A vertical stick 10 cm long casts a shadow 8 cm long.
Question: A vertical stick 10 cm long casts a shadow 8 cm long. At the same time a tower casts a shadow 30 m long. Determine the height of the tower. Solution: Now, $\triangle \mathrm{ABC} \sim \Delta \mathrm{PQR}$ (AA Similarity) $\frac{A B}{B C}=\frac{P Q}{Q R}$ $\frac{10 \mathrm{~cm}}{8 \mathrm{~cm}}=\frac{P Q}{30 \mathrm{~m}}$ $P Q=\frac{30 \mathrm{~m} \times 10 \mathrm{~cm}}{8 \mathrm{~cm}}$ $P Q=\frac{3000 \mathrm{~cm} \times 10 \mathrm{~cm}}{8 \mathrm{~cm}}$ $P Q=\frac{30000 \mathrm{~cm}}...
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Question: Prove that: (i) $\sin \left(\frac{\pi}{3}-x\right) \cos \left(\frac{\pi}{6}+x\right)+\cos \left(\frac{\pi}{3}-x\right) \sin \left(\frac{\pi}{6}+x\right)=1$ (ii) $\sin \left(\frac{4 \pi}{9}+7\right) \cos \left(\frac{\pi}{9}+7\right)-\cos \left(\frac{4 \pi}{9}+7\right) \sin \left(\frac{\pi}{9}+7\right)=\frac{\sqrt{3}}{2}$ (iii) $\sin \left(\frac{3 \pi}{8}-5\right) \cos \left(\frac{\pi}{8}+5\right)+\cos \left(\frac{3 \pi}{8}-5\right) \sin \left(\frac{\pi}{8}+5\right)=1$ Solution: (i) $\fr...
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Question: Prove that: (i) $\sin \left(\frac{\pi}{3}-x\right) \cos \left(\frac{\pi}{6}+x\right)+\cos \left(\frac{\pi}{3}-x\right) \sin \left(\frac{\pi}{6}+x\right)=1$ (ii) $\sin \left(\frac{4 \pi}{9}+7\right) \cos \left(\frac{\pi}{9}+7\right)-\cos \left(\frac{4 \pi}{9}+7\right) \sin \left(\frac{\pi}{9}+7\right)=\frac{\sqrt{3}}{2}$ (iii) $\sin \left(\frac{3 \pi}{8}-5\right) \cos \left(\frac{\pi}{8}+5\right)+\cos \left(\frac{3 \pi}{8}-5\right) \sin \left(\frac{\pi}{8}+5\right)=1$ Solution: (i) $\fr...
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Question: $\frac{d y}{d x}+3 y=e^{-2 x}$ Solution: The given differential equation is $\frac{d y}{d x}+p y=Q$ (where $p=3$ and $Q=e^{-2 x}$ ). Now, I.F $=e^{\int p d x}=e^{\int 3 d x}=e^{3 x}$. The solution of the given differential equation is given by the relation, $y(\mathrm{I} . \mathrm{F} .)=\int(\mathrm{Q} \times \mathrm{I} . \mathrm{F} .) d x+\mathrm{C}$ $\Rightarrow y e^{3 x}=\int\left(e^{-2 x} \times e^{3 x}\right)+\mathrm{C}$ $\Rightarrow y e^{3 x}=\int e^{x} d x+\mathrm{C}$ $\Rightarr...
Read More →In the given figure, ∆ACB ∼ ∆APQ. If BC = 8 cm,
Question: In the given figure, ∆ACB ∆APQ. If BC = 8 cm, PQ = 4 cm, BA = 6.5 cm and AP = 2.8 cm, find CA and AQ. Solution: It is given that $\triangle A C B \sim \triangle A P Q$. $B C=8 \mathrm{~cm}, P Q=4 \mathrm{~cm}, B A=6.5 \mathrm{~cm}$ and $A P=2.8 \mathrm{~cm}$ We have to find $C A$ and $A Q$. Since $\triangle A C B \sim \triangle A P Q$ $\Rightarrow \frac{B A}{A Q}=\frac{C A}{A P}=\frac{B C}{P Q}$ So $\frac{6.5 \mathrm{~cm}}{A Q}=\frac{8 \mathrm{~cm}}{4 \mathrm{~cm}}$ $A Q=\frac{6.5 \mat...
Read More →Find the weight of a solid cone whose base is of diameter 14 cm and vertical height 51 cm,
Question: Find the weight of a solid cone whose base is of diameter 14 cm and vertical height 51 cm, supposing the material of which it is made weighs 10 grams per cubic cm. Solution: It is given that: Diameter (d) = 14 cm Height of the cone (h) = 51cm Radius of the cone(r) = d/2 = 14/2 = 7cm Therefore, Volume of cone $(v)=1 / 3 \pi r^{2} h$ $=1 / 3 * 3.14 * 7 * 5 * 51=2618 \mathrm{~cm}^{3}$ Now it is given that $1 \mathrm{~cm}^{3}$ material weighs $10 \mathrm{gm}$. Therefore, $2618 \mathrm{~cm}...
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Question: $\frac{d y}{d x}+2 y=\sin x$ Solution: The given differential equation is $\frac{d y}{d x}+2 y=\sin x$. This is in the form of $\frac{d y}{d x}+p y=Q($ where $p=2$ and $Q=\sin x)$. Now, I.F $=e^{\int p d x}=e^{\int 2 d x}=e^{2 x}$. The solution of the given differential equation is given by the relation, $y(\mathrm{IF} .)=\int(\mathrm{Q} \times \mathrm{I} . \mathrm{F} .) d x+\mathrm{C}$ $\Rightarrow y e^{2 x}=\int \sin x \cdot e^{2 x} d x+\mathrm{C}$ ...(1) Let $I=\int \sin x \cdot e^{...
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Question: Prove that (i) $\frac{\cos 11^{\circ}+\sin 11^{\circ}}{\cos 11^{\circ}-\sin 11^{\circ}}=\tan 56^{\circ}$. (ii) $\frac{\cos 9^{\circ}+\sin 9^{\circ}}{\cos 9^{\circ}-\sin 9^{\circ}}=\tan 54^{\circ}$ (ii) $\frac{\cos 8^{\circ}-\sin 8^{\circ}}{\cos 8^{\circ}+\sin 8^{\circ}}=\tan 37^{\circ}$ Solution: (i) $\mathrm{LHS}=\frac{\cos 11^{\circ}+\sin 11^{\circ}}{\cos 11^{\circ}-\sin 11^{\circ}}$ $=\frac{\frac{\cos 11^{*}}{\cos 11^{*}}+\frac{\sin 11^{\circ}}{\cos 11^{\circ}}}{\frac{\cos 11^{*}}{\...
Read More →A heap of wheat is in the form of a cone of diameter 9 m and height 3.5 m.
Question: A heap of wheat is in the form of a cone of diameter 9 m and height 3.5 m. Find its volume. How much is canvas cloth required to just cover the heap?(Use = 3.14). Solution: It is given that Diameter of heap (d) = 9 m Therefore, Radius of the heap (r) =d/2 = 9/2 = 4.5 m Height of the heap (h) = 3.5 m Therefore, Volume of the heap $=1 / 3 \pi r^{2} h$ $=1 / 3 * 3.14 * 4.5^{2} * 3.5$ $=74.18 \mathrm{~m}^{3}$ Now, $\mathrm{l}=\sqrt{\mathrm{r}^{2}+\mathrm{h}^{2}}$ $=\sqrt{4.5^{2}+3.5^{2}}=5...
Read More →(i) In the given figure, if AB || CD, find the value of x.
Question: (i) In the given figure, if AB || CD, find the value ofx. (ii) In the given figure, If AB || CD, find the value ofx. (iii) In the given figure, AB || CD, If OA = 3x 19, OB =x 4, OC =x 3 and OD = 4, findx. Solution: (i) It is given that $A B \| C D$. We have to find the value of $x$. Diagonals of the para Now $\frac{D O}{O A}=\frac{C O}{O B}$ $\Rightarrow \frac{4 x-2}{4}=\frac{2 x+4}{x+1}$ $4(2 x+4)=(4 x-2)(x+1)$ $8 x+16=x(4 x-2)+1(4 x-2)$ $8 x+16=4 x^{2}-2 x+4 x-2$ $-4 x^{2}+8 x+16+2-2...
Read More →Which of the following is a homogeneous differential equation?
Question: Which of the following is a homogeneous differential equation? A. $(4 x+6 y+5) d y-(3 y+2 x+4) d x=0$ B. $(x y) d x-\left(x^{3}+y^{3}\right) d y=0$ C. $\left(x^{3}+2 y^{2}\right) d x+2 x y d y=0$ D. $y^{2} d x+\left(x^{2}-x y-y^{2}\right) d y=0$ Solution: Function $\mathrm{F}(x, y)$ is said to be the homogenous function of degree $n$, if $\mathrm{F}(\lambda x, \lambda y)=\lambda^{n} \mathrm{~F}(x, y)$ for any non-zero constant $(\lambda)$. Consider the equation given in alternativeD: $...
Read More →If the radius of the base of a cone is halved, keeping the height same,
Question: If the radius of the base of a cone is halved, keeping the height same, what is the ratio of the volume of the reduced cone to that of the original cone? Solution: Let the radius of the cone berxand height be hx Then the volume of cone $\mathrm{v}_{\mathrm{x}}=\frac{1}{3} \pi \mathrm{r}_{\mathrm{X}}^{2} \mathrm{~h}_{\mathrm{x}}$ Now, Radius of the reduced cone = rx/2 Therefore volume of reduced cone vy $=\frac{1}{3} \pi\left(\frac{r}{2}\right)^{2} * h_{\mathrm{x}}$ $=\frac{1}{12} * \pi...
Read More →A homogeneous differential equation of the form
Question: A homogeneous differential equation of the form $\frac{d x}{d y}=h\left(\frac{x}{y}\right)$ can be solved by making the substitution A. $y=v x$ B. $v=y x$ C. $x=v y$ D. $x=v$ Solution: For solving the homogeneous equation of the form $\frac{d x}{d y}=h\left(\frac{x}{y}\right)$, we need to make the substitution as $x=v y$. Hence, the correct answer is C....
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Question: Prove that $\frac{\tan A+\tan B}{\tan A-\tan B}=\frac{\sin (A+B)}{\sin (A-B)}$. Solution: $\mathrm{LHS}=\frac{\tan A+\tan B}{\tan A-\tan B}$ $=\frac{\frac{\sin A}{\cos A}+\frac{\sin B}{\cos B}}{\frac{\sin A}{\cos A}-\frac{\sin B}{\cos B}}$ $=\frac{\frac{\sin A \cos B+\cos A \sin B}{\cos A \cos B}}{\frac{\sin A \cos B-\cos A \sin B}{\cos A \cos B}}$ $=\frac{\sin A \cos B+\cos A \sin B}{\sin A \cos B-\cos A \sin B}$ $=\frac{\sin (A+B)}{\sin (A-B)}$ = RHS Hemce proved....
Read More →A cylinder and a cone have equal radii of their bases and equal heights.
Question: A cylinder and a cone have equal radii of their bases and equal heights. Show that their volumes are in the ratio 3:1. Solution: It's given that A cylinder and a cone are having equal radii of their bases and heights Let the radius of the cone =radius of the cylinder = r Height of the cone=height of the cylinder = h Let the volume of cone = vx Volume of cylinder = vy $\Rightarrow \frac{v_{x}}{v_{y}}=\frac{\frac{1}{3} \pi r^{2} h}{\pi r^{2} h}=\frac{1}{3}$ ⟹vy/vx= 3/1 Therefore the rati...
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Question: Prove that: $\frac{7 \pi}{12}+\cos \frac{\pi}{12}=\sin \frac{5 \pi}{12}-\sin \frac{\pi}{12}$ Solution: $105^{\circ}=\frac{7 \pi}{12}, 15^{\circ}=\frac{\pi}{12}, 75^{\circ}=\frac{5 \pi}{12}, 15^{\circ}=\frac{\pi}{12}$ LHS = cos105o+ cos15o = cos(90o+ 15o) + cos(90o-75o) = - sin 15o+ sin 75o [As cos(90o+A) =-sinAand cos(90o-B) = sinB] = sin 75o-sin 15o = RHS Hence proved....
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Question: $2 x y+y^{2}-2 x^{2} \frac{d y}{d x}=0 ; y=2$ when $x=1$ Solution: $2 x y+y^{2}-2 x^{2} \frac{d y}{d x}=0$ $\Rightarrow 2 x^{2} \frac{d y}{d x}=2 x y+y^{2}$ $\Rightarrow \frac{d y}{d x}=\frac{2 x y+y^{2}}{2 x^{2}}$ ..(1) Let $F(x, y)=\frac{2 x y+y^{2}}{2 x^{2}}$ $\therefore F(\lambda x, \lambda y)=\frac{2(\lambda x)(\lambda y)+(\lambda y)^{2}}{2(\lambda x)^{2}}=\frac{2 x y+y^{2}}{2 x^{2}}=\lambda^{0} \cdot F(x, y)$ Therefore, the given differential equation is a homogeneous equation. T...
Read More →The ratio of volumes of two cones is 4:5 and the ratio of the radii of their bases is 2:3.
Question: The ratio of volumes of two cones is 4:5 and the ratio of the radii of their bases is 2:3. Find the ratio of their vertical heights. Solution: Let the ratio of the radius be x Radius of $1^{5 \mathrm{st}}$ cone $=2 \mathrm{x}$ Radius of $2^{\text {nd }}$ cone $=3 x$ Let the ratio of the volume be y Volume of $1^{\text {st }}$ cone $=4 \mathrm{y}$ Volume of $2^{\text {nd }}$ cone $=5 y$ $y_{1} / y_{2}=4 y / 5 y$ $=4 / 5$ $\Rightarrow \frac{\frac{1}{3} \pi r_{a}^{2} h_{a}}{\frac{1}{3} \p...
Read More →If cos A =
Question: If $\cos A=-\frac{12}{13}$ and $\cot B=\frac{24}{7}$, where $A$ lies in the second quadrant and $B$ in the third quadrant, find the values of the following: (i) sin (A+B) (ii) cos (A+B) (iii) tan (A+B) Solution: Given: $\cos A=-\frac{12}{13}$ and $\cot B=\frac{24}{7}$ $A$ lies in the second quadrant and $B$ lies in the third quadrant. We know that sine function is positive in the second quadrant and in the third quadrant, both $\sin \mathrm{e}$ and $\cos$ ine functions are negative. Th...
Read More →In Fig. 4.60, AD bisects ∠A, AB = 12 cm, AC = 20 cm and BD = 5 cm, determine CD.
Question: In Fig. 4.60, AD bisects A, AB = 12 cm, AC = 20 cm and BD = 5 cm, determine CD. Solution: It is given that $A D$ bisect $\angle A$. Also, $A B=12 \mathrm{~cm}, A C=20 \mathrm{~cm}$ and $B D=5 \mathrm{~cm}$. We have to find $C D$. Since $A D$ is the bisector of $\angle A$ Then $\frac{A B}{A C}=\frac{B D}{D C}$ $\frac{12 \mathrm{~cm}}{20 \mathrm{~cm}}=\frac{5 \mathrm{~cm}}{D C}$ $12 \mathrm{~cm} \times D C=20 \mathrm{~cm} \times 5 \mathrm{~cm}$ $D C=\frac{100}{12} \mathrm{~cm}$ $=8.33 \m...
Read More →In Fig. 4.60, check whether AD is the bisector of ∠A of ∆ABC in each of the following:
Question: In Fig. 4.60, check whether AD is the bisector of A of ∆ABC in each of the following: (i) AB = 5 cm, AC = 10 cm, BD = 1.5 cm and CD = 3.5 cm(ii) AB = 4 cm, AC = 6 cm, BD = 1.6 cm and CD = 2.4 cm(iii) AB = 8 cm, AC = 24 cm, BD = 6 cm and BC = 24 cm(iv) AB = 6 cm, AC = 8 cm, BD = 1.5 cm and CD = 2 cm(v) AB = 5 cm, AC = 12 cm, BD = 2.5 cm and BC = 9 cm Solution: (i) It is given that $A B=5 \mathrm{~cm}, A C=10 \mathrm{~cm}, B D=1.5 \mathrm{~cm}$ and $C D=3.5 \mathrm{~cm}$. We have to chec...
Read More →The radius and height of a right circular cone are in the ratio 5:12 and its volume is 2512 cubic cm.
Question: The radius and height of a right circular cone are in the ratio 5:12 and its volume is 2512 cubic cm. Find the slant height and radius of the cone. (Use = 3.14). Solution: Let the ratio be y The radius of the cone(r) = 5y Height of the cone = 12y Now we know, Slant height $(\mathrm{l})=\sqrt{\mathrm{r}^{2}+\mathrm{h}^{2}}$ $=\sqrt{5 y^{2}+12 y^{2}}$ = 13y Now the volume of the cone is given $2512 \mathrm{~cm}^{3}$ $\Rightarrow 1 / 3 \pi r^{2} h=2512$ $\Rightarrow 1 / 3 * 3.14 * 5 y^{2}...
Read More →Evaluate the following:
Question: Evaluate the following: (i) sin 78 cos 18 cos 78 sin 18 (ii) cos 47 cos 13 sin 47 sin 13 (iii) sin 36 cos 9 + cos 36 sin 9 (iv) cos 80 cos 20 + sin 80 sin 20 Solution: (i) $\sin 78^{\circ} \cos 18^{\circ}-\cos 78^{\circ} \sin 18^{\circ}$ $=\sin \left(78^{\circ}-18^{\circ}\right)$ $[\mathrm{U} \operatorname{sing} \sin A \cos B-\cos A \sin B=\sin (A-B)]$ $=\sin 60^{\circ}=\frac{\sqrt{3}}{2}$ (ii) $\cos 47^{\circ} \cos 13^{\circ}-\sin 47^{\circ} \sin 13^{\circ}$ $=\cos \left(47^{\circ}+13...
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Question: $\frac{d y}{d x}-\frac{y}{x}+\operatorname{cosec}\left(\frac{y}{x}\right)=0 ; y=0$ when $x=1$ Solution: $\frac{d y}{d x}-\frac{y}{x}+\operatorname{cosec}\left(\frac{y}{x}\right)=0$ $\Rightarrow \frac{d y}{d x}=\frac{y}{x}-\operatorname{cosec}\left(\frac{y}{x}\right)$ ...(1) Let $F(x, y)=\frac{y}{x}-\operatorname{cosec}\left(\frac{y}{x}\right)$. $\therefore F(\lambda x, \lambda y)=\frac{\lambda y}{\lambda x}-\operatorname{cosec}\left(\frac{\lambda y}{\lambda x}\right)$ $\Rightarrow F(\l...
Read More →The radius and the height of a right circular cone are in the ratio 5:12.
Question: The radius and the height of a right circular cone are in the ratio 5:12. If its volume is 314 cubic meter, find the slant height and the radius. (Use = 3.14). Solution: Let us assume the ratio to be y Radius (r) = 5y Height (h) = 12y We know that $1^{2}=r^{2}+h^{2}$ $=5 y^{2}+12 y^{2}$ $=25 y^{2}+144 y^{2}$ $=169^{2}=13 y$ Now it is given that volume $=314 \mathrm{~m}^{3}$ $\Rightarrow 1 / 3 \pi r^{2} h=314 m^{3}$ $\Rightarrow 1 / 3 * 3.14 * 25 y^{2} * 12 y=314 m^{3}$ $\Rightarrow y^{...
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