Answer each of the following questions in one word or one sentence or as per exact requirement of the question.
Question: Answer each of the following questions in one word or one sentence or as per exact requirement of the question. In a $\triangle \mathrm{ABC}$, if $b=\sqrt{3}, c=1$ and $\angle A=30^{\circ}$, find $a$. Solution: In $\triangle \mathrm{ABC}, b=\sqrt{3}, c=1$ and $\angle A=30^{\circ}$. Using cosine formula, we have $\cos A=\frac{b^{2}+c^{2}-a^{2}}{2 b c}$ $\Rightarrow \cos 30^{\circ}=\frac{(\sqrt{3})^{2}+(1)^{2}-a^{2}}{2 \times \sqrt{3} \times 1}$ $\Rightarrow \frac{\sqrt{3}}{2}=\frac{4-a^...
Read More →Answer each of the following questions in one word or one sentence or as per exact requirement of the question.
Question: Answer each of the following questions in one word or one sentence or as per exact requirement of the question. In a $\triangle \mathrm{ABC}$, if $b=\sqrt{3}, c=1$ and $\angle A=30^{\circ}$, find $a$. Solution: In $\triangle \mathrm{ABC}, b=\sqrt{3}, c=1$ and $\angle A=30^{\circ}$. Using cosine formula, we have $\cos A=\frac{b^{2}+c^{2}-a^{2}}{2 b c}$ $\Rightarrow \cos 30^{\circ}=\frac{(\sqrt{3})^{2}+(1)^{2}-a^{2}}{2 \times \sqrt{3} \times 1}$ $\Rightarrow \frac{\sqrt{3}}{2}=\frac{4-a^...
Read More →There are three coins. One is two headed coin (having head on both faces),
Question: There are three coins. One is two headed coin (having head on both faces), another is a biased coin that comes up heads 75% of the time and third is an unbiased coin. One of the three coins is chosen at random and tossed, it shows heads, what is the probability that it was the two headed coin? Solution: Let E1, E2, and E3be the respective events of choosing a two headed coin, a biased coin, and an unbiased coin. $\therefore P\left(E_{1}\right)=P\left(E_{2}\right)=P\left(E_{3}\right)=\f...
Read More →Prove the following trigonometric identities.
Question: Prove the following trigonometric identities. $\frac{1+\cos A}{\sin ^{2} A}=\frac{1}{1-\cos A}$ Solution: We need to prove $\frac{1+\cos A}{\sin ^{2} A}=\frac{1}{1-\cos A}$ Using the property $\cos ^{2} \theta+\sin ^{2} \theta=1$, we get $\mathrm{LHS}=\frac{1+\cos A}{\sin ^{2} A}=\frac{1+\cos A}{1-\cos ^{2} A}$ Further using the identity, $a^{2}-b^{2}=(a+b)(a-b)$, we get $\frac{1+\cos A}{1-\cos ^{2} A}=\frac{1+\cos A}{(1-\cos A)(1+\cos A)}$ $=\frac{1}{1-\cos A}$ = RHS Hence proved....
Read More →Answer each of the following questions in one word or one sentence or as per exact requirement of the question.
Question: Answer each of the following questions in one word or one sentence or as per exact requirement of the question. Find the area of the triangle $\triangle \mathrm{ABC}$ in which $a=1, b=2$ and $\angle C=60^{\circ}$. Solution: In $\triangle \mathrm{ABC}, a=1, b=2$ and $\angle C=60^{\circ}$. Area of the ∆ABC $=\frac{1}{2} a b \sin C$ $=\frac{1}{2} \times 1 \times 2 \times \sin 60^{\circ}$ $=\frac{1}{2} \times 2 \times \frac{\sqrt{3}}{2}$ $=\frac{\sqrt{3}}{2}$ square units...
Read More →Answer each of the following questions in one word or one sentence or as per exact requirement of the question.
Question: Answer each of the following questions in one word or one sentence or as per exact requirement of the question. Find the area of the triangle $\triangle \mathrm{ABC}$ in which $a=1, b=2$ and $\angle C=60^{\circ}$. Solution: In $\triangle \mathrm{ABC}, a=1, b=2$ and $\angle C=60^{\circ}$. Area of the ∆ABC $=\frac{1}{2} a b \sin C$ $=\frac{1}{2} \times 1 \times 2 \times \sin 60^{\circ}$ $=\frac{1}{2} \times 2 \times \frac{\sqrt{3}}{2}$ $=\frac{\sqrt{3}}{2}$ square units...
Read More →Prove the following trigonometric identities.
Question: Prove the following trigonometric identities. $\frac{\left(1+\tan ^{2} \theta\right) \cot \theta}{\operatorname{cosec}^{2} \theta}=\tan \theta$ Solution: We need to prove $\frac{\left(1+\tan ^{2} \theta\right) \cot \theta}{\operatorname{cosec}^{2} \theta}=\tan \theta$ Solving the L.H.S, we get $\frac{\left(1+\tan ^{2} \theta\right) \cot \theta}{\operatorname{cosec}^{2} \theta}=\frac{\sec ^{2} \theta(\cot \theta)}{\operatorname{cosec}^{2} \theta}$ Using $\sec \theta=\frac{1}{\cos \theta...
Read More →In a ∆ABC, if a = 4, b = 3,
Question: In a $\triangle A B C$, if $a=4, b=3, A=\frac{\pi}{3}$. Then side $C$ is given by Solution: In ∆ABC if $a=4, b=3, A=\frac{\pi}{3}$ Since $\cos A=\frac{b^{2}+c^{2}-a^{2}}{2 b c}$ i. e $\cos \frac{\pi}{3}=\frac{9+c^{2}-16}{2 \times 3 \times c}$ i. e $\frac{1}{2}=\frac{c^{2}-7}{6 c}$ i. e $3 c=c^{2}-7$ i. e $c^{2}-3 c-7=0$...
Read More →Simplify
Question: Simplify (i) $\frac{4+\sqrt{5}}{4-\sqrt{5}}+\frac{4-\sqrt{5}}{4+\sqrt{5}}$ (ii) $\frac{1}{\sqrt{3}+\sqrt{2}}-\frac{2}{\sqrt{5}-\sqrt{3}}-\frac{3}{\sqrt{2}-\sqrt{5}}$ (iii) $\frac{2+\sqrt{3}}{2-\sqrt{3}}+\frac{2-\sqrt{3}}{2+\sqrt{3}}+\frac{\sqrt{3}-1}{\sqrt{3}+1}$ (iv) $\frac{2 \sqrt{6}}{\sqrt{2}+\sqrt{3}}+\frac{6 \sqrt{2}}{\sqrt{6}+\sqrt{3}}-\frac{8 \sqrt{3}}{\sqrt{6}+\sqrt{2}}$ Solution: (i) $\frac{4+\sqrt{5}}{4-\sqrt{5}}+\frac{4-\sqrt{5}}{4+\sqrt{5}}$ $=\frac{4+\sqrt{5}}{4-\sqrt{5}} ...
Read More →A laboratory blood test is 99% effective in detecting a certain disease when it is in fact,
Question: A laboratory blood test is 99% effective in detecting a certain disease when it is in fact, present. However, the test also yields a false positive result for 0.5% of the healthy person tested (that is, if a healthy person is tested, then, with probability 0.005, the test will imply he has the disease). If 0.1 percent of the population actually has the disease, what is the probability that a person has the disease given that his test result is positive? Solution: Let E1and E2be the res...
Read More →Prove the following trigonometric identities
Question: Prove the following trigonometric identities $\operatorname{cosec}^{6} \theta=\cot ^{6} \theta+3 \cot ^{2} \theta \operatorname{cosec}^{2} \theta+1$ Solution: We need to prove $\operatorname{cosec}^{6} \theta=\cot ^{6} \theta+3 \cot ^{2} \theta \operatorname{cosec}^{2} \theta+1$ Solving the L.H.S, we get $\operatorname{cosec}^{6} \theta=\left(\operatorname{cosec}^{2} \theta\right)^{3}$ $=\left(1+\cot ^{2} \theta\right)^{3} \quad\left(1+\cot ^{2} \theta=\operatorname{cosec}^{2} \theta\r...
Read More →In a ∆ABC, if
Question: In a ∆ABC, ifa4+b4+c4= 2a2b2+ 2b2c2, thenB= __________. Solution: In a ∆ABC, Given $a^{4}+b^{4}+c^{4}=2 a^{2} b^{2}+2 b^{2} c^{2}$ i.e $a^{4}+b^{4}-2 a^{2} b^{2}+c^{2}-2 b^{2} c^{2}=0$ i.e $\left(a^{2}+c^{2}\right)^{2}+b^{4}-2 a^{2} b^{2}-2 b^{2} c^{2}-2 a^{2} c^{2}=0$ i.e $\left(\underline{a}^{2}+c^{2}\right)^{2}+b^{4}-2 b^{2}\left(\underline{a}^{2}+c^{2}\right)-2 a^{2} c^{2}=0$ i.e $\left(a^{2}+c^{2}-b^{2}\right)^{2}=2 a^{2} c^{2}$ i.e $a^{2}+c^{2}-b^{2}=\sqrt{2} a c$ Using cosine fo...
Read More →In a ∆ABC, if
Question: In a ∆ABC, ifa4+b4+c4= 2a2b2+ 2b2c2, thenB= __________. Solution: In a ∆ABC, Given $a^{4}+b^{4}+c^{4}=2 a^{2} b^{2}+2 b^{2} c^{2}$ i.e $a^{4}+b^{4}-2 a^{2} b^{2}+c^{2}-2 b^{2} c^{2}=0$ i.e $\left(a^{2}+c^{2}\right)^{2}+b^{4}-2 a^{2} b^{2}-2 b^{2} c^{2}-2 a^{2} c^{2}=0$ i.e $\left(\underline{a}^{2}+c^{2}\right)^{2}+b^{4}-2 b^{2}\left(\underline{a}^{2}+c^{2}\right)-2 a^{2} c^{2}=0$ i.e $\left(a^{2}+c^{2}-b^{2}\right)^{2}=2 a^{2} c^{2}$ i.e $a^{2}+c^{2}-b^{2}=\sqrt{2} a c$ Using cosine fo...
Read More →Prove the following trigonometric identities.
Question: Prove the following trigonometric identities. $\sec ^{6} \theta=\tan ^{6} \theta+3 \tan ^{2} \theta \sec ^{2} \theta+1$ Solution: We need to prove $\sec ^{6} \theta=\tan ^{6} \theta+3 \tan ^{2} \theta \sec ^{2} \theta+1$ Solving the L.H.S, we get $\sec ^{6} \theta=\left(\sec ^{2} \theta\right)^{3}$ $=\left(1+\tan ^{2} \theta\right)^{3}$ Further using the identity $(a+b)^{3}=a^{3}+b^{3}+3 a^{2} b+3 a b^{2}$, we get $\left(1+\tan ^{2} \theta\right)^{3}=1+\tan ^{6} \theta+3(1)^{2}\left(\t...
Read More →If the sides of a ∆ABC are a, b,
Question: If the sides of a $\triangle A B C$ are $a, b, \sqrt{a^{2}+a b+b^{2}}$, then the measure of the largest angle is _________________ Solution: Let us suppose the greatest angle isc Using cosine formula, $\cos C=\frac{a^{2}+b^{2}-c^{2}}{2 a b}$ Since $C=\sqrt{a^{2}+a b+b^{2}}$ $\Rightarrow C^{2}=a^{2}+a b+b^{2}$ ie $\cos C=\frac{a^{2}+b^{2}-a^{2}-a b-b^{2}}{2 a b}=\frac{-1}{2}$ ie $\cos C=\frac{1}{2}$ ie $C=\frac{2 \pi}{3}$...
Read More →In answering a question on a multiple choice test, a student either knows the answer or guesses.
Question: In answering a question on a multiple choice test, a student either knows the answer or guesses. Let $\frac{3}{4}$ be the probability that he knows the answer and $\frac{1}{4}$ be the probability that he guesses. Assuming that a student who guesses at the answer will be correct with probabilityWhat is the probability that the student knows the answer given that he answered it correctly? Solution: Let E1and E2be the respective events that the student knows the answer and he guesses the ...
Read More →If the sides of a ∆ABC are a, b,
Question: If the sides of a $\triangle A B C$ are $a, b, \sqrt{a^{2}+a b+b^{2}}$, then the measure of the largest angle is _________________ Solution: Let us suppose the greatest angle isc Using cosine formula, $\cos C=\frac{a^{2}+b^{2}-c^{2}}{2 a b}$ Since $C=\sqrt{a^{2}+a b+b^{2}}$ $\Rightarrow C^{2}=a^{2}+a b+b^{2}$ ie $\cos C=\frac{a^{2}+b^{2}-a^{2}-a b-b^{2}}{2 a b}=\frac{-1}{2}$ ie $\cos C=\frac{1}{2}$ ie $C=\frac{2 \pi}{3}$...
Read More →If angles of a triangle are in A.P. and b :
Question: If angles of a triangle are in A.P. and $b: c=\sqrt{3}: \sqrt{2}$, then $C=$ ________________ Solution: If angle of a triangleABCare inA.P $\Rightarrow 2 \angle B=\angle A+\angle C$ and $\frac{b}{c}=\frac{\sqrt{3}}{\sqrt{2}}$ By angle sum property $\angle A+\angle B+\angle C=\pi$ $\Rightarrow 2 \angle B+\angle B=\pi$ $\Rightarrow \angle B=\frac{\pi}{3}$ also, Using Sine formula $\frac{a}{\sin A}=\frac{b}{\sin B}=\frac{c}{\sin C}$ i. e $\frac{b}{\sin B}=\frac{c}{\sin C}$ i. e $\frac{b}{...
Read More →Prove the following trigonometric identities.
Question: Prove the following trigonometric identities. $\frac{\tan \theta}{1-\cot \theta}+\frac{\cot \theta}{1-\tan \theta}=1+\tan \theta+\cot \theta$ Solution: We need to prove $\frac{\tan \theta}{1-\cot \theta}+\frac{\cot \theta}{1-\tan \theta}=1+\tan \theta+\cot \theta$ Now, using $\cot \theta=\frac{1}{\tan \theta}$ in the L.H.S, we get $\frac{\tan \theta}{1-\cot \theta}+\frac{\cot \theta}{1-\tan \theta}=\frac{\tan \theta}{\left(1-\frac{1}{\tan \theta}\right)}+\frac{\left(\frac{1}{\tan \thet...
Read More →If angles of a triangle are in A.P. and b :
Question: If angles of a triangle are in A.P. and $b: c=\sqrt{3}: \sqrt{2}$, then $C=$ ________________ Solution: If angle of a triangleABCare inA.P $\Rightarrow 2 \angle B=\angle A+\angle C$ and $\frac{b}{c}=\frac{\sqrt{3}}{\sqrt{2}}$ By angle sum property $\angle A+\angle B+\angle C=\pi$ $\Rightarrow 2 \angle B+\angle B=\pi$ $\Rightarrow \angle B=\frac{\pi}{3}$ also, Using Sine formula $\frac{a}{\sin A}=\frac{b}{\sin B}=\frac{c}{\sin C}$ i. e $\frac{b}{\sin B}=\frac{c}{\sin C}$ i. e $\frac{b}{...
Read More →Simplify by rationalising the denominator.
Question: Simplify by rationalising the denominator. (i) $\frac{7 \sqrt{3}-5 \sqrt{2}}{\sqrt{48}+\sqrt{18}}$ (ii) $\frac{2 \sqrt{6}-\sqrt{5}}{3 \sqrt{5}-2 \sqrt{6}}$ Solution: (i) $\frac{7 \sqrt{3}-5 \sqrt{2}}{\sqrt{48}+\sqrt{18}}$ $=\frac{7 \sqrt{3}-5 \sqrt{2}}{\sqrt{16 \times 3}+\sqrt{9 \times 2}}$ $=\frac{7 \sqrt{3}-5 \sqrt{2}}{4 \sqrt{3}+3 \sqrt{2}}$ $=\frac{7 \sqrt{3}-5 \sqrt{2}}{4 \sqrt{3}+3 \sqrt{2}} \times \frac{4 \sqrt{3}-3 \sqrt{2}}{4 \sqrt{3}-3 \sqrt{2}}$ $=\frac{7 \sqrt{3} \times 4 \...
Read More →Prove the following trigonometric identities.
Question: Prove the following trigonometric identities. $\frac{1+\sec \theta}{\sec \theta}=\frac{\sin ^{2} \theta}{1-\cos \theta}$ Solution: We have to prove $\frac{1+\sec \theta}{\sec \theta}=\frac{\sin ^{2} \theta}{1-\cos \theta}$ We know that, $\sin ^{2} \theta+\cos ^{2} \theta=1$ $\frac{1+\sec \theta}{\sec \theta}=\frac{1+\frac{1}{\cos \theta}}{\frac{1}{\cos \theta}}$ $=\frac{\frac{\cos \theta+1}{\cos \theta}}{\frac{1}{\cos \theta}}$ $=\frac{1+\cos \theta}{1}$ Multiplying the numerator and d...
Read More →Of the students in a college, it is known that 60% reside in hostel and 40% are day scholars (not residing in hostel).
Question: Of the students in a college, it is known that 60% reside in hostel and 40% are day scholars (not residing in hostel). Previous year results report that 30% of all students who reside in hostel attain A grade and 20% of day scholars attain A grade in their annual examination. At the end of the year, one student is chosen at random from the college and he has an A grade, what is the probability that the student is hostler? Solution: Let E1and E2be the events that the student is a hostle...
Read More →In a ∆ABC,
Question: In a $\triangle A B C$, if $b=\sqrt{3}, c=1$ and $B-C=\frac{\pi}{2}$, then $A=$ Solution: Given $b=\sqrt{3}, c=1$ and $B-C=\frac{\pi}{2}$ In a triangle $A B C$ By angle sum property, Since $A+B+C=\frac{\pi}{2}$ ...(1) and by Sine formula, $\frac{a}{\sin A}=\frac{b}{\sin B}=\frac{c}{\sin C}$ $\Rightarrow \frac{b}{\sin B}=\frac{c}{\sin C}$ $\Rightarrow \frac{\sqrt{3}}{\sin \left(\frac{\pi}{2}+c\right)}=\frac{c=1}{\sin C}$ i. e $\frac{\sqrt{3}}{\cos C}=\frac{1}{\sin C} \quad\left(\because...
Read More →In a ∆ABC,
Question: In a $\triangle A B C$, if $b=\sqrt{3}, c=1$ and $B-C=\frac{\pi}{2}$, then $A=$ Solution: Given $b=\sqrt{3}, c=1$ and $B-C=\frac{\pi}{2}$ In a triangle $A B C$ By angle sum property, Since $A+B+C=\frac{\pi}{2}$ ...(1) and by Sine formula, $\frac{a}{\sin A}=\frac{b}{\sin B}=\frac{c}{\sin C}$ $\Rightarrow \frac{b}{\sin B}=\frac{c}{\sin C}$ $\Rightarrow \frac{\sqrt{3}}{\sin \left(\frac{\pi}{2}+c\right)}=\frac{c=1}{\sin C}$ i. e $\frac{\sqrt{3}}{\cos C}=\frac{1}{\sin C} \quad\left(\because...
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