Express 2.36¯¯¯¯+0.23¯¯¯¯
Question: Express $2 . \overline{36}+0.23$ as a fraction in simplest form. Solution: Given: $2 . \overline{36}+0 . \overline{23}$ Let $x=2 . \overline{36} \quad \ldots(\mathrm{i})$ $\begin{array}{ll}y=0 . \overline{23} \ldots \text { (ii) }\end{array}$ First we take $x$ and convert it into $\frac{p}{q}$ 100x= 236.3636... ...(iii)Subtracting (i) from (iii) we get $99 x=234$ $\Rightarrow x=\frac{234}{99}$ Similarly, multiplyywith 100 as there are 2 decimal places which are repeating themselves. $1...
Read More →In any triangle ABC, prove the following:
Question: In any triangle ABC, prove the following: $\sin \left(\frac{B-C}{2}\right)=\frac{b-c}{a} \cos \frac{A}{2}$ Solution: Let $\frac{a}{\sin A}=\frac{b}{\sin B}=\frac{c}{\sin C}=k$ Then, Consider the RHS of the equation $\sin \left(\frac{B-C}{2}\right)=\frac{b-c}{a} \cos \frac{A}{2}$ $\mathrm{RHS}=\frac{b-c}{a} \cos \frac{A}{2}$ $=\frac{k(\sin B-\sin C)}{k \sin A} \cos \left(\frac{\pi-(\mathrm{B}+\mathrm{C})}{2}\right) \quad(\because A+B+C=\pi)$ $=\frac{2 \sin \left(\frac{B-C}{2}\right) \co...
Read More →If A + B = 90° and tan A=34, what is cot B?
Question: If $A+B=90^{\circ}$ and $\tan A=\frac{3}{4}$, what is $\cot B ?$ Solution: Given in question: $A+B=90^{\circ}$ $\tan A=\frac{3}{4}$ $A+B=90^{\circ}$ $\Rightarrow B=90^{\circ}-A$ $\Rightarrow \cot B=\cot \left(90^{\circ}-A\right)$ $\Rightarrow \cot B=\tan A$ $\Rightarrow \cot B=\frac{3}{4}\left[\cot \left(90^{\circ}-A\right)=\tan A\right]$ Hence the value of $\cot B$ is $\frac{3}{4}$...
Read More →A manufacturer makes two types of toys A and B.
Question: A manufacturer makes two types of toys A and B. Three machines are needed for this purpose and the time (in minutes) required for each toy on the machines is given below: Each machine is available for a maximum of 6 hours per day. If the profit on each toy of type A is Rs 7.50 and that on each toy of type B is Rs 5, show that 15 toys of type A and 30 of type B should be manufactured in a day to get maximum profit. Solution: Letxandytoys of type A and type B respectively be manufactured...
Read More →Write the value of tan 10° tan 15° tan 75° tan 80°?
Question: Write the value of $\tan 10^{\circ} \tan 15^{\circ} \tan 75^{\circ} \tan 80^{\circ} ?$ Solution: We have to find: $\tan 10^{\circ} \tan 15^{\circ} \tan 75^{\circ} \tan 80^{\circ}$ $=\tan 10^{\circ} \tan 15^{\circ} \tan 75^{\circ} \tan 80^{\circ}$ $=\tan \left(90^{\circ}-80^{\circ}\right) \tan \left(90^{\circ}-75^{\circ}\right) \tan 75^{\circ} \tan 80^{\circ}$ $\left[\tan \left(90^{\circ}-\theta\right)=\cot \theta\right]$ $=\cot 80^{\circ} \cot 75^{\circ} \tan 75^{\circ} \tan 80^{\circ}...
Read More →In triangle ABC, prove the following:
Question: In triangle ABC, prove the following: $\frac{a+b}{c}=\frac{\cos \left(\frac{A-B}{2}\right)}{\sin \frac{C}{2}}$ Solution: Let $\frac{a}{\sin A}=\frac{b}{\sin B}=\frac{c}{\sin C}=k$ ...(1) Then, Consider the LHS of the equation $\frac{a+b}{c}=\frac{\cos \left(\frac{A-B}{2}\right)}{\sin \frac{C}{2}}$. $\mathrm{LHS}=\frac{a+b}{c}$ $=\frac{k \sin A+k \sin B}{k \sin C} \quad($ using $(1))$ $=\frac{2 \sin \left(\frac{A+B}{2}\right) \cos \left(\frac{A-B}{2}\right)}{2 \sin \frac{C}{2} \cos \fra...
Read More →In triangle ABC, prove the following:
Question: In triangle ABC, prove the following: $\frac{c}{a+b}=\frac{1-\tan \left(\frac{A}{2}\right) \tan \left(\frac{B}{2}\right)}{1+\tan \left(\frac{A}{2}\right) \tan \left(\frac{B}{2}\right)}$ Solution: Let $\frac{a}{\sin A}=\frac{b}{\sin B}=\frac{c}{\sin C}=k$ We need to prove: $\frac{c}{a+b}=\frac{1-\tan \left(\frac{A}{2}\right) \tan \left(\frac{B}{2}\right)}{1+\tan \left(\frac{A}{2}\right) \tan \left(\frac{B}{2}\right)}$ Consider $\mathrm{LHS}=\frac{\mathrm{c}}{\mathrm{a}+\mathrm{b}}$ $=\f...
Read More →Write the value of cos 1° cos 2° cos 3° ....... cos 179° cos 180°.
Question: Write the value of $\cos 1^{\circ} \cos 2^{\circ} \cos 3^{\circ}$ $\cos 179^{\circ} \cos 180^{\circ}$ Solution: Given that: $\cos 1^{\circ} \cos 2^{\circ} \cos 3^{\circ} \ldots \cos 179^{\circ} \cos 180^{\circ}$ $=\cos 1^{\circ} \cos 2^{\circ} \cos 3^{\circ} \ldots \cos 179^{\circ} \cos 180^{\circ}$ $=\cos 1^{\circ} \cos 2^{\prime} \cos 3^{\circ} \ldots \cos 89^{\circ} \cos 90^{\circ} \cos 91^{\circ} \ldots \cos 179^{\circ} \cos 180^{\circ}$ $\left[\cos 90^{\circ}=0\right]$ $=\cos 1^{\...
Read More →Express each of the following decimals in the form
Question: Express each of the following decimals in the form $\frac{p}{a}$, where $p, q$ are integers and $q \neq 0$. (i) $0 . \overline{2}$ (ii) $0 . \overline{53}$ (iii) $2 . \overline{93}$ (iv) $18 . \overline{48}$ (v) $0 . \overline{235}$ (vi) $0.00 \overline{32}$ (vii) 1. $3 \overline{23}$ (viii) $0.3 \overline{178}$ (ix) $32.12 \overline{35}$ (x) $0.40 \overline{7}$ Solution: (i) $0 . \overline{2}$ Letx= 0.222... .....(i)Only one digit is repeated so, we multiplyxby 10.10x= 2.222... .....(...
Read More →Write the acute angle θ satisfying 3–√ sin θ=cos θ.
Question: Write the acute angle $\theta$ satisfying $\sqrt{3} \sin \theta=\cos \theta$. Solution: We have: $\sqrt{3} \sin \theta=\cos \theta$ $\Rightarrow \sqrt{3} \sin \theta=\cos \theta$ $\Rightarrow \frac{\sin \theta}{\cos \theta}=\frac{1}{\sqrt{3}}$ $\Rightarrow \tan \theta=\tan 30^{\circ}$ $\Rightarrow \theta=30^{\circ}$ Hence the acute angle is...
Read More →A dietician wishes to mix together two kinds of food X and Y in such a way that the mixture contains at least 10 units of vitamin A,
Question: A dietician wishes to mix together two kinds of food X and Y in such a way that the mixture contains at least 10 units of vitamin A, 12 units of vitamin B and 8 units of vitamin C. The vitamin content of one kg food is given below: One kg of food X costs Rs 16 and one kg of food Y costs Rs 20. Find the least cost of the mixture which will produce the required diet? Solution: Let the mixture containxkg of food X andykg of food Y. The mathematical formulation of the given problem is as f...
Read More →If A + B = 90° and cos B=35, what is the value of sin A?
Question: If $A+B=90^{\circ}$ and $\cos B=\frac{3}{5}$, what is the value of $\sin A$ ? Solution: We have: $A+B=90^{\circ}$ $\cos B=\frac{3}{5}$ $A+B=90^{\circ}$ $\Rightarrow A=90^{\circ}-B$ $\Rightarrow \sin A=\sin \left(90^{\circ}-B\right)$ $\Rightarrow \sin A=\cos B$ $\Rightarrow \sin A=\frac{3}{5}\left[\sin \left(90^{\circ}-B\right)=\cos B\right]$ Hence the value of $\sin A$ is $\frac{3}{5}$...
Read More →In triangle ABC, prove the following:
Question: In triangle ABC, prove the following: $\frac{c}{a-b}=\frac{\tan \left(\frac{A}{2}\right)+\tan \left(\frac{B}{2}\right)}{\tan \left(\frac{A}{2}\right)-\tan \left(\frac{B}{2}\right)}$ Solution: Let $\frac{a}{\sin A}=\frac{b}{\sin B}=\frac{c}{\sin C}=k$ ...(1) We need to prove: $\frac{c}{a-b}=\frac{\tan \left(\frac{A}{2}\right)+\tan \left(\frac{B}{2}\right)}{\tan \left(\frac{A}{2}\right)-\tan \left(\frac{B}{2}\right)}$ Consider $\mathrm{LHS}=\frac{c}{a-b}$ $=\frac{k \sin C}{k(\sin A-\sin ...
Read More →In triangle ABC, prove the following:
Question: In triangle ABC, prove the following: $\frac{c}{a-b}=\frac{\tan \left(\frac{A}{2}\right)+\tan \left(\frac{B}{2}\right)}{\tan \left(\frac{A}{2}\right)-\tan \left(\frac{B}{2}\right)}$ Solution: Let $\frac{a}{\sin A}=\frac{b}{\sin B}=\frac{c}{\sin C}=k$ ...(1) We need to prove: $\frac{c}{a-b}=\frac{\tan \left(\frac{A}{2}\right)+\tan \left(\frac{B}{2}\right)}{\tan \left(\frac{A}{2}\right)-\tan \left(\frac{B}{2}\right)}$ Consider $\mathrm{LHS}=\frac{c}{a-b}$ $=\frac{k \sin C}{k(\sin A-\sin ...
Read More →If tan A=34 and A+B=90°, then what is the value of cot B?
Question: If $\tan A=\frac{3}{4}$ and $A+B=90^{\circ}$, then what is the value of $\cot \mathrm{B} ?$ Solution: Given that: $A+B=90^{\circ}$ $\tan A=\frac{3}{4}$ $A+B=90^{\circ}$ $\Rightarrow B=90^{\circ}-A$ $\Rightarrow \cot B=\cot \left(90^{\circ}-A\right)$ $\Rightarrow \cot B=\tan A$ $\Rightarrow \cot B=\frac{3}{4}\left[\cot \left(90^{\circ}-A\right)=\tan A\right]$ Hence the value of $\cot B$ is $\frac{3}{4}$...
Read More →If cot θ=13√, write the value of 1−cos2 θ2−sin2 θ.
Question: If $\cot \theta=\frac{1}{\sqrt{3}}$, write the value of $\frac{1-\cos ^{2} \theta}{2-\sin ^{2} \theta}$. Solution: Given: $\cot \theta=\frac{1}{\sqrt{3}}$ $\frac{\text { Base }}{\text { Perpendicular }}=\frac{1}{\sqrt{3}}$ Base $=1$ Perpendicular $=\sqrt{3}$ Hypotenuse $=\sqrt{(\text { Perpendicular })^{2}+(\text { Base })^{2}}$ Hypotenuse $=2$ Now we find, $\frac{1-\cos ^{2} \theta}{2-\sin ^{2} \theta}$ $=\frac{1-\frac{(\text { Base })^{2}}{(\text { hypotenuse })^{2}}}{2-\frac{(\text ...
Read More →In triangle ABC, prove the following:
Question: In triangle ABC, prove the following: $(a-b) \cos \frac{C}{2}=c \sin \left(\frac{A-B}{2}\right)$ Solution: Let $\frac{a}{\sin A}=\frac{b}{\sin B}=\frac{c}{\sin C}=k$ ....(1) Consider the LHS of the equation $(a-b) \cos \frac{C}{2}=c \sin \left(\frac{A-B}{2}\right)$ $\mathrm{LHS}=(a-b) \cos \frac{C}{2}$ $=k(\sin A-\sin B) \cos \frac{C}{2} \quad($ using $(1))$ $=k \times 2 \sin \left(\frac{A-B}{2}\right) \cos \left(\frac{A+B}{2}\right) \cos \frac{C}{2}$ $=2 k \sin \left(\frac{A-B}{2}\rig...
Read More →A farmer mixes two brands P and Q of cattle feed. Brand P,
Question: A farmer mixes two brands P and Q of cattle feed. Brand P, costing Rs 250 per bag contains 3 units of nutritional element A, 2.5 units of element B and 2 units of element C. Brand Q costing Rs 200 per bag contains 1.5 units of nutritional elements A, 11.25 units of element B, and 3 units of element C. The minimum requirements of nutrients A, B and C are 18 units, 45 units and 24 units respectively. Determine the number of bags of each brand which should be mixed in order to produce a m...
Read More →Given tan θ=15√, what is the value of cosec2 θ−sec2 θcosec2 θ+sec2 θ?
Question: Given $\tan \theta=\frac{1}{\sqrt{5}}$, what is the value of $\frac{\operatorname{cosec}^{2} \theta-\sec ^{2} \theta}{\operatorname{cosec}^{2} \theta+\sec ^{2} \theta}$ ? Solution: Given: $\tan \theta=\frac{1}{\sqrt{5}}$ We know that: $\tan \theta=\frac{\text { Perpendicular }}{\text { Base }}$ $\frac{\text { Perpendicular }}{\text { Base }}=\frac{1}{\sqrt{5}}$ Hypotenuse $=\sqrt{(\text { Perpendicular })^{2}+(\text { Base })^{2}}$ Hypotenuse $=\sqrt{1+5}$ Hypotenuse $=\sqrt{6}$ Now we...
Read More →In triangle ABC, prove the following:
Question: In triangle ABC, prove the following: $\frac{a-b}{a+b}=\frac{\tan \left(\frac{A-B}{2}\right)}{\tan \left(\frac{A+B}{2}\right)}$ Solution: Assume $\frac{a}{\sin A}=\frac{b}{\sin B}=\frac{c}{\sin C}=k$ Consider the LHS of the equation $\frac{a-b}{a+b}=\frac{\tan \left(\frac{A-B}{2}\right)}{\tan \left(\frac{A+B}{2}\right)}$. $\mathrm{LHS}=\frac{a-b}{a+b}$ $=\frac{k(\sin A-\sin B)}{k(\sin A+\sin B)}$ $\because \sin A-\sin B=2 \sin \left(\frac{A-B}{2}\right) \cos \left(\frac{A+B}{2}\right),...
Read More →If 3 cot θ = 4, find the value of 4 cos θ−sin θ2 cos θ+sin θ.
Question: If $3 \cot \theta=4$, find the value of $\frac{4 \cos \theta-\sin \theta}{2 \cos \theta+\sin \theta}$. Solution: We have: $3 \cot \theta=4$ $\cot \theta=\frac{4}{3}$ Since we know that in right angle triangle $\cot \theta=\frac{\text { Base }}{\text { perpendicular }}$ $\cos \theta=\frac{\text { Base }}{\text { Hypotenuse }}$ $\sin \theta=\frac{\text { Perpendicular }}{\text { Hypotenuse }}$ Hypotenuse $=\sqrt{(\text { Perpendicular })^{2}+(\text { Base })^{2}}$ Hypotenuse $=\sqrt{(3)^...
Read More →In ∆ABC, if a = 18,
Question: In ∆ABC, ifa= 18,b= 24 andc= 30 and c= 90, find sinA, sinBand sinC. Solution: Given,C= 90, a= 18,b= 24 andc= 30 According to sine rule, $\frac{a}{\sin A}=\frac{b}{\sin B}=\frac{c}{\sin C}$. $\Rightarrow \frac{c}{\sin C}=\frac{a}{\sin A}$ $\Rightarrow \sin A=\frac{a \sin C}{c}$ $=\frac{18 \times \sin \left(90^{\circ}\right)}{30}$ $=\frac{18}{30}$ $=\frac{3}{5}$ Also, $\frac{b}{\sin B}=\frac{c}{\sin C}$ $\Rightarrow \sin B=\frac{b \sin C}{c}$ $=\frac{24 \sin 90^{\circ}}{30}$ $=\frac{24}{...
Read More →In ∆ABC, if a = 18,
Question: In ∆ABC, ifa= 18,b= 24 andc= 30 and c= 90, find sinA, sinBand sinC. Solution: Given,C= 90, a= 18,b= 24 andc= 30 According to sine rule, $\frac{a}{\sin A}=\frac{b}{\sin B}=\frac{c}{\sin C}$. $\Rightarrow \frac{c}{\sin C}=\frac{a}{\sin A}$ $\Rightarrow \sin A=\frac{a \sin C}{c}$ $=\frac{18 \times \sin \left(90^{\circ}\right)}{30}$ $=\frac{18}{30}$ $=\frac{3}{5}$ Also, $\frac{b}{\sin B}=\frac{c}{\sin C}$ $\Rightarrow \sin B=\frac{b \sin C}{c}$ $=\frac{24 \sin 90^{\circ}}{30}$ $=\frac{24}{...
Read More →If cos θ=23, find the value of sec θ−1sec θ+1.
Question: If $\cos \theta=\frac{2}{3}$, find the value of $\frac{\sec \theta-1}{\sec \theta+1}$ Solution: Given in question: $\cos \theta=\frac{2}{3}$ We have to find $\frac{\sec \theta-1}{\sec \theta+1}$ $\Rightarrow \frac{\sec \theta-1}{\sec \theta+1}=\frac{\frac{1}{\cos \theta}-1}{\frac{1}{\cos \theta}+1}$ $\Rightarrow \frac{\sec \theta-1}{\sec \theta+1}=\frac{\frac{3}{2}-1}{\frac{3}{2}+1}$ $\Rightarrow \frac{\sec \theta-1}{\sec \theta+1}=\frac{1}{5}$ Hence the value of $\frac{\sec \theta-1}{...
Read More →If tan θ=45, find the value of cos θ−sin θcos θ+sin θ.
Question: If $\tan \theta=\frac{4}{5}$, find the value of $\frac{\cos \theta-\sin \theta}{\cos \theta+\sin \theta}$. Solution: It is given that $\tan \theta=\frac{4}{5}$. We have to find $\frac{\cos \theta-\sin \theta}{\cos \theta+\sin \theta}$. $\frac{\cos \theta-\sin \theta}{\cos \theta+\sin \theta}$ $=\frac{1-\frac{\sin \theta}{\cos \theta}}{1+\frac{\sin \theta}{\cos \theta}}$ [Dividing both numerator and denominator by $\cos \theta$ ] $=\frac{1-\tan \theta}{1+\tan \theta}$ $=\frac{1}{9}$...
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