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Question Number 18755    Answers: 1   Comments: 0

In a △ABC, if c = 2, A = 120°, a = (√6) , then C =

$$\mathrm{In}\:\mathrm{a}\:\bigtriangleup{ABC},\:\mathrm{if}\:{c}\:=\:\mathrm{2},\:{A}\:=\:\mathrm{120}°,\:{a}\:=\:\sqrt{\mathrm{6}}\:, \\ $$$$\mathrm{then}\:\:{C}\:= \\ $$

Question Number 18803    Answers: 1   Comments: 1

Question Number 18752    Answers: 0   Comments: 0

$$ \\ $$

Question Number 18749    Answers: 1   Comments: 1

Motion in two dimensions, in a plane can be studied by expressing position, velocity and acceleration as vectors in Cartesian co-ordinates A^→ = A_x i^∧ + A_y j^∧ where i^∧ and j^∧ are unit vector along x and y directions, respectively and A_x and A_y are corresponding components of A^→ (Figure). Motion can also be studied by expressing vectors in circular polar co-ordinates as A^→ = A_r r^∧ + A_θ θ^∧ where r^∧ = (r^→ /r) = cos θ i^∧ + sin θ j^∧ and θ^∧ = −sin θ i^∧ + cos θ j^∧ are unit vectors along direction in which ′r′ and ′θ′ are increasing. (a) Express i^∧ and j^∧ in terms of r^∧ and θ^∧ (b) Show that both r^∧ and θ^∧ are unit vectors and are perpendicular to each other. (c) Show that (d/dt)(r^∧ ) = ωθ^∧ where ω = (dθ/dt) and (d/dt)(θ^∧ ) = −ωr^∧ (d) For a particle moving along a spiral given by r^→ = αθr^∧ , where α = 1 (unit), find dimensions of ′α′. (e) Find velocity and acceleration in polar vector representation for particle moving along spiral described in (d) above.

$$\mathrm{Motion}\:\mathrm{in}\:\mathrm{two}\:\mathrm{dimensions},\:\mathrm{in}\:\mathrm{a}\:\mathrm{plane} \\ $$$$\mathrm{can}\:\mathrm{be}\:\mathrm{studied}\:\mathrm{by}\:\mathrm{expressing}\:\mathrm{position}, \\ $$$$\mathrm{velocity}\:\mathrm{and}\:\mathrm{acceleration}\:\mathrm{as}\:\mathrm{vectors}\:\mathrm{in} \\ $$$$\mathrm{Cartesian}\:\mathrm{co}-\mathrm{ordinates}\:\overset{\rightarrow} {{A}}\:=\:{A}_{{x}} \overset{\wedge} {{i}}\:+\:{A}_{{y}} \overset{\wedge} {{j}} \\ $$$$\mathrm{where}\:\overset{\wedge} {{i}}\:\mathrm{and}\:\overset{\wedge} {{j}}\:\mathrm{are}\:\mathrm{unit}\:\mathrm{vector}\:\mathrm{along}\:{x} \\ $$$$\mathrm{and}\:{y}\:\mathrm{directions},\:\mathrm{respectively}\:\mathrm{and}\:{A}_{{x}} \\ $$$$\mathrm{and}\:{A}_{{y}} \:\mathrm{are}\:\mathrm{corresponding}\:\mathrm{components} \\ $$$$\mathrm{of}\:\overset{\rightarrow} {{A}}\:\left(\mathrm{Figure}\right).\:\mathrm{Motion}\:\mathrm{can}\:\mathrm{also}\:\mathrm{be} \\ $$$$\mathrm{studied}\:\mathrm{by}\:\mathrm{expressing}\:\mathrm{vectors}\:\mathrm{in}\:\mathrm{circular} \\ $$$$\mathrm{polar}\:\mathrm{co}-\mathrm{ordinates}\:\mathrm{as}\:\overset{\rightarrow} {{A}}\:=\:{A}_{{r}} \overset{\wedge} {{r}}\:+\:{A}_{\theta} \overset{\wedge} {\theta} \\ $$$$\mathrm{where}\:\overset{\wedge} {{r}}\:=\:\frac{\overset{\rightarrow} {{r}}}{{r}}\:=\:\mathrm{cos}\:\theta\:\overset{\wedge} {{i}}\:+\:\mathrm{sin}\:\theta\:\overset{\wedge} {{j}}\:\mathrm{and}\:\overset{\wedge} {\theta}\:= \\ $$$$−\mathrm{sin}\:\theta\:\overset{\wedge} {{i}}\:+\:\mathrm{cos}\:\theta\:\overset{\wedge} {{j}}\:\mathrm{are}\:\mathrm{unit}\:\mathrm{vectors}\:\mathrm{along} \\ $$$$\mathrm{direction}\:\mathrm{in}\:\mathrm{which}\:'{r}'\:\mathrm{and}\:'\theta'\:\mathrm{are} \\ $$$$\mathrm{increasing}. \\ $$$$\left({a}\right)\:\mathrm{Express}\:\overset{\wedge} {{i}}\:\mathrm{and}\:\overset{\wedge} {{j}}\:\mathrm{in}\:\mathrm{terms}\:\mathrm{of}\:\overset{\wedge} {{r}}\:\mathrm{and}\:\overset{\wedge} {\theta} \\ $$$$\left({b}\right)\:\mathrm{Show}\:\mathrm{that}\:\mathrm{both}\:\overset{\wedge} {{r}}\:\mathrm{and}\:\overset{\wedge} {\theta}\:\mathrm{are}\:\mathrm{unit} \\ $$$$\mathrm{vectors}\:\mathrm{and}\:\mathrm{are}\:\mathrm{perpendicular}\:\mathrm{to}\:\mathrm{each} \\ $$$$\mathrm{other}. \\ $$$$\left({c}\right)\:\mathrm{Show}\:\mathrm{that}\:\frac{{d}}{{dt}}\left(\overset{\wedge} {{r}}\right)\:=\:\omega\overset{\wedge} {\theta}\:\mathrm{where} \\ $$$$\omega\:=\:\frac{{d}\theta}{{dt}}\:\mathrm{and}\:\frac{{d}}{{dt}}\left(\overset{\wedge} {\theta}\right)\:=\:−\omega\overset{\wedge} {{r}} \\ $$$$\left({d}\right)\:\mathrm{For}\:\mathrm{a}\:\mathrm{particle}\:\mathrm{moving}\:\mathrm{along}\:\mathrm{a}\:\mathrm{spiral} \\ $$$$\mathrm{given}\:\mathrm{by}\:\overset{\rightarrow} {{r}}\:=\:\alpha\theta\overset{\wedge} {{r}},\:\mathrm{where}\:\alpha\:=\:\mathrm{1}\:\left(\mathrm{unit}\right), \\ $$$$\mathrm{find}\:\mathrm{dimensions}\:\mathrm{of}\:'\alpha'. \\ $$$$\left({e}\right)\:\mathrm{Find}\:\mathrm{velocity}\:\mathrm{and}\:\mathrm{acceleration}\:\mathrm{in} \\ $$$$\mathrm{polar}\:\mathrm{vector}\:\mathrm{representation}\:\mathrm{for}\:\mathrm{particle} \\ $$$$\mathrm{moving}\:\mathrm{along}\:\mathrm{spiral}\:\mathrm{described}\:\mathrm{in}\:\left({d}\right) \\ $$$$\mathrm{above}. \\ $$

Question Number 18748    Answers: 1   Comments: 0

Question Number 18747    Answers: 2   Comments: 0

Question Number 18746    Answers: 2   Comments: 0

Question Number 18744    Answers: 0   Comments: 1

Question Number 18742    Answers: 1   Comments: 0

The value of cot16°cot44° + cot44°cot76° − cot76°cot16° is

$$\mathrm{The}\:\mathrm{value}\:\mathrm{of}\:\mathrm{cot16}°\mathrm{cot44}°\:+\:\mathrm{cot44}°\mathrm{cot76}° \\ $$$$−\:\mathrm{cot76}°\mathrm{cot16}°\:\mathrm{is} \\ $$

Question Number 18731    Answers: 3   Comments: 0

tan ((2π)/5) − tan (π/(15)) − (√3) tan ((2π)/5) tan (π/(15)) =

$$\mathrm{tan}\:\frac{\mathrm{2}\pi}{\mathrm{5}}\:−\:\:\mathrm{tan}\:\frac{\pi}{\mathrm{15}}\:−\:\sqrt{\mathrm{3}}\:\mathrm{tan}\:\frac{\mathrm{2}\pi}{\mathrm{5}}\:\mathrm{tan}\:\frac{\pi}{\mathrm{15}}\:= \\ $$

Question Number 18730    Answers: 0   Comments: 3

If [((((2/3))^5 ))^(1/9) ]^(√(x−5)) =a^0 , find the value of x.

$$\mathrm{If}\:\:\left[\sqrt[{\mathrm{9}}]{\left(\frac{\mathrm{2}}{\mathrm{3}}\right)^{\mathrm{5}} }\right]^{\sqrt{{x}−\mathrm{5}}} ={a}^{\mathrm{0}} ,\:\mathrm{find}\:\mathrm{the}\:\mathrm{value}\:\mathrm{of}\:{x}. \\ $$

Question Number 18723    Answers: 1   Comments: 0

Question Number 18722    Answers: 0   Comments: 0

Question Number 18715    Answers: 1   Comments: 0

∣x + 1∣

$$\mid\mathrm{x}\:+\:\mathrm{1}\mid \\ $$

Question Number 18712    Answers: 0   Comments: 0

Question Number 18704    Answers: 1   Comments: 0

Question Number 18702    Answers: 1   Comments: 1

If a sin^2 x+b cos^2 x = c, b sin^2 y+a cos^2 y = d and a tan^2 x = b tan y then (a^2 /b^2 ) is equal to

$$\mathrm{If}\:\:\:{a}\:\mathrm{sin}^{\mathrm{2}} {x}+{b}\:\mathrm{cos}^{\mathrm{2}} {x}\:=\:{c},\: \\ $$$$\:\:\:\:\:\:\:{b}\:\mathrm{sin}^{\mathrm{2}} {y}+{a}\:\mathrm{cos}^{\mathrm{2}} {y}\:=\:{d}\:\:\:\mathrm{and} \\ $$$$\:\:\:\:\:\:\:{a}\:\mathrm{tan}^{\mathrm{2}} {x}\:=\:{b}\:\mathrm{tan}\:{y}\:\: \\ $$$$\mathrm{then}\:\frac{{a}^{\mathrm{2}} }{{b}^{\mathrm{2}} }\:\mathrm{is}\:\mathrm{equal}\:\mathrm{to} \\ $$

Question Number 18701    Answers: 1   Comments: 1

The value of tan 6° tan 42° tan 66° tan 78° is

$$\mathrm{The}\:\mathrm{value}\:\mathrm{of} \\ $$$$\:\mathrm{tan}\:\mathrm{6}°\:\mathrm{tan}\:\mathrm{42}°\:\mathrm{tan}\:\mathrm{66}°\:\mathrm{tan}\:\mathrm{78}°\:\:\mathrm{is} \\ $$

Question Number 18695    Answers: 1   Comments: 0

if ((n tanθ)/(cos^2 (α−θ)))=((m tan(α−θ))/(cos^2 θ)) then show that 2θ=α−tan^(−1) (((n−m)/(n+m))tanα)

$$\mathrm{if}\:\:\frac{\mathrm{n}\:\mathrm{tan}\theta}{\mathrm{cos}^{\mathrm{2}} \left(\alpha−\theta\right)}=\frac{\mathrm{m}\:\mathrm{tan}\left(\alpha−\theta\right)}{\mathrm{cos}^{\mathrm{2}} \theta} \\ $$$$\mathrm{then}\:\mathrm{show}\:\mathrm{that} \\ $$$$\mathrm{2}\theta=\alpha−\mathrm{tan}^{−\mathrm{1}} \left(\frac{\mathrm{n}−\mathrm{m}}{\mathrm{n}+\mathrm{m}}\mathrm{tan}\alpha\right) \\ $$

Question Number 18881    Answers: 0   Comments: 0

If (6 (√6) +14)^(2n+1) = m and if f is the fractional part of m, then f m is equal to

$$\mathrm{If}\:\left(\mathrm{6}\:\sqrt{\mathrm{6}}\:+\mathrm{14}\right)^{\mathrm{2}{n}+\mathrm{1}} =\:{m}\:\mathrm{and}\:\mathrm{if}\:{f}\:\:\mathrm{is}\:\mathrm{the} \\ $$$$\mathrm{fractional}\:\mathrm{part}\:\mathrm{of}\:{m},\:\mathrm{then}\:{f}\:{m}\:\mathrm{is}\:\mathrm{equal}\:\mathrm{to} \\ $$

Question Number 18693    Answers: 1   Comments: 0

prove that 2tan^(−1) [tan(α/2)tan((π/4)−(β/2))] =tan^(−1) (((sinα cosβ)/(cosα +sinβ)))

$$\mathrm{prove}\:\mathrm{that}\:\mathrm{2tan}^{−\mathrm{1}} \left[\mathrm{tan}\frac{\alpha}{\mathrm{2}}\mathrm{tan}\left(\frac{\pi}{\mathrm{4}}−\frac{\beta}{\mathrm{2}}\right)\right] \\ $$$$=\mathrm{tan}^{−\mathrm{1}} \left(\frac{\mathrm{sin}\alpha\:\mathrm{cos}\beta}{\mathrm{cos}\alpha\:+\mathrm{sin}\beta}\right) \\ $$

Question Number 18967    Answers: 0   Comments: 3

Let PQRS be a rectangle such that PQ = a and QR = b. Suppose r_1 is the radius of the circle passing through P and Q and touching RS and r_2 is the radius of the circle passing through Q and R and touching PS. Show that : 5(a + b) ≤ 8(r_1 + r_2 )

$$\mathrm{Let}\:\mathrm{PQRS}\:\mathrm{be}\:\mathrm{a}\:\mathrm{rectangle}\:\mathrm{such}\:\mathrm{that} \\ $$$$\mathrm{PQ}\:=\:{a}\:\mathrm{and}\:\mathrm{QR}\:=\:{b}.\:\mathrm{Suppose}\:\mathrm{r}_{\mathrm{1}} \:\mathrm{is}\:\mathrm{the} \\ $$$$\mathrm{radius}\:\mathrm{of}\:\mathrm{the}\:\mathrm{circle}\:\mathrm{passing}\:\mathrm{through}\:\mathrm{P} \\ $$$$\mathrm{and}\:\mathrm{Q}\:\mathrm{and}\:\mathrm{touching}\:\mathrm{RS}\:\mathrm{and}\:\mathrm{r}_{\mathrm{2}} \:\mathrm{is}\:\mathrm{the} \\ $$$$\mathrm{radius}\:\mathrm{of}\:\mathrm{the}\:\mathrm{circle}\:\mathrm{passing}\:\mathrm{through}\:\mathrm{Q} \\ $$$$\mathrm{and}\:\mathrm{R}\:\mathrm{and}\:\mathrm{touching}\:\mathrm{PS}.\:\mathrm{Show}\:\mathrm{that}\:: \\ $$$$\mathrm{5}\left({a}\:+\:{b}\right)\:\leqslant\:\mathrm{8}\left(\mathrm{r}_{\mathrm{1}} \:+\:\mathrm{r}_{\mathrm{2}} \right) \\ $$

Question Number 18681    Answers: 1   Comments: 0

y = ∣sin x∣ + 2 y = ∣x∣ + 2 −π −π ≤ x ≤ π Find the area that have created from the equations above

$${y}\:=\:\mid\mathrm{sin}\:{x}\mid\:+\:\mathrm{2} \\ $$$${y}\:=\:\mid{x}\mid\:+\:\mathrm{2}\:−\pi \\ $$$$−\pi\:\leqslant\:{x}\:\leqslant\:\pi \\ $$$$\mathrm{Find}\:\mathrm{the}\:\mathrm{area}\:\mathrm{that}\:\mathrm{have}\:\mathrm{created} \\ $$$$\mathrm{from}\:\mathrm{the}\:\mathrm{equations}\:\mathrm{above} \\ $$

Question Number 18678    Answers: 2   Comments: 0

lim_(x→∞) ((4^(x + 1) + 2^(x +1) − 3^(x + 1) )/(4^(x − 1) + 2^(x − 1 ) + 3^(x + 1) ))

$$\underset{{x}\rightarrow\infty} {\mathrm{lim}}\:\:\frac{\mathrm{4}^{{x}\:+\:\mathrm{1}} \:+\:\mathrm{2}^{{x}\:+\mathrm{1}} \:−\:\mathrm{3}^{{x}\:+\:\mathrm{1}} }{\mathrm{4}^{{x}\:−\:\mathrm{1}} \:+\:\mathrm{2}^{{x}\:−\:\mathrm{1}\:} +\:\mathrm{3}^{{x}\:+\:\mathrm{1}} \:} \\ $$

Question Number 18667    Answers: 2   Comments: 0

If x=2^(1/3) − 2^(−1/3) , find the value of 2x^3 +6x.

$$\mathrm{If}\:{x}=\mathrm{2}^{\mathrm{1}/\mathrm{3}} −\:\mathrm{2}^{−\mathrm{1}/\mathrm{3}} ,\:\mathrm{find}\:\mathrm{the}\:\mathrm{value}\:\mathrm{of} \\ $$$$\:\:\:\:\mathrm{2}{x}^{\mathrm{3}} +\mathrm{6}{x}. \\ $$

Question Number 18666    Answers: 1   Comments: 0

An elastic material has a length of 36cm when a load of 40N is hung on it and a length of 45cm when a load of 60N is hung on it. what is the Original length of the string ?

$$\mathrm{An}\:\mathrm{elastic}\:\mathrm{material}\:\mathrm{has}\:\mathrm{a}\:\mathrm{length}\:\mathrm{of}\:\:\mathrm{36cm}\:\mathrm{when}\:\mathrm{a}\:\mathrm{load}\:\mathrm{of}\:\mathrm{40N}\:\mathrm{is}\:\mathrm{hung}\:\mathrm{on}\:\mathrm{it}\:\mathrm{and} \\ $$$$\mathrm{a}\:\mathrm{length}\:\mathrm{of}\:\mathrm{45cm}\:\mathrm{when}\:\mathrm{a}\:\mathrm{load}\:\mathrm{of}\:\mathrm{60N}\:\mathrm{is}\:\mathrm{hung}\:\mathrm{on}\:\mathrm{it}.\:\mathrm{what}\:\mathrm{is}\:\mathrm{the}\:\mathrm{Original}\: \\ $$$$\mathrm{length}\:\mathrm{of}\:\mathrm{the}\:\mathrm{string}\:? \\ $$

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