if the
Question: If $P(A)=\frac{3}{5}$ and $P(B)=\frac{1}{5}$, find $P(A \cap B)$ if $A$ and $B$ are independent events. Solution: It is given that $\mathrm{P}(\mathrm{A})=\frac{3}{5}$ and $\mathrm{P}(\mathrm{B})=\frac{1}{5}$ A and B are independent events. Therefore, $P(A \cap B)=P(A) \cdot P(B)=\frac{3}{5} \cdot \frac{1}{5}=\frac{3}{25}$...
Read More →Locate 10−−√ on the number line.
Question: Locate $\sqrt{10}$ on the number line Solution: To represent $\sqrt{10}$ on the number line, follow the following steps of construction: (i) Mark points 0 and 3 as $O$ and $B$, respectively. (ii) At point $A$, draw $A B \perp O A$ such that $A B=1$ units. (iii) Join OA. (iv) With $O$ as centre and radius $O A$, draw an arc intersecting the number line at point $P$. Thus, point P represents $\sqrt{10}$ on the number line. Justification: In right $\triangle O A B$, Using Pythagoras theor...
Read More →If A and B are events such that P (A|B) = P(B|A), then
Question: If A and B are events such that P (A|B) = P(B|A), then (A) $A \subset B$ but $A \neq B$ (B) $A=B$ (C) $A \cap B=\Phi$ (D) $P(A)=P(B)$ Solution: It is given that, P(A|B) = P(B|A) $\Rightarrow \frac{P(A \cap B)}{P(B)}=\frac{P(A \cap B)}{P(A)}$ $\Rightarrow P(A)=P(B)$ Thus, the correct answer is D....
Read More →a cos A + b cos B + c cos C = 2b sin A sin C
Question: acosA+bcosB+ccosC= 2bsinAsinC Solution: By sine rule, we know that $\frac{a}{\sin A}=\frac{b}{\sin B}=\frac{c}{\sin C}=k$ (say) $\Rightarrow a=k \sin A, b=k \sin B, c=k \sin C$ Now, LHS $=a \cos A+b \cos B+c \cos C$ $=k \sin A \cos A+k \sin B \cos B+k \sin C \cos C$ $=\frac{k}{2}(2 \sin A \cos A+2 \sin B \cos B+2 \sin C \cos C)$ $=\frac{k}{2}(\sin 2 A+\sin 2 B+2 \sin C \cos C)$ $=\frac{k}{2}\left(2 \sin \frac{2 A+2 B}{2} \cos \frac{2 A-2 B}{2}+2 \sin C \cos C\right)$ $=\frac{k}{2}(2 \s...
Read More →Locate 3–√ on the number line.
Question: Locate $\sqrt{3}$ on the number line. Solution: To represent $\sqrt{3}$ on the number line, follow the following steps of construction: (i) Mark points 0 and 1 as $\mathrm{O}$ and $\mathrm{A}$, respectively. (ii) At point $A$, draw $A B \perp O A$ such that $A B=1$ units. (iii) Join OB. (iv) At point $B$, draw $D B \perp O A$ such that $D B=1$ units. (v) Join OD. (vi) With $O$ as centre and radius $O D$, draw an arc intersecting the number line at point $Q$. Thus, point $Q$ represents ...
Read More →In ∆ABC, prove that:
Question: In ∆ABC, prove that: $a(\cos B+\cos C-1)+b(\cos C+\cos A-1)+c(\cos A+\cos B-1)=0$ Solution: Consider the LHS of the given equation. LHS $=a(\cos B+\cos C-1)+b(\cos C+\cos A-1)+c(\cos A+\cos B-1)$ $=a \cos B+b \cos C+a \cos C+b \cos A+c \cos A+c \cos B-(a+b+c)$ $=(a \cos B+b \cos A)+(b \cos C+c \cos B)+(a \cos C+c \cos A)-(a+b+c)$ $=c+a+b-(a+b+c) \quad$. (Using projection formula : $a=b \cos C+c \cos B, b=a \cos C+c \cos A, c=a \cos B+b \cos A$ ) $=0=\mathrm{RHS}$ Hence proved....
Read More →if the
Question: If $\mathrm{P}(\mathrm{A})=\frac{1}{2}, \mathrm{P}(\mathrm{B})=0$, then $\mathrm{P}(\mathrm{A} \mid \mathrm{B})$ is (A) 0 (B) $\frac{1}{2}$ (C) not defined (D) 1 Solution: It is given that $\mathrm{P}(\mathrm{A})=\frac{1}{2}$ and $\mathrm{P}(\mathrm{B})=0$ $P(A \mid B)=\frac{P(A \cap B)}{P(B)}=\frac{P(A \cap B)}{0}$ Therefore, P (A|B) is not defined. Thus, the correct answer is C....
Read More →In ∆ABC, prove the following:
Question: In ∆ABC, prove the following: $\frac{c-b \cos A}{b-c \cos A}=\frac{\cos B}{\cos C}$ Solution: LetABCbe any triangle. $\mathrm{LHS}=\frac{c-b \cos A}{b-c \cos A}$ $=\frac{a \cos B+b \cos A-b \cos A}{a \cos C+c \cos A-c \cos A} \quad$ [Using projection formula : $c=a \cos B+b \cos A, b=a \cos C+c \cos A]$ $=\frac{a \cos B}{a \cos C}$ $=\frac{\cos B}{\cos C}=\mathrm{RHS}$ Hence proved....
Read More →Consider the experiment of throwing a die,
Question: Consider the experiment of throwing a die, if a multiple of 3 comes up, throw the die again and if any other number comes, toss a coin. Find the conditional probability of the event the coin shows a tail, given that at least one die shows a 3. Solution: The outcomes of the given experiment can be represented by the following tree diagram. The sample space of the experiment is, $S=\left\{\begin{array}{l}(1, \mathrm{H}),(1, \mathrm{~T}),(2, \mathrm{H}),(2, \mathrm{~T}),(3,1)(3,2),(3,3),(...
Read More →In ∆ABC, prove the following:
Question: In ∆ABC, prove the following: $\left(c^{2}-a^{2}+b^{2}\right) \tan A=\left(a^{2}-b^{2}+c^{2}\right) \tan B=\left(b^{2}-c^{2}+a^{2}\right) \tan C$ Solution: We know that $\cos A=\frac{b^{2}+c^{2}-a^{2}}{2 b c}, \frac{\sin A}{a}=\frac{\sin B}{b}=\frac{\sin C}{c}=k$ So, $\left(c^{2}-a^{2}+b^{2}\right) \tan A=\left(c^{2}-a^{2}+b^{2}\right) \frac{\sin A}{\cos A}$ $=\left(c^{2}-a^{2}+b^{2}\right) \sin A \frac{2 b c}{b^{2}+c^{2}-a^{2}}$ $=2 b c \sin A$ $=2 k a b c \quad \ldots(1)$ $\left(a^{2...
Read More →In Fig. 5.48, AD = 4 cm, BD = 3 cm and CB = 12 cm, find the cot θ.
Question: In Fig. $5.48, A D=4 \mathrm{~cm}, B D=3 \mathrm{~cm}$ and $C B=12 \mathrm{~cm}$, find the $\cot \theta$. (a) $\frac{12}{5}$ (b) $\frac{5}{12}$ (C) $\frac{13}{12}$ (d) $\frac{12}{13}$ Solution: We have the following given data in the figure, $A D=4 \mathrm{~cm}, B D=3 \mathrm{~cm}, C B=12 \mathrm{~cm}$ Now we will use Pythagoras theorem in $\triangle A B D$, $A B=\sqrt{3^{2}+4^{2}}$ $=5 \mathrm{~cm}$ Therefore, $\cot \theta=\frac{C B}{A B}$ $=\frac{12}{5}$ So the answer is $(a)$...
Read More →In Fig. 5.47, the value of cos ϕ is
Question: In Fig. 5.47, the value of $\cos \phi$ is (a) $\frac{5}{4}$ (b) $\frac{5}{3}$ (c) $\frac{3}{5}$ (d) $\frac{4}{5}$ Solution: We should proceed with the fact that sum of angles on one side of a straight line is $180^{\circ}$. So from the given figure, $\theta+\phi+90^{\circ}=180^{\circ}$ So, $\theta=90^{\circ}-\phi$......(1) Now from the triangle $\triangle A B C$, $\sin \theta=\frac{4}{5}$ Now we will use equation (1) in the above, $\sin \left(90^{\circ}-\phi\right)=\frac{4}{5}$ Therefo...
Read More →Represent 5–√ on the number line.
Question: Represent $\sqrt{5}$ on the number line. Solution: To represent $\sqrt{5}$ on the number line, follow the following steps of construction: (i) Mark points 0 and 2 as O and P, respectively.(ii) At point A, draw ABOA such that AB = 1 units.(iii) Join OB.(iv) With O as centre and radius OB, draw an arc intersecting the number line at point P. Thus, point represents $\sqrt{5}$ on the number line. Justification:In rightΔ∆OAB,Using Pythagoras theorem, $\mathrm{OB}=\sqrt{\mathrm{OA}^{2}+\math...
Read More →Given that the two numbers appearing on throwing the two dice are different.
Question: Given that the two numbers appearing on throwing the two dice are different. Find the probability of the event the sum of numbers on the dice is 4. Solution: When dice is thrown, number of observations in the sample space = 6 6 = 36 Let A be the event that the sum of the numbers on the dice is 4 and B be the event that the two numbers appearing on throwing the two dice are different. $\therefore \mathrm{A}=\{(1,3),(2,2),(3,1)\}$ $\mathrm{B}=\left\{\begin{array}{lllll}(1,2) (1,3) (1,4) ...
Read More →In ∆ABC, prove the following:
Question: In ∆ABC, prove the following: $2(b c \cos A+c a \cos B+a b \cos C)=a^{2}+b^{2}+c^{2}$ Solution: $\mathrm{LHS}=2(b c \cos A+c a \cos B+a b \cos C)$ On using the cosine law, we get: $\mathrm{LHS}=2\left[b c\left(\frac{b^{2}+c^{2}-a^{2}}{2 b c}\right)+c a\left(\frac{a^{2}+c^{2}-b^{2}}{2 a c}\right)+a b\left(\frac{a^{2}+b^{2}-c^{2}}{2 a b}\right)\right]$ $=b^{2}+c^{2}-a^{2}+a^{2}+c^{2}-b^{2}+a^{2}+b^{2}-c^{2}$ $=a^{2}+b^{2}+c^{2}=\mathrm{RHS}$ Hence proved....
Read More →In ∆ABC, prove the following:
Question: In ∆ABC, prove the following: $2(b c \cos A+c a \cos B+a b \cos C)=a^{2}+b^{2}+c^{2}$ Solution: $\mathrm{LHS}=2(b c \cos A+c a \cos B+a b \cos C)$ On using the cosine law, we get: $\mathrm{LHS}=2\left[b c\left(\frac{b^{2}+c^{2}-a^{2}}{2 b c}\right)+c a\left(\frac{a^{2}+c^{2}-b^{2}}{2 a c}\right)+a b\left(\frac{a^{2}+b^{2}-c^{2}}{2 a b}\right)\right]$ $=b^{2}+c^{2}-a^{2}+a^{2}+c^{2}-b^{2}+a^{2}+b^{2}-c^{2}$ $=a^{2}+b^{2}+c^{2}=\mathrm{RHS}$ Hence proved....
Read More →Simplify
Question: Simplify (i) $3 \sqrt{45}-\sqrt{125}+\sqrt{200}-\sqrt{50}$ (ii) $\frac{2 \sqrt{30}}{\sqrt{6}}-\frac{3 \sqrt{140}}{\sqrt{28}}+\frac{\sqrt{55}}{\sqrt{99}}$ (iii) $\sqrt{72}+\sqrt{800}-\sqrt{18}$ Solution: (i) $3 \sqrt{45}-\sqrt{125}+\sqrt{200}-\sqrt{50}$ $=3 \sqrt{9 \times 5}-\sqrt{25 \times 5}+\sqrt{100 \times 2}-\sqrt{25 \times 2}$ $=3 \times 3 \sqrt{5}-5 \sqrt{5}+10 \sqrt{2}-5 \sqrt{2}$ $=9 \sqrt{5}-5 \sqrt{5}+5 \sqrt{2}$ $=4 \sqrt{5}+5 \sqrt{2}$ (ii) $\frac{2 \sqrt{30}}{\sqrt{6}}-\fr...
Read More →In ∆ABC, prove the following:
Question: In ∆ABC, prove the following: $c(a \cos B-b \cos A)=a^{2}-b^{2}$ Solution: Consider $c(a \cos B-b \cos A)=c a \cos B-c b \cos A$ $=c a\left(\frac{a^{2}+c^{2}-b^{2}}{2 a c}\right)-c b\left(\frac{b^{2}+c^{2}-a^{2}}{2 b c}\right)$ $=\frac{a^{2}+c^{2}-b^{2}-b^{2}-c^{2}+a^{2}}{2}$ $=\frac{2\left(a^{2}-b^{2}\right)}{2}$ $=a^{2}-b^{2}$ Hence proved....
Read More →The value of tan 55°cot 35°+ cot 1° cot 2° cot 3° .... cot 90°, is
Question: The value of $\frac{\tan 55^{\circ}}{\cot 35^{\circ}}+\cot 1^{\circ} \cot 2^{\circ} \cot 3^{\circ}$ $\cot 90^{\circ}$, is (a) 2(b) 2(c) 1(d) 0 Solution: We have to find the value of the following expression $\frac{\tan 55^{\circ}}{\cot 35^{\circ}}+\cot 1^{\circ} \cot 2^{\circ} \cot 3^{\circ} \ldots \cot 90^{\circ}$ $=\frac{\tan 55^{\circ}}{\cot 35^{\circ}}+\cot 1^{\circ} \cot 2^{\circ} \cot 3 \ldots \cot 90^{\circ}$ $=\frac{\tan \left(90^{\circ}-35^{\circ}\right)}{\cot 35^{\circ}}+\cot...
Read More →In ∆ABC, prove the following:
Question: In ∆ABC, prove the following: $c(a \cos B-b \cos A)=a^{2}-b^{2}$ Solution: Consider $c(a \cos B-b \cos A)=c a \cos B-c b \cos A$ $=c a\left(\frac{a^{2}+c^{2}-b^{2}}{2 a c}\right)-c b\left(\frac{b^{2}+c^{2}-a^{2}}{2 b c}\right)$ $=\frac{a^{2}+c^{2}-b^{2}-b^{2}-c^{2}+a^{2}}{2}$ $=\frac{2\left(a^{2}-b^{2}\right)}{2}$ $=a^{2}-b^{2}$ Hence proved....
Read More →An instructor has a question bank consisting of 300 easy True/False questions,
Question: An instructor has a question bank consisting of 300 easy True/False questions, 200 difficult True/False questions, 500 easy multiple choice questions and 400 difficult multiple choice questions. If a question is selected at random from the question bank, what is the probability that it will be an easy question given that it is a multiple choice question? Solution: The given data can be tabulated as Let us denote E = easy questions, M = multiple choice questions, D = difficult questions...
Read More →In ∆ABC, prove the following:
Question: In ∆ABC, prove the following: $b(c \cos A-a \cos C)=c^{2}-a^{2}$ Solution: LetABCbe any triangle. Consider $b(c \cos A-a \cos C)=b c \cos A-a b \cos C$ $=b c\left(\frac{b^{2}+c^{2}-a^{2}}{2 b c}\right)-a b\left(\frac{a^{2}+b^{2}-c^{2}}{2 a b}\right)$ $=\frac{b^{2}+c^{2}-a^{2}-a^{2}-b^{2}+c^{2}}{2}$ $=\frac{2\left(c^{2}-a^{2}\right)}{2}$ $=c^{2}-a^{2}$ Hence proved....
Read More →On her birthday Reema distributed chocolates in an orphanage.
Question: On her birthday Reema distributed chocolates in an orphanage. The total number of chocolates she distributed is given by $(5+\sqrt{11})(5-\sqrt{11})$. (i) Find the number of chocolates distributed by her.(ii) Write the moral values depicted here by Reema. Solution: (i) As, $(5+\sqrt{11})(5-\sqrt{11})$ $=5^{2}-(\sqrt{11})^{2} \quad\left[(a+b)(a-b)=a^{2}-b^{2}\right]$ $=25-11$ $=14$ Hence, the number of chocolates distributed by Reema is 14.(ii) The moral values depicted here by Reema is...
Read More →In ∆ ABC, if a = 18, b = 24 and c = 30,
Question: In ∆ABC, ifa= 18,b= 24 andc= 30, find cosA, cosBand cosC. Solution: GIven : $a=18, b=24$ and $c=30$. $\cos A=\frac{b^{2}+c^{2}-a^{2}}{2 b c}=\frac{576+900-324}{2 \times 24 \times 30}=\frac{1152}{1140}=\frac{4}{5}$ $\cos B=\frac{a^{2}+c^{2}-b^{2}}{2 a c}=\frac{324+900-576}{2 \times 18 \times 30}=\frac{648}{1080}=\frac{3}{5}$ $\cos C=\frac{a^{2}+b^{2}-c^{2}}{2 a b}=\frac{576+324-900}{2 \times 24 \times 18}=0$ Hence, $\cos A=\frac{4}{5}, \cos B=\frac{3}{5}, \cos C=0$...
Read More →tan 5° ✕ tan 30° ✕ 4 tan 85° is equal to
Question: $\tan 5^{\circ} \times \tan 30^{\circ} \times 4 \tan 85^{\circ}$ is equal to (a) $\frac{4}{\sqrt{3}}$ (b) $4 \sqrt{3}$ (c) 1 (d) 4 Solution: We have to find $\tan 5^{\circ} \times \tan 30^{\circ} \times 4 \tan 85^{\circ}$ We know that $\tan \left(90^{\circ}-\theta\right)=\cot \theta$ $\tan \theta \cot \theta=1$ $\tan 30^{\circ}=\frac{1}{\sqrt{3}}$ So $\tan 5 \times \tan 30^{\circ} \times 4 \tan 85^{\circ}$ $=\tan \left(90^{\circ}-85^{\circ}\right) \times \tan 30^{\circ} \times 4 \tan 8...
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