if the
Question: If $\vec{a} \cdot \vec{a}=0$ and $\vec{a} \cdot \vec{b}=0$, then what can be concluded about the vector $\vec{b} ?$ Solution: It is given that $\vec{a} \cdot \vec{a}=0$ and $\vec{a} \cdot \vec{b}=0$. Now, $\vec{a} \cdot \vec{a}=0 \Rightarrow|\vec{a}|^{2}=0 \Rightarrow|\vec{a}|=0$ $\therefore \vec{a}$ is a zero vector. Hence, vector $\vec{b}$ satisfying $\vec{a} \cdot \vec{b}=0$ can be any vector....
Read More →sin 47° + sin 61° − sin 11° − sin 25° is equal to
Question: sin 47 + sin 61 sin 11 sin 25 is equal to (a) sin 36 (b) cos 36 (c) sin 7 (d) cos 7 Solution: (d) cos 7 $\sin 47^{\circ}+\sin 61^{\circ}-\sin 11^{\circ}-\sin 25^{\circ}$ $=\sin 47^{\circ}-\sin 25^{\circ}+\sin 61^{\circ}-\sin 11^{\circ}$ $=2 \sin \left(\frac{47^{\circ}-25^{\circ}}{2}\right) \cos \left(\frac{47^{\circ}+25^{\circ}}{2}\right)+2 \sin \left(\frac{61^{\circ}-11^{\circ}}{2}\right) \cos \left(\frac{61^{\circ}+11^{\circ}}{2}\right)$ $=2 \sin 11^{\circ} \cos 36^{\circ}+2 \sin 25^...
Read More →sin 47° + sin 61° − sin 11° − sin 25° is equal to
Question: sin 47 + sin 61 sin 11 sin 25 is equal to (a) sin 36 (b) cos 36 (c) sin 7 (d) cos 7 Solution: (d) cos 7 $\sin 47^{\circ}+\sin 61^{\circ}-\sin 11^{\circ}-\sin 25^{\circ}$ $=\sin 47^{\circ}-\sin 25^{\circ}+\sin 61^{\circ}-\sin 11^{\circ}$ $=2 \sin \left(\frac{47^{\circ}-25^{\circ}}{2}\right) \cos \left(\frac{47^{\circ}+25^{\circ}}{2}\right)+2 \sin \left(\frac{61^{\circ}-11^{\circ}}{2}\right) \cos \left(\frac{61^{\circ}+11^{\circ}}{2}\right)$ $=2 \sin 11^{\circ} \cos 36^{\circ}+2 \sin 25^...
Read More →If cot θ=78, evaluate :
Question: If $\cot \theta=\frac{7}{8}$, evaluate : (i) $\frac{(1+\sin \theta)(1-\sin \theta)}{(1+\cos \theta)(1-\cos \theta)}$ (ii) $\cot ^{2} \theta$ Solution: (i) Given: $\cot \theta=\frac{7}{8}$ To evaluate: $\frac{(1+\sin \theta)(1-\sin \theta)}{(1+\cos \theta)(1-\cos \theta)}$ $\frac{(1+\sin \theta)(1-\sin \theta)}{(1+\cos \theta)(1-\cos \theta)}$....(1) We know the following formula $(a+b)(a-b)=a^{2}-b^{2}$ By applying the above formula in the numerator of equation (1) , We get, $(1+\sin \...
Read More →Show that
Question: Show that $|\vec{a}| \vec{b}+|\vec{b}| \vec{a}$ is perpendicular to $|\vec{a}| \vec{b}-|\vec{b}| \vec{a}$, for any two nonzero vectors $\vec{a}$ and $\vec{b}$ Solution: $(|\vec{a}| \vec{b}+|\vec{b}| \vec{a}) \cdot(|\vec{a}| \vec{b}-|\vec{b}| \vec{a})$ $=|\vec{a}|^{2} \vec{b} \cdot \vec{b}-|\vec{a}||\vec{b}| \vec{b} \cdot \vec{a}+|\vec{b}||\vec{a}| \vec{a} \cdot \vec{b}-|\vec{b}|^{2} \vec{a} \cdot \vec{a}$ $=|\vec{a}|^{2}|\vec{b}|^{2}-|\vec{b}|^{2}|\vec{a}|^{2}$ $=0$ Hence, $|\vec{a}| \...
Read More →The value of sin 50° − sin 70° + sin 10° is equal to
Question: The value of sin 50 sin 70 + sin 10 is equal to (a) 1 (b) 0 (c) 1/2 (d) 2 Solution: (b) 0 $\sin 50^{\circ}-\sin 70^{\circ}+\sin 10^{\circ}$ $=2 \sin \left(\frac{50^{\circ}-70^{\circ}}{2}\right) \cos \left(\frac{50^{\circ}+70^{\circ}}{2}\right)+\sin 10^{\circ} \quad\left[\because \sin A-\sin B=2 \sin \left(\frac{A-B}{2}\right) \cos \left(\frac{A+B}{2}\right)\right]$ $=2 \sin \left(-10^{\circ}\right) \cos 60^{\circ}+\sin 10^{\circ}$ $=2 \times \frac{1}{2} \sin \left(-10^{\circ}\right)+\s...
Read More →The value of sin 50° − sin 70° + sin 10° is equal to
Question: The value of sin 50 sin 70 + sin 10 is equal to (a) 1 (b) 0 (c) 1/2 (d) 2 Solution: (b) 0 $\sin 50^{\circ}-\sin 70^{\circ}+\sin 10^{\circ}$ $=2 \sin \left(\frac{50^{\circ}-70^{\circ}}{2}\right) \cos \left(\frac{50^{\circ}+70^{\circ}}{2}\right)+\sin 10^{\circ} \quad\left[\because \sin A-\sin B=2 \sin \left(\frac{A-B}{2}\right) \cos \left(\frac{A+B}{2}\right)\right]$ $=2 \sin \left(-10^{\circ}\right) \cos 60^{\circ}+\sin 10^{\circ}$ $=2 \times \frac{1}{2} \sin \left(-10^{\circ}\right)+\s...
Read More →The class size of distribution is 25 and the first class-interval is 200 - 224. There are seven class-intervals.
Question: The class size of distribution is 25 and the first class-interval is 200 - 224. There are seven class-intervals. (i) Write the class-intervals. (ii) Write the class marks of each interval. Solution: Given Class size = 25 First class interval = 200 - 224 (i) Seven class interval are: 200 - 240, 225 - 249, 250 - 274, 275 - 299, 300 - 324, 325 - 349,3 50 - 374. (ii) Class mark = 200 - 224 $=\frac{200+224}{2}$ =424/2 = 212 Class mark = 225 - 249 $=\frac{225+249}{2}$ =479/2 = 237 Class mark...
Read More →If
Question: If $\vec{a}=2 \hat{i}+2 \hat{j}+3 \hat{k}, \vec{b}=-\hat{i}+2 \hat{j}+\hat{k}$ and $\vec{c}=3 \hat{i}+\hat{j}$ are such that $\vec{a}+\lambda \vec{b}$ is perpendicular to $\vec{c}$, then find the value of $\lambda$. Solution: The given vectors are $\vec{a}=2 \hat{i}+2 \hat{j}+3 \hat{k}, \vec{b}=-\hat{i}+2 \hat{j}+\hat{k}$, and $\vec{c}=3 \hat{i}+\hat{j}$. Now, $\vec{a}+\lambda \vec{b}=(2 \hat{i}+2 \hat{j}+3 \hat{k})+\lambda(-\hat{i}+2 \hat{j}+\hat{k})=(2-\lambda) \hat{i}+(2+2 \lambda) ...
Read More →cos 35° + cos 85° + cos 155° =
Question: cos 35 + cos 85 + cos 155 = (a) 0 (b) $\frac{1}{\sqrt{3}}$ (c) $\frac{1}{\sqrt{2}}$ (d) cos 275 Solution: (a) 0 $\cos 35^{\circ}+\cos 85^{\circ}+\cos 155^{\circ}$ $=2 \cos \left(\frac{35^{\circ}+85^{\circ}}{2}\right) \cos \left(\frac{35^{\circ}-85^{\circ}}{2}\right)+\cos 155^{\circ} \quad\left[\because \cos A+\cos B=2 \cos \left(\frac{A+B}{2}\right) \cos \left(\frac{A-B}{2}\right)\right]$ $=2 \cos 60^{\circ} \cos \left(-25^{\circ}\right)+\cos 155^{\circ}$ $=2 \times \frac{1}{2} \cos 25...
Read More →cos 35° + cos 85° + cos 155° =
Question: cos 35 + cos 85 + cos 155 = (a) 0 (b) $\frac{1}{\sqrt{3}}$ (c) $\frac{1}{\sqrt{2}}$ (d) cos 275 Solution: (a) 0 $\cos 35^{\circ}+\cos 85^{\circ}+\cos 155^{\circ}$ $=2 \cos \left(\frac{35^{\circ}+85^{\circ}}{2}\right) \cos \left(\frac{35^{\circ}-85^{\circ}}{2}\right)+\cos 155^{\circ} \quad\left[\because \cos A+\cos B=2 \cos \left(\frac{A+B}{2}\right) \cos \left(\frac{A-B}{2}\right)\right]$ $=2 \cos 60^{\circ} \cos \left(-25^{\circ}\right)+\cos 155^{\circ}$ $=2 \times \frac{1}{2} \cos 25...
Read More →If sin α + sin β = a and cos α − cos β = b,
Question: If $\sin \alpha+\sin \beta=a$ and $\cos \alpha-\cos \beta=b$, then $\tan \frac{\alpha-\beta}{2}=$ (a) $-\frac{a}{b}$ (b) $-\frac{b}{a}$ (c) $\sqrt{a^{2}+b^{2}}$ (d) None of these Solution: (b) $-\frac{b}{a}$ Given: sin + sin =a .....(i) cos cos =b .....(ii) Dividing (i) by (ii): $\Rightarrow \frac{\sin \alpha+\sin B}{\cos \alpha-\cos B}=\frac{a}{b}$ $\Rightarrow \frac{2 \sin \left(\frac{\alpha+\beta}{2}\right) \cos \left(\frac{\alpha-\beta}{2}\right)}{-2 \sin \left(\frac{\alpha+\beta}{...
Read More →Find
Question: Find $|\vec{x}|$, if for a unit vector $\vec{a},(\vec{x}-\vec{a}) \cdot(\vec{x}+\vec{a})=12$. Solution: $(\vec{x}-\vec{a}) \cdot(\vec{x}+\vec{a})=12$ $\Rightarrow \vec{x} \cdot \vec{x}+\vec{x} \cdot \vec{a}-\vec{a} \cdot \vec{x}-\bar{a} \cdot \vec{a}=12$ $\Rightarrow|\vec{x}|^{2}-|\vec{a}|^{2}=12$ $\Rightarrow|\vec{x}|^{2}-1=12 \quad[|\vec{a}|=1$ as $\vec{a}$ is a unit vector $]$ $\Rightarrow|\vec{x}|^{2}=13$ $\therefore|\vec{x}|=\sqrt{13}$...
Read More →The number of runs scored by a cricket player in 25 innings is as follows:
Question: The number of runs scored by a cricket player in 25 innings is as follows: 26, 35, 94, 48, 82, 105, 53, 0, 39, 42, 71, 0, 64, 15, 34, 15, 34, 6, 71, 0, 64, 15, 34, 15, 34, 67, 0, 42, 124, 84, 54, 48, 139, 64, 47 (i) Rearrange these runs in ascending order. (ii) Determine the player, is highest score. (iii) How many times did the player not score a run? (iv) How many centuries did he score? (v) How many times did he score more than 50 runs? Solution: The numbers of runs scored by a play...
Read More →If sin α + sin β = a and cos α − cos β = b,
Question: If $\sin \alpha+\sin \beta=a$ and $\cos \alpha-\cos \beta=b$, then $\tan \frac{\alpha-\beta}{2}=$ (a) $-\frac{a}{b}$ (b) $-\frac{b}{a}$ (c) $\sqrt{a^{2}+b^{2}}$ (d) None of these Solution: (b) $-\frac{b}{a}$ Given: sin + sin =a .....(i) cos cos =b .....(ii) Dividing (i) by (ii): $\Rightarrow \frac{\sin \alpha+\sin B}{\cos \alpha-\cos B}=\frac{a}{b}$ $\Rightarrow \frac{2 \sin \left(\frac{\alpha+\beta}{2}\right) \cos \left(\frac{\alpha-\beta}{2}\right)}{-2 \sin \left(\frac{\alpha+\beta}{...
Read More →The weights of new born babies are as follows:
Question: The weights of new born babies are as follows:2.3, 2.2, 2.1, 2.7, 2.6, 2.5, 3.0, 2.8, 2.8, 2.9, 3.1, 2.5, 2.8, 2.7, 2.9, 2. 4. (i) Rearrange the weights in descending order. (ii) What is the highest weight? (iii) What is the lowest weight? (iv) Determine the range? (v) How many babies were born on that day? (vi) How many babies weigh below 2.5 Kg? (vii) How many babies weigh more than 2.8 Kg? (viii) How many babies weigh 2.8 Kg? Solution: The weights of new born babies(in kg) are as fo...
Read More →Find the magnitude of two vectors
Question: Find the magnitude of two vectors $\vec{a}$ and $\vec{b}$, having the same magnitude and such that the angle between them is $60^{\circ}$ and their scalar product is $\frac{1}{2}$. Solution: Let $\theta$ be the angle between the vectors $\vec{a}$ and $\vec{b}$. It is given that $|\vec{a}|=|\vec{b}|, \vec{a} \cdot \vec{b}=\frac{1}{2}$, and $\theta=60^{\circ}$. $\ldots(1)$ We know that $\vec{a} \cdot \vec{b}=|\vec{a}||\vec{b}| \cos \theta$. $\therefore \frac{1}{2}=|\vec{a}||\vec{a}| \cos...
Read More →In ∆PQR, right angled at Q, PQ = 4 cm and RQ = 3 cm.
Question: In ∆PQR, right angled at Q, PQ = 4 cm and RQ = 3 cm. Find the values of sin P, sin R, sec P and sec R. Solution: Given: $\triangle P Q R$ is right angled at vertex $Q$. $P Q=4 \mathrm{~cm}$ $R Q=3 \mathrm{~cm}$ To find: $\sin P, \sin R, \sec P, \sec R$ Given $\triangle P Q R$ is as shown below Hypotenuse sidePRis unknown. Therefore, we find side $P R$ of $\triangle P Q R$ by Pythagoras theorem By applying Pythagoras theorem to $\triangle P Q R$ We get, $P R^{2}=P Q^{2}+R Q^{2}$ Substit...
Read More →The value of sin 78° − sin 66° − sin 42° + sin 60° is
Question: The value of sin 78 sin 66 sin 42 + sin 60 is (a) $\frac{1}{2}$ (b) $-\frac{1}{2}$ (c) 1 (d) None of these Solution: (d) None of these $\sin 78^{\circ}-\sin 66^{\circ}-\sin 42^{\circ}+\sin 60^{\circ}$ $=\sin 78^{\circ}-\sin 42^{\circ}-\sin 66^{\circ}+\sin 60^{\circ}$ $=2 \sin \left(\frac{78^{\circ}-42^{\circ}}{2}\right) \cos \left(\frac{78^{\circ}+42}{2}\right)-\sin 66^{\circ}+\sin 60^{\circ}\left[\because \sin \mathrm{A}-\sin \mathrm{B}=2 \sin \left(\frac{\mathrm{A}-\mathrm{B}}{2}\rig...
Read More →The value of sin 78° − sin 66° − sin 42° + sin 60° is
Question: The value of sin 78 sin 66 sin 42 + sin 60 is (a) $\frac{1}{2}$ (b) $-\frac{1}{2}$ (c) 1 (d) None of these Solution: (d) None of these $\sin 78^{\circ}-\sin 66^{\circ}-\sin 42^{\circ}+\sin 60^{\circ}$ $=\sin 78^{\circ}-\sin 42^{\circ}-\sin 66^{\circ}+\sin 60^{\circ}$ $=2 \sin \left(\frac{78^{\circ}-42^{\circ}}{2}\right) \cos \left(\frac{78^{\circ}+42}{2}\right)-\sin 66^{\circ}+\sin 60^{\circ}\left[\because \sin \mathrm{A}-\sin \mathrm{B}=2 \sin \left(\frac{\mathrm{A}-\mathrm{B}}{2}\rig...
Read More →Evaluate the product
Question: Evaluate the product $(3 \vec{a}-5 \vec{b}) \cdot(2 \vec{a}+7 \vec{b})$. Solution: $(3 \vec{a}-5 \vec{b}) \cdot(2 \vec{a}+7 \vec{b})$ $=3 \vec{a} \cdot 2 \vec{a}+3 \vec{a} \cdot 7 \vec{b}-5 \vec{b} \cdot 2 \vec{a}-5 \vec{b} \cdot 7 \vec{b}$ $=6 \vec{a} \cdot \vec{a}+21 \vec{a} \cdot \vec{b}-10 \vec{a} \cdot \vec{b}-35 \vec{b} \cdot \vec{b}$ $=6|\vec{a}|^{2}+11 \vec{a} \cdot \vec{b}-35|\vec{b}|^{2}$...
Read More →The final marks in mathematics of 30 students are as follows:
Question: The final marks in mathematics of 30 students are as follows: 53, 61, 48, 60, 78, 68, 55, 100, 67, 0, 75, 88, 77, 37, 84, 58, 60, 48, 62, 56, 44, 58, 52, 64, 98, 59, 70, 39, 50, 60. (i) Arrange these marks in ascending order 30 to 39 one group 40 to 49 second group etc. Now answer the following: (ii) What is the lowest score? (iii) What is the highest score? (iv) What is the range? (v) If 40 is the pass mark how many failed? (vi) How many have scored 75 or more? (vii) Which observation...
Read More →The value of cos 52° + cos 68° + cos 172° is
Question: The value of cos 52 + cos 68 + cos 172 is (a) 0 (b) 1 (c) 2 (d) 3/2 Solution: (a) 0 $\cos 52^{\circ}+\cos 68^{\circ}+\cos 172^{\circ}$ $=2 \cos \left(\frac{52^{\circ}+68^{\circ}}{2}\right) \cos \left(\frac{52^{\circ}-68^{\circ}}{2}\right)+\cos 172^{\circ} \quad\left[\because \cos A+\cos B=2 \cos \left(\frac{A+B}{2}\right) \cos \left(\frac{A-B}{2}\right)\right]$ $=2 \cos 60^{\circ} \cos \left(-8^{\circ}\right)+\cos 172^{\circ}$ $=2 \times \frac{1}{2} \cos 8^{\circ}+\cos 172^{\circ}$ $=\...
Read More →The value of cos 52° + cos 68° + cos 172° is
Question: The value of cos 52 + cos 68 + cos 172 is (a) 0 (b) 1 (c) 2 (d) 3/2 Solution: (a) 0 $\cos 52^{\circ}+\cos 68^{\circ}+\cos 172^{\circ}$ $=2 \cos \left(\frac{52^{\circ}+68^{\circ}}{2}\right) \cos \left(\frac{52^{\circ}-68^{\circ}}{2}\right)+\cos 172^{\circ} \quad\left[\because \cos A+\cos B=2 \cos \left(\frac{A+B}{2}\right) \cos \left(\frac{A-B}{2}\right)\right]$ $=2 \cos 60^{\circ} \cos \left(-8^{\circ}\right)+\cos 172^{\circ}$ $=2 \times \frac{1}{2} \cos 8^{\circ}+\cos 172^{\circ}$ $=\...
Read More →Find
Question: Find $|\vec{a}|$ and $|\vec{b}|$, if $(\vec{a}+\vec{b}) \cdot(\vec{a}-\vec{b})=8$ and $|\vec{a}|=8|\vec{b}|$. Solution: $(\vec{a} \cdot \vec{b}) \cdot(\vec{a}-\vec{b})=8$ $\Rightarrow \vec{a} \cdot \vec{a}-\vec{a} \cdot \vec{b}+\vec{b} \cdot \vec{a}-\vec{b} \cdot \vec{b}=8$ $\Rightarrow|\vec{a}|^{2}-|\vec{b}|^{2}=8$ $\Rightarrow(8|\vec{b}|)^{2}-|\vec{b}|^{2}=8 \quad[|\vec{a}|=8|\vec{b}|]$ $\Rightarrow 64|\vec{b}|^{2}-|\vec{b}|^{2}=8$ $\Rightarrow 63|\vec{b}|^{2}=8$ $\Rightarrow|\vec{b}...
Read More →