If
Question: If $\sin A=\frac{1}{2}, \cos B=\frac{\sqrt{3}}{2}$, where $\frac{\pi}{2}A\pi$ and $0B\frac{\pi}{2}$, find the following: (i) $\tan (A+B)$ (ii) $\tan (A-B)$ Solution: Given: $\sin A=\frac{1}{2}$ and $\cos B=\frac{\sqrt{3}}{2}$ Here, $\frac{\pi}{2}A\pi$ and $0B\frac{\pi}{2}$. That is, $A$ is in the second quadrant and $B$ is in the first quadrant. We know that in the second quadrant, sine function is positive and cosine and tan functions are negative In the first quadrant, all $\mathrm{T...
Read More →Two cones have their heights in the ratio 1:3 and the radii of their bases in the ratio 3:1.
Question: Two cones have their heights in the ratio 1:3 and the radii of their bases in the ratio 3:1.Find the ratio of their volumes. Solution: Let ratio of the height of the cone be $h_{x}$ height of the $1^{\text {st }}$ cone $=h_{x}$ height of the of the $2^{\text {nd }}$ cone $=3 h_{x}$ Let the ratio of the radius of the of the cone =rx radius of the $1^{\text {st }}$ cone $=3 r_{x}$ radius of the $2^{\text {nd }}$ cone $=r_{x}$ The ratio of the volume $=\mathrm{v}_{1} / \mathrm{v}_{2}$ Whe...
Read More →D, E and F are the points on sides BC,
Question: D, E and F are the points on sides BC, CA and AB respectively of ∆ABC such that AD bisects A, BE bisects B and CF bisects C. If AB = 5 cm, BC = 8 cm and CA = 4 cm, determine AF, CE and BD. Solution: It is given that $A B=5 \mathrm{~cm}, B C=8 \mathrm{~cm}$ and $C A=4 \mathrm{~cm}$. We have to find $A F, C E$ and $B D$. Since $A D$ is bisector of $\angle A$ So $\frac{A B}{A C}=\frac{B D}{C D}$ Then, 54=BDBC-BD⇒54=BD8-BD⇒40-5BD=4BD⇒9BD=40 So, $B D=\frac{40}{9}$ Since $B E$ is the bisecto...
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Question: $\left[x \sin ^{2}\left(\frac{y}{x}-y\right)\right] d x+x d y=0 ; y=\frac{\pi}{4}$ when $x=1$ Solution: $\left[x \sin ^{2}\left(\frac{y}{x}\right)-y\right] d x+x d y=0$ $\Rightarrow \frac{d y}{d x}=\frac{-\left[x \sin ^{2}\left(\frac{y}{x}\right)-y\right]}{x}$ ...(1) Let $F(x, y)=\frac{-\left[x \sin ^{2}\left(\frac{y}{x}\right)-y\right]}{x}$. $\therefore F(\lambda x, \lambda y)=\frac{-\left[\lambda x \cdot \sin ^{2}\left(\frac{\lambda x}{\lambda y}\right)-\lambda y\right]}{\lambda x}=\...
Read More →If sin A =
Question: If $\sin A=\frac{1}{2}, \cos B=\frac{12}{13}$, where $\frac{\pi}{2}A\pi$ and $\frac{3 \pi}{2}B2 \pi$, find $\tan (A-B)$. Solution: Given: $\sin A=\frac{1}{2} \quad$ and $\quad \cos B=\frac{12}{13}$ Here, $\frac{\pi}{2}A\pi$ and $\frac{3 \pi}{2}B2 \pi$. That is, $A$ is in the second quadrant and $B$ is in the fourth quadrant. We know that in the second quadrant, $\sin e$ function is positive and $\cos$ ine and tan functions are negative. In the fourth quadran $t, \sin e$ and tan functio...
Read More →Find the volume of a conical tank with the following dimensions in liters:
Question: Find the volume of a conical tank with the following dimensions in liters: (a) Radius is 7cm and the slant height of the cone is 25 cm (b) Slant height is 12cm and height is 13 cm Solution: (a)It is given that Radius of the cone(r) = 7 cm Slant height of the cone (l) = 25 cm As we know that, $1^{2}=r^{2}+h^{2}$ $\mathrm{h}=\sqrt{1^{2}-\mathrm{r}^{2}}$ $\mathrm{~h}=\sqrt{25^{2}-7^{2}}$ $=\mathrm{r}=\sqrt{625-49}$ = 24 cm Volume of a right circular cone $=1 / 3 \pi r^{2} h$ $=1 / 3 * 3.1...
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Question: If $\tan A=\frac{3}{4}, \cos B=\frac{9}{41}$, where $\piA\frac{3 \pi}{2}$ and $0B\frac{\pi}{2}$, find $\tan (A+B)$. Solution: Given: $\tan A=\frac{3}{4}$ and $\cos B=\frac{9}{41}$ Here, $\piA\frac{3 \pi}{2}$ and $0B\frac{\pi}{2}$. That is, $A$ is in third quadrant and $B$ is in first qudrant. We know that tan function is positive in first and third quadrant $s$, and in the first quadrant, $\sin e$ function is also positive. Therefore, $\sin B=\sqrt{1-\cos ^{2} B}$ $=\sqrt{1-\left(\frac...
Read More →In ∆ABC (Fig. 4.59), if ∠1 = ∠2, prove that ABAC=BDDC.
Question: In $\triangle \mathrm{ABC}$ (Fig. 4.59), if $\angle 1=\angle 2$, prove that $\mathrm{ABAC}=\mathrm{BDDC}$. Solution: We have to prove that $\frac{A B}{A C}=\frac{B D}{D C}$. In ∆ABC, $\angle \mathrm{l}=\angle 2$ (Given) So, $A D$ is the bisector of $\angle A$ Therefore, $\frac{A B}{A C}=\frac{B D}{D C}$...
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Question: $x^{2} d y+\left(x y+y^{2}\right) d x=0 ; y=1$ when $x=1$ Solution: $x^{2} d y+\left(x y+y^{2}\right) d x=0$ $\Rightarrow x^{2} d y=-\left(x y+y^{2}\right) d x$ $\Rightarrow \frac{d y}{d x}=\frac{-\left(x y+y^{2}\right)}{x^{2}}$ ...(1) Let $F(x, y)=\frac{-\left(x y+y^{2}\right)}{x^{2}}$. $\therefore F(\lambda x, \lambda y)=\frac{\left[\lambda x \cdot \lambda y+(\lambda y)^{2}\right]}{(\lambda x)^{2}}=\frac{-\left(x y+y^{2}\right)}{x^{2}}=\lambda^{0} \cdot F(x, y)$ Therefore, the given ...
Read More →Find the volume of the right circular cone with the following dimensions:
Question: Find the volume of the right circular cone with the following dimensions: (a) Radius is 6 cm and the height of the cone is 7cm (b) Radius is 3.5 cm and height is 12 cm (c) Slant height is 21 cm and height is 28 cm Solution: (a)It is given that Radius of the cone (r) = 6 cm Height of the cone (h) = 7 cm Volume of a right circular cone $=1 / 3 \pi r^{2} h$ $=1 / 3 * 3.14 * 6^{2} * 7=264 \mathrm{~cm}^{3}$ (b)It is given that: Radius of the cone (r) = 3.5 cm Height of the cone (h) = 12 cm ...
Read More →In Fig. 4.58, ∆ABC is a triangle such that ABAC=BDDC,
Question: In Fig. 4.58, $\triangle \mathrm{ABC}$ is a triangle such that $\mathrm{ABAC}=\mathrm{BDDC}, \angle \mathrm{B}=70^{\circ}, \angle \mathrm{C}=50^{\circ}$. Find the $\angle \mathrm{BAD}$. Solution: It is given that in $\triangle A B C, \frac{A B}{A C}=\frac{B D}{D C}, \angle B=70^{\circ}$ and $\angle C=50^{\circ}$. We have to find $\angle B A D$. In $\triangle A B C$, $\angle A=180^{\circ}-\left(70^{\circ}+50^{\circ}\right)$ $=180^{\circ}-120^{\circ}$ $=60^{\circ}$ Since $\frac{A B}{A C}...
Read More →find the following:
Question: If $\cos A=-\frac{24}{25}$ and $\cos B=\frac{3}{5}$, where $\piA\frac{3 \pi}{2}$ and $\frac{3 \pi}{2}B2 \pi$, find the following: (i) $\sin (A+B)$ (ii) $\cos (A+B)$ Solution: Given: $\cos A=-\frac{24}{25} \quad$ and $\quad \cos B=\frac{3}{5}$ and $\piA\frac{3 \pi}{2} \quad$ and $\quad \frac{3 \pi}{2}B2 \pi$. That is, $A$ is in third quadrant and $\mathrm{B}$ is in fourth qudrant. We know that $\sin e$ function is negative in third and fourth quadrant $s$. Therefore, $\sin A=-\sqrt{1-\c...
Read More →In Fig. 4.57, AE is the bisector of the exterior ∠CAD meeting BC produced in
Question: In Fig. 4.57, AE is the bisector of the exterior CAD meeting BC produced in E. If AB = 10 cm, AC = 6 cm and BC = 12 cm, find CE. Solution: It is given that $A E$ is the bisector of the exterior $\angle C A D$ Meeting $B C$ produced $E$ and $A B=10 \mathrm{~cm}, A C=6 \mathrm{~cm}$ and $B C=12 \mathrm{~cm}$ Since $A E$ is the bisector of the exterior $\angle C A D$ So $\frac{B E}{C E}=\frac{A B}{A C}$ $\frac{12+x}{x}=\frac{10}{6}$ $72+6 x=10 x$ $4 x=72$ $x=18$ Hence $C E=18 \mathrm{~cm}...
Read More →A cylinder and a cone have equal radii of their base and equal heights.
Question: A cylinder and a cone have equal radii of their base and equal heights. If their curved surface area are in the ratio 8:5, show that the radius of each is to the height of each as 3:4. Solution: It is given that the base radius and the height of the cone and the cylinder are the same. So let the base radius of each is 'r' and the vertical height of each is 'h'. Let the slant height of the cone be 'l'. The curved surface area of the cone =rl The curved surface area of the cylinder = It ...
Read More →If a ∆ ABC, AD is the bisector of ∠A, meeting side BC at D.
Question: If a ∆ ABC, AD is the bisector of A, meeting side BC at D. (i) If BD = 2.5 cm, AB = 5 cm and AC = 4.2 cm, find DC.(ii) If BD = 2 cm, AB = 5 cm and DC = 3 cm, find AC.(iii) If AB = 3.5 cm, AC = 4.2 cm and DC = 2.8 cm, find BD.(iv) If AB = 10 cm, AC = 14 cm and BC = 6 cm, find BD and DC.(v) If AC = 4.2 cm, DC = 6 cm and BC = 10 cm. find AB.(vi) If AB = 5.6 cm, AC = 6 cm and DC = 3 cm, find BC.(vii) If AD = 5.6 cm, BC = 6 cm and BD = 3.2 cm, find AC.(viii) If AB = 10 cm, AC = 6 cm and BC ...
Read More →(a) If $sin mid A= rac{12}{13}$ and $sin B= rac{4}{5}$, where $ rac{pi}{2}<A<pi$ and $0<B< rac{pi}{2}$, find the following:
Question: (a) If $\sin \mid A=\frac{12}{13}$ and $\sin B=\frac{4}{5}$, where $\frac{\pi}{2}A\pi$ and $0B\frac{\pi}{2}$, find the following: (i) $\sin (A+B)$ (ii) $\cos (A+B)$ (b) If $\sin A=\frac{3}{5}, \cos B=-\frac{12}{13}$, where $A$ and $B$ both lie in second quadrant, find the value of $\sin (A+B)$. Solution: (a) Given : $\sin A=\frac{12}{13}$ and $\sin B=\frac{4}{5}$ When, $\frac{\pi}{2}A\pi$ and $0B\frac{\pi}{2}$, $\cos A=-\sqrt{1-\sin ^{2} A}$ and $\cos B=\sqrt{1-\sin ^{2} B}$ ( As cosin...
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Question: $(x+y) d y+(x-y) d x=0 ; y=1$ when $x=1$ Solution: $(x+y) d y+(x-y) d x=0$ $\Rightarrow(x+y) d y=-(x-y) d x$ $\Rightarrow \frac{d y}{d x}=\frac{-(x-y)}{x+y}$ $\ldots(1)$ Let $F(x, y)=\frac{-(x-y)}{x+y}$. $\therefore F(\lambda x, \lambda y)=\frac{-(\lambda x-\lambda y)}{\lambda x-\lambda y}=\frac{-(x-y)}{x+y}=\lambda^{0} \cdot F(x, y)$ Therefore, the given differential equation is a homogeneous equation. To solve it, we make the substitution as: y=vx $\Rightarrow \frac{d}{d x}(y)=\frac{...
Read More →A bus stop is barricaded from the remaining part of the road,
Question: A bus stop is barricaded from the remaining part of the road, by using 50 hollow cones made of recycled card board. Each cone has a base diameter of $40 \mathrm{~cm}_{2}$ and height $1 \mathrm{~m}$. If the outer side of each of the cones is to be painted and the cost of painting is Rs 12 per $\mathrm{m}^{2}$, what will be the cost of painting all these cones? Solution: The area to be painted is the curved surface area of each cone. The formula of the curved surface area of a cone with ...
Read More →If sin
Question: If $\sin \mathrm{A}=\frac{4}{5}$ and $\cos \mathrm{B}=\frac{5}{13}$, where $0\mathrm{A}, \mathrm{B}\frac{\pi}{2}$, find the values of the following: (i) sin (A + B) (ii) cos (A + B) (iii) sin (A B) (iv) cos (A B) Solution: Given : $\sin A=\frac{4}{5}$ and $\cos B=\frac{5}{13}$ We know that $\cos A=\sqrt{1-\sin ^{2} A} \quad$ and $\quad \sin B=\sqrt{1-\cos ^{2} B} \quad, \quad$ where $0A, B\frac{\pi}{2}$ $\Rightarrow \cos A=\sqrt{1-\left(\frac{4}{5}\right)^{2}} \quad$ and $\quad \sin B=...
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Question: $\left(1+e^{\frac{x}{y}}\right) d x+e^{\frac{x}{y}}\left(1-\frac{x}{y}\right) d y=0$ Solution: $\left(1+e^{\frac{x}{y}}\right) d x+e^{\frac{x}{y}}\left(1-\frac{x}{y}\right) d y=0$ $\Rightarrow\left(1+e^{\frac{x}{y}}\right) d x=-e^{\frac{x}{y}}\left(1-\frac{x}{y}\right) d y$ $\Rightarrow \frac{d x}{d y}=\frac{-e^{\frac{x}{y}}\left(1-\frac{x}{y}\right)}{1+e^{\frac{x}{y}}}$ ...(1) Let $F(x, y)=\frac{-e^{\frac{x}{y}}\left(1-\frac{x}{y}\right)}{1+e^{\frac{x}{y}}}$. $\therefore F(\lambda x, ...
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Question: $y d x+x \log \left(\frac{y}{x}\right) d y-2 x d y=0$ Solution: $y d x+x \log \left(\frac{y}{x}\right) d y-2 x d y=0$ $\Rightarrow y d x=\left[2 x-x \log \left(\frac{y}{x}\right)\right] d y$ $\Rightarrow \frac{d y}{d x}=\frac{y}{2 x-x \log \left(\frac{y}{x}\right)}$ ...(1) Let $F(x, y)=\frac{y}{2 x-x \log \left(\frac{y}{x}\right)}$. $\therefore F(\lambda x, \lambda y)=\frac{\lambda y}{2(\lambda x)-(\lambda x) \log \left(\frac{\lambda y}{\lambda x}\right)}=\frac{y}{2 x-\log \left(\frac{...
Read More →What length of tarpaulin 4 m wide will be required to make a conical tent of height 8 m and base radius 6 m?
Question: What length of tarpaulin 4 m wide will be required to make a conical tent of height 8 m and base radius 6 m? Assume that the extra length of material will be required for stitching margins and wastage in cutting is approximately 20 cm. Solution: Given that, Height of conical tent (h) = 8 m Radius of base of tent (r) = 6 m Slant height (l) (l) $=\sqrt{\mathrm{r}^{2}+\mathrm{h}^{2}}$ $=\sqrt{8^{2}+6^{2}}$ $=\sqrt{100}=\sqrt{10} \mathrm{~m}$ C.S.A of conical tent $=\pi r \mid$ $=(3.14 * 6...
Read More →The circumference of the base of a 10 m height conical tent is 44m,
Question: The circumference of the base of a 10 m height conical tent is 44m, calculate the length of canvas used in making the tent if width of canvas is 2 m. Solution: We know that C.S.A of cone = rl Given circumference = 2r ⟹ 2 227 r = 44 ⟹ r/7 = 1 ⟹ r = 7 m Therefore $\mathrm{l}=\sqrt{\mathrm{r}^{2}+\mathrm{h}^{2}}$ $=\sqrt{7^{2}+10^{2}}$ $\mathrm{l}=\sqrt{149} \mathrm{~m}$ Therefore C.S.A of tent = rl $=\frac{22}{7} * 7 * \sqrt{149}$ $=22 \sqrt{149}$ Therefore the length of canvas used in m...
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Question: $x \frac{d y}{d x}-y+x \sin \left(\frac{y}{x}\right)=0$ Solution: $x \frac{d y}{d x}-y+x \sin \left(\frac{y}{x}\right)=0$ $\Rightarrow x \frac{d y}{d x}=y-x \sin \left(\frac{y}{x}\right)$ $\Rightarrow \frac{d y}{d x}=\frac{y-x \sin \left(\frac{y}{x}\right)}{x}$ ...(1) Let $F(x, y)=\frac{y-x \sin \left(\frac{y}{x}\right)}{x}$. $\therefore F(\lambda x, \lambda y)=\frac{\lambda y-\lambda x \sin \left(\frac{\lambda y}{\lambda x}\right)}{\lambda x}=\frac{y-x \sin \left(\frac{y}{x}\right)}{x...
Read More →Sketch the graphs of the following curves on the same scale and the same axes:
Question: Sketch the graphs of the following curves on the same scale and the same axes: (i) $y=\cos x$ and $y=\cos \left(x-\frac{\pi}{4}\right)$ (ii) $y=\cos 2 x$ and $y=\cos 2\left(x-\frac{\pi}{4}\right)$ (iii) $y=\cos x$ and $y=\cos \frac{x}{2}$ (iv) $y=\cos ^{2} x$ and $y=\cos x$ Solution: (i) First, we draw the graph ofy= cosx. Let us now draw the graph of $y=\cos \left(x-\frac{\pi}{4}\right)$. $y=\cos \left(\mathrm{x}-\frac{\pi}{4}\right)$ $\Rightarrow y-0=\cos \left(x-\frac{\pi}{4}\right)...
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