Find the intercepts cut off by the plane
Question: Find the intercepts cut off by the plane $2 x+y-z=5$ Solution: $2 x+y-z=5$ $\ldots(1)$ Dividing both sides of equation (1) by 5, we obtain $\frac{2}{5} x+\frac{y}{5}-\frac{z}{5}=1$ $\Rightarrow \frac{x}{\frac{5}{2}}+\frac{y}{5}+\frac{z}{-5}=1$ $\ldots(2)$ It is known that the equation of a plane in intercept form is $\frac{x}{a}+\frac{y}{b}+\frac{z}{c}=1$, where $a, b, c$ are the intercepts cut off by the plane at $x, y$, and $z$ axes respectively. Therefore, for the given equation, $a...
Read More →Prove that:
Question: Prove that: $\cos 6^{\circ} \cos 42^{\circ} \cos 66^{\circ} \cos 78^{\circ}=\frac{1}{16}$ Solution: $\mathrm{LHS}=\cos 6^{\circ} \cos 42^{\circ} \cos 66^{\circ} \cos 78^{\circ}$ $=\frac{1}{4}\left(2 \cos 6^{\circ} \cos 66^{\circ}\right)\left(2 \cos 42^{\circ} \cos 78^{\circ}\right)$ $=\frac{1}{4}\left(\cos 72^{\circ}+\cos 60^{\circ}\right)\left(\cos 120^{\circ}+\cos 36^{\circ}\right) \quad[\because 2 \cos A \cos B=\cos (A+B)+\cos (A-B)]$ $=\frac{1}{4}\left\{\cos \left(90^{\circ}-72^{\c...
Read More →Find the median of the data:
Question: Find the median of the data: 31, 38, 27, 28, 36, 25, 35, 40 Solution: Numbers are 31, 38, 27, 28, 36, 25, 35, 40 Arranging the numbers in ascending order 25, 27, 28, 31, 35, 36, 38, 40 n = 8 (even) $\therefore$ Median $=\frac{\overline{2}^{\text {th }} \text { value }+\left(\frac{\mathrm{n}}{2}+1\right)^{\text {th }} \text { value }}{2}$ $=\frac{\frac{8}{2}^{\text {th }} \text { value }+\left(\frac{8}{2}+1\right)^{\text {th }} \text { value }}{2}$ $=\frac{4^{\text {th }} \text { value ...
Read More →Find the equations of the planes that passes through three points.
Question: Find the equations of the planes that passes through three points. (a) (1, 1, 1), (6, 4, 5), (4, 2, 3) (b) (1, 1, 0), (1, 2, 1), (2, 2, 1) Solution: (a)The given points are A (1, 1, 1), B (6, 4, 5), and C (4, 2, 3) $\left|\begin{array}{lll}1 1 -1 \\ 6 4 -5 \\ -4 -2 3\end{array}\right|=(12-10)-(18-20)-(-12+16)$ $=2+2-4$ $=0$ Since A, B, C are collinear points, there will be infinite number of planes passing through the given points. (b)The given points are A (1, 1, 0), B (1, 2, 1), and ...
Read More →Prove that:
Question: Prove that: $\cos \frac{\pi}{15} \cos \frac{2 \pi}{15} \cos \frac{4 \pi}{15} \cos \frac{7 \pi}{15}=\frac{1}{16}$ Solution: $\mathrm{LHS}=\cos \frac{\pi}{15} \cos \frac{2 \pi}{15} \cos \frac{4 \pi}{15} \cos \frac{7 \pi}{15}$ $=\frac{2 \sin \frac{\pi}{15} \cos \frac{\pi}{15}}{2 \sin \frac{\pi}{15}} \cos \frac{2 \pi}{15} \cos \frac{4 \pi}{15} \cos \frac{7 \pi}{15}$ $\left[\right.$ On dividing and multiplying by $\left.2 \sin \frac{\pi}{15}\right]$ $=\frac{2 \sin \frac{2 \pi}{15} \times \c...
Read More →Find the median of the data:
Question: Find the median of the data: 133, 73, 89, 108, 94, 104, 94, 85, 100, 120 Solution: Numbers are 133, 73, 89, 108, 94, 104, 94, 85, 100, 120 Arranging the numbers in ascending order 73, 85, 89, 94,94, 100, 104, 108, 120, 133 n = 10 (even) $\therefore$ Median $=\frac{\frac{\mathrm{n}}{2} \text { th value }+\left(\frac{\mathrm{n}}{2}+1\right) \text { th value }}{2}$ $=\frac{\frac{10^{\text {th }}}{2} \text { value }+\left(\frac{10}{2}+1\right)^{\text {th }} \text { value }}{2}$ $=\frac{5^{...
Read More →Find the median of the data:
Question: Find the median of the data: 83, 37, 70, 29, 45, 63, 41, 70, 34, 54 Solution: Numbers are 83, 37, 70, 29, 45, 63, 41, 70, 34, 54 Arranging the numbers in ascending order 29, 34, 37, 41, 45, 54, 63, 70, 70, 83 n = 10 (even) $\therefore$ Median $=\frac{\frac{\mathrm{n}}{2} \text { th value }+\left(\frac{\mathrm{n}}{2}+1\right) \text { th value }}{2}$ $=\frac{\frac{10}{2} \text { th value }+\left(\frac{10}{2}+1\right) \text { th value }}{2}$ $=\frac{5^{\text {th }} \text { value }+6^{\tex...
Read More →Prove that:
Question: Prove that: $\cos 78^{\circ} \cos 42^{\circ} \cos 36^{\circ}=\frac{1}{8}$ Solution: $\mathrm{LHS}=\cos 78^{\circ} \cos 42^{\circ} \cos 36^{\circ}$ $=\frac{\left(2 \cos 78^{\circ} \cos 42^{\circ}\right)}{2} \cos 36^{\circ}$ $=\frac{\cos \left(78^{\circ}+42^{\circ}\right)+\cos \left(78^{\circ}-42^{\circ}\right)}{2} \times \cos 36^{\circ}$ $[2 \cos A \cos B=\cos (A+B)+\cos (A-B)]$ $=\frac{1}{2}\left(\cos 120^{\circ}+\cos 36^{\circ}\right) \cos 36^{\circ}$ $=\frac{1}{2}\left(-\cos \left(18...
Read More →Find the vector and Cartesian equation of the planes
Question: Find the vector and Cartesian equation of the planes (a) that passes through the point $(1,0,-2)$ and the normal to the plane is $\hat{i}+\hat{j}-\hat{k}$. (b) that passes through the point $(1,4,6)$ and the normal vector to the plane is $\hat{i}-2 \hat{j}+\hat{k}$. Solution: (a) The position vector of point $(1,0,-2)$ is $\vec{a}=\hat{i}-2 \hat{k}$ The normal vector $\vec{N}$ perpendicular to the plane is $\vec{N}=\hat{i}+\hat{j}-\hat{k}$ The vector equation of the plane is given by, ...
Read More →In a ∆ABC, right angled at A,
Question: In a $\Delta \mathrm{ABC}$, right angled at $\mathrm{A}$, if $\tan \mathrm{C}=\sqrt{3}$, find the value of $\sin \mathrm{B} \cos \mathrm{C}+\cos \mathrm{B} \sin \mathrm{C}$. Solution: Given: $\tan C=\sqrt{3}$ To find: $\sin B \cos C+\cos B \sin C$ The given $\triangle A B C$ is as shown in figure below SideBCis unknown and can be found using Pythagoras theorem Therefore, $B C^{2}=A B^{2}+A C^{2}$ Now by substituting the value of known sides from figure (a) We get, $B C^{2}=(\sqrt{3})^{...
Read More →Find the missing frequencies in the following frequency distribution if it is known that the mean of the distribution is 50.
Question: Find the missing frequencies in the following frequency distribution if it is known that the mean of the distribution is 50. Solution: It is given that Mean = 50 $\Rightarrow \sum f x / N=50$ $\Rightarrow \frac{3480+30 \mathrm{f}_{1}+70 \mathrm{f}_{2}}{\mathrm{~N}}=50$ $\Rightarrow 3480+30 f_{1}+70 f_{2}=50 \times 120$ $\Rightarrow 30 \mathrm{f}_{1}+70 \mathrm{f}_{3}=6000-3480$ $\Rightarrow 10\left(3 f_{1}+7 f_{2}\right)=10(252)$ $\Rightarrow 3 \mathrm{f}_{1}+7 \mathrm{f}_{2}=252 \cdot...
Read More →Prove that:
Question: Prove that: $\sin ^{2} 42^{\circ}-\cos ^{2} 78=\frac{\sqrt{5}+1}{8}$ Solution: $\mathrm{LHS}=\sin ^{2} 42^{\circ}-\cos ^{2} 78^{\circ}$ $=\sin ^{2}\left(90^{\circ}-48^{\circ}\right)-\cos ^{2}\left(90^{\circ}-12^{\circ}\right)$ $=\cos ^{2} 48^{\circ}-\sin ^{2} 12^{\circ}$ $=\cos \left(48^{\circ}+12^{\circ}\right) \cos \left(48^{\circ}-12^{\circ}\right) \quad\left[\cos (\mathrm{A}+\mathrm{B}) \cos (\mathrm{A}-\mathrm{B})=\cos ^{2} \mathrm{~A}-\sin ^{2}\right]$ $=\cos 60^{\circ} \cos 36^{...
Read More →Prove that:
Question: Prove that: $\sin ^{2} 24^{\circ}-\sin ^{2} 6^{\circ}=\frac{\sqrt{5}-1}{8}$ Solution: LHS $=\sin ^{2} 24^{\circ}-\sin ^{2} 6^{\circ}$ $=\sin \left(24^{\circ}+6^{\circ}\right) \sin \left(24^{\circ}-6^{\circ}\right) \quad\left[\sin (\mathrm{A}+\mathrm{B}) \sin (\mathrm{A}-\mathrm{B})=\sin ^{2} \mathrm{~A}-\sin ^{2} \mathrm{~B}\right]$ $=\sin 30^{\circ} \sin 18^{\circ}$ $=\frac{1}{2} \times \frac{\sqrt{5}-1}{4} \quad\left(\because \sin 18^{\circ}=\frac{\sqrt{5}-1}{4}\right)$ $=\frac{\sqrt...
Read More →Prove that:
Question: Prove that: $\sin ^{2} \frac{2 \pi}{5}-\sin ^{2-} \frac{\pi}{3}=\frac{\sqrt{5}-1}{8}$ Solution: $\frac{2 \pi}{5}=72^{\circ}, \frac{\pi}{2}=60^{\circ}$ $\mathrm{LHS}=\sin ^{2} 72^{\circ}-\sin ^{2} 60^{\circ}$ $=\sin ^{2}\left(90^{\circ}-18^{\circ}\right)-\frac{3}{4}$ $=\cos ^{2} 18^{\circ}-\frac{3}{4} \quad\left(\because \sin \left(90^{\circ}-\theta\right)=\cos \theta\right)$ $=\left(\frac{\sqrt{10+2 \sqrt{5}}}{4}\right)^{2}-\frac{3}{4} \quad\left(\because \cos 18^{\circ}=\frac{\sqrt{10...
Read More →In the following cases, find the coordinates of the foot of the perpendicular drawn from the origin.
Question: In the following cases, find the coordinates of the foot of the perpendicular drawn from the origin. (a) $2 x+3 y+4 z-12=0$ (b) $3 y+4 z-6=0$ (c) $x+y+z=1$ (d) $5 y+8=0$ Solution: (a)Let the coordinates of the foot of perpendicular P from the origin to the plane be (x1,y1,z1). $2 x+3 y+4 z-12=0$ $\Rightarrow 2 x+3 y+4 z=12$ The direction ratios of normal are 2, 3, and 4. $\therefore \sqrt{(2)^{2}+(3)^{2}+(4)^{2}}=\sqrt{29}$ Dividing both sides of equation (1) by $\sqrt{29}$, we obtain ...
Read More →Prove that:
Question: Prove that $\left|\cos x \cos \left(\frac{\pi}{3}-x\right) \cos \left(\frac{\pi}{3}+x\right)\right| \leq \frac{1}{4}$ for all values of $x$ Solution: $\frac{\pi}{3}=60^{\circ}$ We have: $\left|\cos x \cos \left(60^{\circ}-x\right) \cos \left(60^{\circ}+x\right)\right|$ $=\left|\cos x\left(\cos ^{2} 60^{\circ}-\sin ^{2} x\right)\right| \quad\left[\cos ^{2} \mathrm{~A}-\sin ^{2} \mathrm{~B}=\cos (\mathrm{A}-\mathrm{B}) \cos (\mathrm{A}+\mathrm{B})\right]$ $=\left|\cos x\left(\frac{1}{4}-...
Read More →If ∠A and ∠P are acute angles such that tan A = tan P, then show that ∠A = ∠P.
Question: If $\angle A$ and $\angle P$ are acute angles such that $\tan A=\tan P$, then show that $\angle A=\angle P$. Solution: Given: $\tan A=\tan P$ To show: $\angle A=\angle P$ Consider two right angled triangles $\mathrm{ABC}$ and $\mathrm{PQR}$ such that $\tan A=\tan P$ Therefore we have, $\tan A=\frac{B C}{A B}$ and $\tan P=\frac{Q R}{P Q}$ Since it is given that $\tan A=\tan P$ Therefore, $\frac{B C}{A B}=\frac{Q R}{P Q}$ Now by interchanging position ofABandQRby cross multiplication We ...
Read More →Five coins were simultaneously tossed 1000 times and at each, toss the number of heads was observed.
Question: Five coins were simultaneously tossed 1000 times and at each, toss the number of heads was observed. The number of tosses during which 0, 1, 2, 3, 4 and 5 heads were obtained are shown in the table below. Find the mean number of heads per toss. Solution: $\therefore$ Mean number of heads per toss $=\sum \mathrm{fx} / \mathrm{N}$ =2470/1000 = 2.47...
Read More →Candidates of four schools appear in a mathematics test. The data were as follows:
Question: Candidates of four schools appear in a mathematics test. The data were as follows: If the average score of the candidates of all four schools is 66, Find the number of candidates that appeared from school III. Solution: Given the average score of all schools = 66 $\Rightarrow \frac{\mathrm{N}_{1} \overline{\mathrm{x}}_{1}+\mathrm{N}_{2} \overline{\mathrm{x}}_{2}+\mathrm{N}_{3} \overline{\mathrm{x}}_{3}+\mathrm{N}_{4} \overline{\mathrm{x}}_{4}}{\mathrm{~N}_{1}+\mathrm{N}_{2}+\mathrm{N}_...
Read More →Prove that:
Question: Prove that $\left|\sin x \sin \left(\frac{\pi}{3}-x\right) \sin \left(\frac{\pi}{3}+x\right)\right| \leq \frac{1}{4}$ for all values of $x$ Solution: $\frac{\pi}{3}=60^{\circ}$ We have: $|\sin x \sin (60-x) \sin (60+x)|$ $=\left|\sin x\left(\sin ^{2} 60-\sin ^{2} x\right)\right|$ $\left[\because \sin (A+B) \sin (A-B)=\sin ^{2} A-\sin ^{2} B\right]$ $=\left|\sin x\left(\frac{3}{4}-\sin ^{2} x\right)\right|$ $=\left|\frac{1}{4} \sin x\left(3-4 \sin ^{2} x\right)\right|$ $=\left|\frac{1}{...
Read More →Find the mean of the following distribution:
Question: Find the mean of the following distribution: Solution: $\therefore$ Mean $\overline{\mathrm{x}}=\frac{\sum \mathrm{fx}}{\mathrm{N}}$ = 800/40 = 20....
Read More →If ∠A and ∠B are acute angles such that cos A = cos B,
Question: If $\angle A$ and $\angle B$ are acute angles such that $\cos A=\cos B$, then show that $\angle A=\angle B$. Solution: Given: $\cos A=\cos B$.....(1) To show: $\angle A=\angle B$ $\triangle A B C$ is as shown in figure below Therefore $\frac{A C}{A B}=\frac{B C}{A B}$ Now observe that denominator of above equality is same that isAB Hence $\frac{A C}{A B}=\frac{B C}{A B}$ only when $A C=B C$ Therefore $A C=B C$.....(2) We know that when two sides of a triangle are equal, then angle oppo...
Read More →sin
Question: $\sin ^{3} x+\sin ^{3}\left(\frac{2 \pi}{3}+x\right)+\sin ^{3}\left(\frac{4 \pi}{3}+x\right)=-\frac{3}{4} \sin 3 x$ Solution: $\mathrm{LHS}=\sin ^{3} x+\sin ^{3}\left(\frac{2 \pi}{3}+x\right)+\sin ^{3}\left(\frac{4 \pi}{3}+x\right)$ $=\frac{3 \sin x-\sin 3 x}{4}+\frac{3 \sin \left(\frac{2 \pi}{3}+x\right)-\sin 3\left(\frac{2 \pi}{3}+x\right)}{4}+\frac{3 \sin \left(\frac{4 \pi}{3}+x\right)-\sin 3\left(\frac{4 \pi}{3}+x\right)}{4}$ $\left[\sin ^{3} \theta=\frac{3 \sin \theta-\sin 3 \thet...
Read More →Find the value of p, if the mean of the following distribution is 20.
Question: Find the value of p, if the mean of the following distribution is 20. Solution: It is given that, Mean = 20 $\Rightarrow \frac{\sum f x}{N}=20$ $\Rightarrow \frac{5 p^{2}+100 p+295}{5 p+15}=20$ $\Rightarrow 5 p^{2}+100 p+295=20(5 p+15)$ $\Rightarrow 5 p^{2}+100 p+295=100 p+300$ $\Rightarrow 5 p^{2}=300-295$ $\Rightarrow 5 p^{2}=5$ $\Rightarrow p^{2}=1$ $\Rightarrow p=\pm 1$ Frequency cant be negative. Hence, value of p is 1....
Read More →If cosec A = 2, find the value of 1tan A+sin A1+cos A.
Question: If $\operatorname{cosec} A=2$, find the value of $\frac{1}{\tan A}+\frac{\sin A}{1+\cos A}$. Solution: Given: $\operatorname{cosec} A=2 \ldots \ldots$ (1) To find: $\frac{1}{\tan A}+\frac{\sin A}{1+\cos A}$ Now we know $\operatorname{cosec} A$ is defined as below $\operatorname{cosec} A=\frac{1}{\sin A}$ Therefore, $\sin A=\frac{1}{\operatorname{cosec} A}$ Now by substituting the value of $\operatorname{cosec} A$ from equation (1) We get, $\sin A=\frac{1}{2}$ Now by substituting the va...
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