\:\mathrm{are}\:\mathrm{three}\:\mathrm{circles}\:\mathrm{all}\: \\ $$$$\mathrm{with}\:\mathrm{radius}\:\mathrm{1}\:\mathrm{unit}.\mathrm{A}\:\mathrm{and}\:\mathrm{E}\:\mathrm{touch} \\ $$$$\mathrm{at}\:\mathrm{P}\:\mathrm{whole}\:\mathrm{B}\:\mathrm{and}\:\mathrm{E}\:\mathrm{touch}\:\mathrm{at}\:\mathrm{Q}. \\ $$$$\angle\mathrm{POQ}=\mathrm{x}°\:\mathrm{where}\:\mathrm{O}\:\mathrm{is}\:\mathrm{the}\:\mathrm{centre}\: \\ $$$$\mathrm{of}\:\mathrm{E}.\mathrm{Find}\:\mathrm{the}\:\mathrm{area}\:\mathrm{of}\:\mathrm{the}\: \\ $$$$\mathrm{overlapping}\:\mathrm{portion}\:\mathrm{of}\:\mathrm{A}\:\mathrm{and}\:\mathrm{B}\:\mathrm{if} \\ $$$$\mathrm{0}\leqslant\mathrm{x}\leqslant\mathrm{60}° \\ $$

Question Number 16572    Answers: 1   Comments: 0

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