A storage tank consists of a circular cylinder with a hemisphere adjoined on either end.
Question: A storage tank consists of a circular cylinder with a hemisphere adjoined on either end. If the external diameter of the cylinder be 1.4 m and its length is 8 m, find the cost of painting it on the outside at the rate of Rs.10 perm2 Solution: Diameter of a cylinder = 1.4 m Therefore radius of cylinder =1.42 = 0.7 m Height of cylinder = 8 m Therefore surface area of tank $=2 \pi r h+2 \pi r^{2}$ $=2 \times 22 / 7 \times 0.7 \times 8+2 \times 22 / 7 \times(0.7)^{2}$ $=176 / 5+77 / 25=38....
Read More →If angle θ is divided into two parts such that the tangents of one part is λ times the tangent of other,
Question: If angle $\theta$ is divided into two parts such that the tangents of one part is $\lambda$ times the tangent of other, and $\phi$ is their difference, then show that $\sin \theta=\frac{\lambda+1}{\lambda-1} \sin \phi$. [NCERT EXEMPLER] Solution: Let $\alpha$ and $\beta$ be the two parts of angle $\theta$. Then, $\theta=\alpha+\beta$ and $\phi=\alpha-\beta$ (Given) Now, $\tan \alpha=\lambda \tan \beta$ (Given) $\Rightarrow \frac{\tan \alpha}{\tan \beta}=\frac{\lambda}{1}$ Applying comp...
Read More →If tan α = x +1, tan β = x − 1,
Question: If tan =x+1, tan =x 1, show that 2 cot ( ) =x2. Solution: $\mathrm{LHS}=2 \cot (\alpha-\beta)$ $=\frac{2(1+\tan \alpha \tan \beta)}{[\tan \alpha-\tan \beta]}$ $=\frac{2+2(x+1)(x-1)}{(x+1-x+1)}$ $=\frac{2+2 x^{2}-2}{2}$ $=\frac{2 x^{2}}{2}$ $=x^{2}$ = RHS Hence proved....
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Question: $\frac{d y}{d x}-3 y \cot x=\sin 2 x ; y=2$ when $x=\frac{\pi}{2}$ Solution: The given differential equation is $\frac{d y}{d x}-3 y \cot x=\sin 2 x$. This is a linear differential equation of the form: $\frac{d y}{d x}+p y=Q($ where $p=-3 \cot x$ and $Q=\sin 2 x)$ Now, I.F $=e^{\int p d x}=e^{-3 \int \cot x d x}=e^{-3 \log |\sin x|}=e^{\log \left|\frac{1}{\mid \sin ^{3} x}\right|}=\frac{1}{\sin ^{3} x} .$ The general solution of the given differential equation is given by the relation...
Read More →A vertical stick of length 6 m casts a shadow 4 m long on the ground and at the same time a tower casts a shadow 28 m long.
Question: A vertical stick of length 6 m casts a shadow 4 m long on the ground and at the same time a tower casts a shadow 28 m long. Find the height of the tower. Solution: It is given that length of vertical stick $=6 \mathrm{~m}$ We have to find the height of the tower. Suppose $A B$ is the height of the tower and $B C$ is its shadow. Now, $\triangle A B C \sim \triangle P C R \quad \angle B=\angle Q$ and $\angle A=\angle P$ $\frac{A B}{B C}=\frac{P Q}{Q R}$ $\frac{A B}{28} \mathrm{~m}=\frac{...
Read More →If sin α sin β − cos α cos β + 1 = 0,
Question: If sin sin cos cos + 1 = 0, prove that 1 + cot tan = 0. Solution: Given: $\sin \alpha s$ in $\beta-\cos \alpha \cos \beta+1=0$ $\Rightarrow-(\cos \alpha \cos \beta-\sin \alpha \sin \beta)+1=0$ $\Rightarrow-\cos (\alpha+\beta)+1=0$ $\Rightarrow \cos (\alpha+\beta)=1$ Therefore, $\sin (\alpha+\beta)=0 \quad \ldots(1)\left(\right.$ Since $\left.\sin \theta=\sqrt{1-\cos ^{2} \theta}\right)$ Hence , $1+\cot \alpha \tan \beta=1+\frac{\cos \alpha \sin \beta}{\sin \alpha \cos \beta}$ $=\frac{\...
Read More →If ∆ABC and ∆AMP are two right triangles, right angled at B and M respectively such that ∠MAP = ∠BAC. Prove that
Question: If ∆ABC and ∆AMP are two right triangles, right angled at B and M respectively such that MAP = BAC. Prove that (i) $\triangle \mathrm{ABC} \sim \triangle \mathrm{AMP}$ (ii) $\mathrm{CAPA}=\mathrm{BCMP}$ Solution: (1) It is given that $\triangle A B C$ and $\triangle A M P$ are two right angle triangles. Now, in $\triangle A B C$ and $\triangle A M P$, we have $\angle M A P=\angle B A C \quad$ (Given) $\angle A M P=\angle B=90^{\circ}$ $\triangle A B C \sim \triangle A M P$ (AA Similari...
Read More →Prove that:
Question: Prove that: (i) $\frac{1}{\sin (x-a) \sin (x-b)}=\frac{\cot (x-a)-\cot (x-b)}{\sin (a-b)}$ (ii) $\frac{1}{\sin (x-a) \cos (x-b)}=\frac{\cot (x-a)+\tan (x-b)}{\cos (a-b)}$ (iii) $\frac{1}{\cos (x-a) \cos (a-b)}=\frac{\tan (x-b)-\tan (x-a)}{\sin (a-b)}$ Solution: (i) $\mathrm{RHS}=\frac{\cot (x-a)-\cot (x-b)}{\sin (a-b)}$ $=\frac{\frac{\cos (x-a)}{\sin (x-a)}-\frac{\cos (x-b)}{\sin (x-b)}}{\sin (a-b)}$ $=\frac{\sin (x-b) \cos (x-a)-\sin (x-a) \cos (x-b)}{\sin (x-a) \sin (x-b) \sin (a-b)}...
Read More →Prove that:
Question: Prove that: (i) $\frac{1}{\sin (x-a) \sin (x-b)}=\frac{\cot (x-a)-\cot (x-b)}{\sin (a-b)}$ (ii) $\frac{1}{\sin (x-a) \cos (x-b)}=\frac{\cot (x-a)+\tan (x-b)}{\cos (a-b)}$ (iii) $\frac{1}{\cos (x-a) \cos (a-b)}=\frac{\tan (x-b)-\tan (x-a)}{\sin (a-b)}$ Solution: (i) $\mathrm{RHS}=\frac{\cot (x-a)-\cot (x-b)}{\sin (a-b)}$ $=\frac{\frac{\cos (x-a)}{\sin (x-a)}-\frac{\cos (x-b)}{\sin (x-b)}}{\sin (a-b)}$ $=\frac{\sin (x-b) \cos (x-a)-\sin (x-a) \cos (x-b)}{\sin (x-a) \sin (x-b) \sin (a-b)}...
Read More →Diagonals AC and BD of a trapezium ABCD with AB || DC intersect each other at the point O.
Question: Diagonals AC and BD of a trapezium ABCD with AB || DC intersect each other at the point O. Using similarity criterion for two triangles, show thatOAOC=OBOD. Solution: It is given that trapezium $A B C D$ with $A B \| D C . O$ is the point of intersection of $A C$ and $B D$. We have to prove that $\frac{O A}{O C}=\frac{O B}{O D}$ Now, in $\triangle A O B$ and $\triangle C O D$ $\angle A O B=\angle C O D \quad$ (Vertically opposite angles) $\angle \mathrm{OAB}=\angle \mathrm{OCD}$ (Alter...
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Question: $\left(1+x^{2}\right) \frac{d y}{d x}+2 x y=\frac{1}{1+x^{2}} ; y=0$ when $x=1$ Solution: $\left(1+x^{2}\right) \frac{d y}{d x}+2 x y=\frac{1}{1+x^{2}}$ $\Rightarrow \frac{d y}{d x}+\frac{2 x y}{1+x^{2}}=\frac{1}{\left(1+x^{2}\right)^{2}}$ This is a linear differential equation of the form: $\frac{d y}{d x}+p y=Q$ (where $p=\frac{2 x}{1+x^{2}}$ and $Q=\frac{1}{\left(1+x^{2}\right)^{2}}$ ) Now, I.F $=e^{\int p d x}=e^{\int_{1+x^{2}}^{2 x d x}}=e^{\log \left(1+x^{2}\right)}=1+x^{2}$. The...
Read More →A wooden toy is in the form of a cone surmounted on a hemisphere.
Question: A wooden toy is in the form of a cone surmounted on a hemisphere. The diameter of the base of the cone is $16 \mathrm{~cm}$ and its height is $15 \mathrm{~cm}$. Find the cost of painting the toy at Rs. 7 per $100 \mathrm{~cm}$ Solution: Diameter of cone = 16 cm Radius of cone = 8 cm Height of cone = 15 cm Slant height of cone $=\sqrt{8^{2}+15^{2}}$ $=\sqrt{64+225}$ $=\sqrt{289}=17 \mathrm{~cm}$ Therefore Total curved surface area of toy $=\pi r l+2 \pi r^{2}$ $=22 / 7 \times 8 \times 1...
Read More →If sin α + sin β = a and cos α + cos β = b, show that
Question: If sin + sin =aand cos + cos =b, show that (i) $\sin (\alpha+\beta)=\frac{2 a b}{a^{2}+b^{2}}$ (ii) $\cos (\alpha+\beta)=\frac{b^{2}-a^{2}}{b^{2}+a^{2}}$ Solution: (i) $a^{2}+b^{2}=(\sin \alpha+\sin \beta)^{2}+(\cos \alpha+\cos \beta)^{2}$ $\Rightarrow a^{2}+b^{2}=\sin ^{2} \alpha+\sin ^{2} \beta+2 \sin \alpha \sin \beta+\cos ^{2} \alpha+\cos ^{2} \beta+2 \cos \alpha \cos \beta$ $\Rightarrow a^{2}+b^{2}=\sin ^{2} \alpha+\cos ^{2} \alpha+\sin ^{2} \beta+\cos ^{2} \beta+2(\sin \alpha \si...
Read More →If sin α + sin β = a and cos α + cos β = b, show that
Question: If sin + sin =aand cos + cos =b, show that (i) $\sin (\alpha+\beta)=\frac{2 a b}{a^{2}+b^{2}}$ (ii) $\cos (\alpha+\beta)=\frac{b^{2}-a^{2}}{b^{2}+a^{2}}$ Solution: (i) $a^{2}+b^{2}=(\sin \alpha+\sin \beta)^{2}+(\cos \alpha+\cos \beta)^{2}$ $\Rightarrow a^{2}+b^{2}=\sin ^{2} \alpha+\sin ^{2} \beta+2 \sin \alpha \sin \beta+\cos ^{2} \alpha+\cos ^{2} \beta+2 \cos \alpha \cos \beta$ $\Rightarrow a^{2}+b^{2}=\sin ^{2} \alpha+\cos ^{2} \alpha+\sin ^{2} \beta+\cos ^{2} \beta+2(\sin \alpha \si...
Read More →A girl of heigh 90 cm is walking away from the base of a lamp-post at a speed of 1.2 m/sec.
Question: A girl of heigh 90 cm is walking away from the base of a lamp-post at a speed of 1.2 m/sec. If the lamp is 3.6 m above the ground, find the length of her shadow after 4 seconds. Solution: It is given that, girl height $=90 \mathrm{~cm}$, speed $=1.2 \mathrm{~m} / \mathrm{sec}$ and height of lamp $=3.6 \mathrm{~m}$. We have to find the length of her shadow after $4 \mathrm{sec}$ Let $A B$ be the lamp post and $C D$ be the girl. Suppose $D E$ is the length of her shadow. Let $D E=x$ And ...
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Question: $\frac{d y}{d x}+2 y \tan x=\sin x ; y=0$ when $x=\frac{\pi}{3}$ Solution: The given differential equation is $\frac{d y}{d x}+2 y \tan x=\sin x$. This is a linear equation of the form: $\frac{d y}{d x}+p y=Q$ (where $p=2 \tan x$ and $Q=\sin x$ ) $\frac{d y}{d x}+p y=Q($ where $p=2 \tan x$ and $Q=\sin x)$ Now, I.F $=e^{\int p d x}=e^{\int 2 \tan x d x}=e^{2 \log |\sec x|}=e^{\log \left(\sec ^{2} x\right)}=\sec ^{2} x$ The general solution of the given differential equation is given by ...
Read More →A cylinder of same height and radius is placed on top of a hemisphere.
Question: A cylinder of same height and radius is placed on top of a hemisphere. Find the curved surface area of the shape if the length of the shape is 7cm. Solution: Given length of the shape = 7cm But length = r + r 2r = 7cm r = 3.5 cm Also; h = r Total surface area of shape $=2 \pi r h+2 \pi r^{2}$ $=2 \pi r r+2 \pi r^{2}$ $=2 \pi r^{2}+2 \pi r^{2}$ $=4 \times 22 / 7 \times(3.5)^{2}=154 \mathrm{~cm}^{2}$...
Read More →In an isosceles ∆ABC,
Question: In an isosceles $\triangle \mathrm{ABC}$, the base $\mathrm{AB}$ is produced both the ways to $\mathrm{P}$ and $\mathrm{Q}$ such that $\mathrm{AP} \times \mathrm{BQ}=\mathrm{AC}^{2}$. Prove that $\triangle \mathrm{APC} \sim \triangle \mathrm{BCQ}$. Solution: It is given that $\triangle A B C$ is isosceles and $A P \times B Q=A C^{2}$. We have to prove that $\triangle A P C \sim \triangle B C Q$. It is given that $\triangle A B C$ is an isosceles triangle, so $\mathrm{AC}=\mathrm{BC}$. ...
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Question: $\frac{d y}{d x}+2 y \tan x=\sin x ; y=0$ when $x=\frac{\pi}{3}$ Solution: The given differential equation is $\frac{d y}{d x}+2 y \tan x=\sin x$. This is a linear equation of the form: $\frac{d y}{d x}+p y=Q$ (where $p=2 \tan x$ and $Q=\sin x$ ) $\frac{d y}{d x}+p y=Q($ where $p=2 \tan x$ and $Q=\sin x)$ Now, I.F $=e^{\int p d x}=e^{\int 2 \tan x d x}=e^{2 \log |\sec x|}=e^{\log \left(\sec ^{2} x\right)}=\sec ^{2} x$ The general solution of the given differential equation is given by ...
Read More →If α, β are two different values of x lying between 0 and 2π,
Question: If , are two different values ofxlying between 0 and 2,which satisfy the equation 6 cosx+ 8 sinx= 9, find the value of sin ( + ). Solution: Given: $6 \cos x+8 \sin x=9$ $\Rightarrow 6 \cos x=9-8 \sin x$ $\Rightarrow 36 \cos ^{2} x=(9-8 \sin x)^{2}$ $\Rightarrow 36\left(1-\sin ^{2} x\right)=81+64 \sin ^{2} x-144 \sin x$ $\Rightarrow 100 \sin ^{2} x-144 \sin x+45=0$ Now, $\alpha$ and $\beta$ are the roots of the given equation; therefore, $\cos \alpha$ and $\cos \beta$ are the roots of t...
Read More →Assuming the earth to be a sphere of radius 6370 km,
Question: Assuming the earth to be a sphere of radius 6370 km, how many square kilometers is the area of the land if three-fourths of the earths surface is covered by water? Solution: $3 / 4^{\text {th }}$ of earth surface is covered by water Therefore $1 / 4^{\text {th }}$ of earth surface is covered by land Therefore Surface area covered by land $=1 / 4 \times 4 \pi r^{2}$ $=1 / 4 \times 4 \times 22 / 7 \times(6370)^{2}$ $=127527400 \mathrm{~km}^{2}$...
Read More →In Fig. 4.145, if AB ⊥ BC, DC ⊥ BC and DE ⊥ AC, prove that ∆CED ∼ ∆ABC.
Question: In Fig. 4.145, if $\mathrm{AB} \perp \mathrm{BC}, \mathrm{DC} \perp \mathrm{BC}$ and $\mathrm{DE} \perp \mathrm{AC}$, prove that $\triangle \mathrm{CED} \sim \triangle \mathrm{ABC}$. Solution: It is given that $A B \perp B C, D C \perp B C$ and $D E \perp A C$. We have to prove that $\triangle C E D \sim \triangle A B C$. Now, $A B \perp B C, D C \perp B C$, so $A B \| D C$. In $\triangle \mathrm{ABC}$ and $\triangle \mathrm{CED}$, $\angle B=\angle E=90^{\circ}$ (Given) $\angle \mathrm...
Read More →The dome of a building is in the form of a hemisphere. Its radius is 63dm.
Question: The dome of a building is in the form of a hemisphere. Its radius is 63dm. Find the cost of painting it at the rate of Rs.2 per sq m. Solution: Dome radius = 63 dm = 6.3 m Inner surface area of dome $=2 \pi r^{2}$ $=2 \times 3.14 \times(6.3)^{2}$ $=249.48 \mathrm{~m}^{2}$ Now, cost of $1 \mathrm{~m}^{2}=$ Rs. 2 Therefore cost of $249.48 \mathrm{~m}^{2}=$ Rs. $(249.48 \times 2)=$ Rs. $498.96$...
Read More →If sin (α + β) = 1 and sin (α − β) =
Question: If $\sin (\alpha+\beta)=1$ and $\sin (\alpha-\beta)=\frac{1}{2}$, where $0 \leq \alpha, \beta \leq \frac{\pi}{2}$, then find the values of $\tan (\alpha+2 \beta)$ and $\tan (2 \alpha+\beta)$. Solution: Given: $\sin (\alpha+\beta)=1$ and $\sin (\alpha-\beta)=\frac{1}{2}$ $\Rightarrow \alpha+\beta=90^{\circ} \quad \ldots(1)$ and $\alpha-\beta=30^{\circ} \quad \ldots(2)$ By adding eq $(1)$ and eq $(2)$ we get : $2 \alpha=120^{\circ}$ $\Rightarrow \alpha=60^{\circ}$ By subtracting eq $(2)$...
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Question: $\left(x+3 y^{2}\right) \frac{d y}{d x}=y(y0)$ Solution: $\left(x+3 y^{2}\right) \frac{d y}{d x}=y$ $\Rightarrow \frac{d y}{d x}=\frac{y}{x+3 y^{2}}$ $\Rightarrow \frac{d x}{d y}=\frac{x+3 y^{2}}{y}=\frac{x}{y}+3 y$ $\Rightarrow \frac{d x}{d y}-\frac{x}{y}=3 y$ This is a linear differential equation of the form: $\frac{d x}{d y}+p x=Q$ (where $p=-\frac{1}{y}$ and $Q=3 y$ ) The general solution of the given differential equation is given by the relation, $x($ I.F. $)=\int($ Q $\times$ I...
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