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Question: If $\sin ^{-1} x+\sin ^{-1} y=\frac{\pi}{3}$ and $\cos ^{-1} x-\cos ^{-1} y=\frac{\pi}{6}$, find the values of $x$ and $y .$ Solution: $\cos ^{-1} x-\cos ^{-1} y=\frac{\pi}{6}$ $\Rightarrow \frac{\pi}{2}-\sin ^{-1} x-\frac{\pi}{2}+\sin ^{-1} y=\frac{\pi}{6} \quad\left[\because \cos ^{-1} x=\frac{\pi}{2}-\sin ^{-1} x\right]$ $\Rightarrow-\left(\sin ^{-1} x-\sin ^{-1} y\right)=\frac{\pi}{6}$ $\Rightarrow \sin ^{-1} x-\sin ^{-1} y=-\frac{\pi}{6}$ Solving $\sin ^{-1} x+\sin ^{-1} y=\frac{\...
Read More →A balloon is connected to a meteorological ground station by a cable of length
Question: A balloon is connected to a meteorological ground station by a cable of length 215 m inclined at 60 to the horizontal. Determine the height of the balloon from the ground. Assume that there is no slack in the cable. Solution: LetABbe the balloon of heighth. And the balloon is connected to the metrological ground station by a cable of length 215 m. LetAC= 215 and Here we have to find height of balloon. We have the following corresponding figure So we use trigonometric ratios In a triang...
Read More →A piece of equipment cost a certain factory Rs 600,000.
Question: A piece of equipment cost a certain factory Rs 600,000. If it depreciates in value, 15% the first, 13.5% the next year, 12% the third year, and so on. What will be its value at the end of 10 years, all percentages applying to the original cost? Solution: Cost of the piece of equipment at the end of the first year = Rs 600000 15% of 600000 = Rs 600000 Rs 90000 = Rs 510000 Cost of the piece of equipment at the end of the second year = Rs 600000 13.5% of 600000 = Rs 600000 Rs 81000 = Rs 5...
Read More →A tree breaks due to the storm and the broken part bends so that the top
Question: A tree breaks due to the storm and the broken part bends so that the top of the tree touches the ground making an angle of 30 with the ground. The distance from the foot of the tree to the point where the top touches the ground is 10 metres. Find the height of the tree. Solution: LetABbe the tree of heighth. And the top of tree makes an angle 30 with ground. The distance between foot of tree to the point where the top touches the ground ism. LetBC= 10. And. Here we have to find height ...
Read More →if
Question: If $\cos ^{-1} x+\cos ^{-1} y=\frac{\pi}{4}$, find the value of $\sin ^{-1} x+\sin ^{-1} y$ Solution: $\cos ^{-1} x+\cos ^{-1} y=\frac{\pi}{4}$ $\Rightarrow \frac{\pi}{2}-\sin ^{-1} x+\frac{\pi}{2}-\sin ^{-1} y=\frac{\pi}{4} \quad\left[\because \cos ^{-1} x=\frac{\pi}{2}-\sin ^{-1} x\right]$ $\Rightarrow \pi-\left(\sin ^{-1} x+\sin ^{-1} y\right)=\frac{\pi}{4}$ $\Rightarrow \sin ^{-1} x+\sin ^{-1} y=\frac{3 \pi}{4}$...
Read More →There are 25 trees at equal distances of 5 metres in a line with a well, the distance of the well from the nearest tree being 10 metres.
Question: There are 25 trees at equal distances of 5 metres in a line with a well, the distance of the well from the nearest tree being 10 metres. A gardener waters all the trees separately starting from the well and he returns to the well after watering each tree to get water for the next. Find the total distance the gardener will cover in order to water all the trees. Solution: Let $S_{n}$ be the total distance travelled by the gardener. Letdbe the common difference (distance) between two tree...
Read More →The length of the shadow of a tower standing on level plane is found to be 2x
Question: The length of the shadow of a tower standing on level plane is found to be $2 x$ metres longer when the sun's altitude is $30^{\circ}$ than when it was $45^{\circ}$. Prove that the height of tower is $x$ ( $\sqrt{3}+1$ ) metres. Solution: LetABbe the tower of heighthm. the length of shadow of tower to be found 2xmeters at the plane longer when suns altitude is 30 than when it was 45. LetBC=ym, $C D=2 \times \mathrm{m}$ and $\angle A D B=30^{\circ}, \angle A C B=45^{\circ}$ We have to f...
Read More →Evaluate:
Question: Evaluate: (i) $\cot \left(\sin ^{-1} \frac{3}{4}+\sec ^{-1} \frac{4}{3}\right)$ (ii) $\sin \left(\tan ^{-1} x+\tan ^{-1} \frac{1}{x}\right)$ for $x0$ (iii) $\sin \left(\tan ^{-1} x+\tan ^{-1} \frac{1}{x}\right)$ for $x0$ (iv) $\cot \left(\tan ^{-1} a+\cot ^{-1} a\right)$ (v) $\cos \left(\sec ^{-1} x+\operatorname{cosec}^{-1} x\right),|x| \geq 1$ Solution: (i) $\cot \left(\sin ^{-1} \frac{3}{4}+\sec ^{-1} \frac{4}{3}\right)$ $=\cot \left(\sin ^{-1} \frac{3}{4}+\cos ^{-1} \frac{3}{4}\rig...
Read More →A manufacturer of radio sets produced 600 units in the third year and 700 units in the seventh year.
Question: A manufacturer of radio sets produced 600 units in the third year and 700 units in the seventh year. Assuming that the product increases uniformly by a fixed number every year, find (i) the production in the first year (ii) the total product in 7 years and (iii) the product in the 10th year. Solution: Let $a_{n}$ denote the production of radio sets in the $n$th year. Here, $a_{3}=600, a_{7}=700$ We know: $a_{n}=a+(n-1) d$ $a_{3}=a+2 d$ $\Rightarrow 600=a+2 d \quad \ldots \ldots(1)$ And...
Read More →A vertical tower stands on a horizontal plane and is surmounted by a flag-staff of height 7 m
Question: A vertical tower stands on a horizontal plane and is surmounted by a flag-staff of height 7 m. From a point on the plane, the angle of elevation of the bottom of the flag-staff is 30 and that of the top of the flag-staff is 45. Find the height of the tower. Solution: LetBCbe the tower of heighthm.ABbe the flag staff of height 7 m on tower andDbe the point on the plane making an angle of elevation of the top of the flag staff is 45 and angle of elevation of the bottom of the flag staff ...
Read More →A man arranges to pay off a debt of Rs 3600 by 40 annual instalments which form an arithmetic series.
Question: A man arranges to pay off a debt of Rs 3600 by 40 annual instalments which form an arithmetic series. When 30 of the instalments are paid, he dies leaving one-third of the debt unpaid, find the value of the first instalment. Solution: Let $S_{40}$ denote the total loan amount to be paid in 40 annual instalments. $\therefore S_{40}=3600$ Let Rsabe the value of the first instalment and Rsdbe the common difference. We know: $S_{n}=\frac{n}{2}\{2 a+(n-1) d\}$ $\Rightarrow \frac{40}{2}\{2 a...
Read More →A man arranges to pay off a debt of Rs 3600 by 40 annual instalments which form an arithmetic series.
Question: A man arranges to pay off a debt of Rs 3600 by 40 annual instalments which form an arithmetic series. When 30 of the instalments are paid, he dies leaving one-third of the debt unpaid, find the value of the first instalment. Solution: Let $S_{40}$ denote the total loan amount to be paid in 40 annual instalments. $\therefore S_{40}=3600$ Let Rsabe the value of the first instalment and Rsdbe the common difference. We know: $S_{n}=\frac{n}{2}\{2 a+(n-1) d\}$ $\Rightarrow \frac{40}{2}\{2 a...
Read More →Find the domain of each of the following functions:
Question: Find the domain of each of the following functions: (i) $f(x)=\sin ^{-1} x^{2}$ (ii) $f(x)=\sin ^{-1} x+\sin x$ (iii) $f(x) \sin ^{-1} \sqrt{x^{2}-1}$ (iv) $f(x)=\sin ^{-1} x+\sin ^{-1} 2 x$ Solution: (i) To the domain of $\sin ^{-1} y$ which is $[-1,1]$ $\therefore x^{2} \in[0,1]$ as $x^{2}$ can not be negative $\therefore x \in[-1,1]$ Hence, the domain is $[-1,1]$ (ii) Let $f(x)=g(x)+h(x)$, where Therefore, the domain of $f(x)$ is given by the intersection of the domain of $g(x)$ and...
Read More →A man sitting at a height of 20 m on a tall tree on
Question: A man sitting at a height of 20 m on a tall tree on a small island in the middle of a river observes two poles directly opposite to each other on the two banks of the river and in line with the foot of tree. If the angles of depression of the feet of the poles from a point at which the man is sitting on the tree on either side of the river are 60 and 30respectively. Find the width of the river. Solution: Letbe the width of river. And the angles of depression on either side of the river...
Read More →A man saves Rs 32 during the first year.
Question: A man saves Rs 32 during the first year. Rs 36 in the second year and in this way he increases his savings by Rs 4 every year. Find in what time his saving will be Rs 200. Solution: Letnbe the time in which the man saved Rs 200. Here,d= 4,a= 32 We know: $S_{n}=\frac{n}{2}\{2 a+(n-1) d\}$ $\Rightarrow 200=\frac{n}{2}\{2 \times 32+(n-1) 4\}$ $\Rightarrow 400=64 n+4 n^{2}-4 n$ $\Rightarrow 4 n^{2}+60 n-400=0$ $\Rightarrow n^{2}+15 n-100=0$ $\Rightarrow n^{2}+20 n-5 n-100=0$ $\Rightarrow(n...
Read More →Solve this
Question: (i) $\sin ^{-1} \frac{1}{2}-2 \sin ^{-1} \frac{1}{\sqrt{2}}$ (ii) $\sin ^{-1}\left\{\cos \left(\sin ^{-1} \frac{\sqrt{3}}{2}\right)\right\}$ Solution: (i) $\sin ^{-1} \frac{1}{2}-2 \sin ^{-1} \frac{1}{\sqrt{2}}=\sin ^{-1} \frac{1}{2}-\sin ^{-1} 2 \times \frac{1}{\sqrt{2}} \sqrt{1-\left(\frac{1}{\sqrt{2}}\right)^{2}}$ $=\sin ^{-1} \frac{1}{2}-\sin ^{-1} \sqrt{2} \times \frac{1}{\sqrt{2}}$ $=\sin ^{-1} \frac{1}{2}-\sin ^{-1} 1$ $=\sin ^{-1}\left(\sin \frac{\pi}{6}\right)-\sin ^{-1}\left(...
Read More →A man saved Rs 16500 in ten years.
Question: A man saved Rs 16500 in ten years. In each year after the first he saved Rs 100 more than he did in the receding year. How much did he save in the first year? Solution: Let the amount saved by the man in the first year be Rs A. Letdbe the common difference. Let $S_{10}$ denote the amount he saves in ten years. Here,n=10,d=100 We know: $S_{n}=\frac{n}{2}\{2 A+(n-1) d\}$ $\therefore S_{10}=\frac{10}{2}\{2 A+(10-1) 100\}$ $\Rightarrow 16500=5\{2 A+900\}$ $\Rightarrow 3300=2 A+900$ $\Right...
Read More →In a trapezium ABCD, if E and F be the mid-point of the diagonals AC and BD respectively
Question: In a trapeziumABCD, ifEandFbe the mid-point of the diagonalsACandBDrespectively. Then,EF= ? (a) $\frac{1}{2} A B$ (b) $\frac{1}{2} C D$ (c) $\frac{1}{2}(A B+C D)$ (d) $\frac{1}{2}(A B-C D)$ Solution: (d) $\frac{1}{2}(A B-C D)$ Explanation: JoinCFand produce it to cutABatG.Then ∆CDF∆GBF [∵DF = BF,DCF =BGFandCDF =GBF]CD = GBThus, in∆CAG, the pointsEandFare the mid points ofACandCG, respectively. $\therefore E F=\frac{1}{2}(A G)=\frac{1}{2}(A B-G B)=\frac{1}{2}(A B-C D)$...
Read More →Find the principal value of each of the following:
Question: Find the principal value of each of the following: (i) $\sin ^{-1}\left(-\frac{\sqrt{3}}{2}\right)$ (ii) $\sin ^{-1}\left(\cos \frac{2 \pi}{3}\right)$ (iii) $\sin ^{-1}\left(\frac{\sqrt{3}-1}{2 \sqrt{2}}\right)$ (iv) $\sin ^{-1}\left(\frac{\sqrt{3}+1}{2 \sqrt{2}}\right)$ (v) $\sin ^{-1}\left(\cos \frac{3 \pi}{4}\right)$ (vi) $\sin ^{-1}\left(\tan \frac{5 \pi}{4}\right)$ Solution: (i) $\sin ^{-1}\left(-\frac{\sqrt{3}}{2}\right)=\sin ^{-1}\left[\sin \left(-\frac{\pi}{3}\right)\right]=-\f...
Read More →Evaluate:
Question: Evaluate: $\sin \left\{\cos ^{-1}\left(-\frac{3}{5}\right)+\cot ^{-1}\left(-\frac{5}{12}\right)\right\}$ Solution: $\sin \left\{\cos ^{-1}\left(-\frac{3}{5}\right)+\cot ^{-1}\left(-\frac{5}{12}\right)\right\}=\sin \left\{\pi-\cos ^{-1}\left(\frac{3}{5}\right)+\pi-\cot ^{-1}\left(\frac{5}{12}\right)\right\}$ $=\sin \left\{2 \pi-\left[\cos ^{-1}\left(\frac{3}{5}\right)+\cot ^{-1}\left(\frac{5}{12}\right)\right]\right\}$ $=-\sin \left\{\cos ^{-1}\left(\frac{3}{5}\right)+\cot ^{-1}\left(\f...
Read More →A man is employed to count Rs 10710. He counts at the rate of Rs 180 per minute for half an hour.
Question: A man is employed to count Rs 10710. He counts at the rate of Rs 180 per minute for half an hour. After this he counts at the rate of Rs 3 less every minute than the preceding minute. Find the time taken by him to count the entire amount. Solution: It is given that the man counts Rs 180 per minute for half an hour. Sum of money the man counts in 30 minutes = Rs 18030 = Rs 5400 Total money counted by the man = Rs 10710 Money left for counting after 30 minutes = Rs (10710 5400) = Rs 5310...
Read More →The parallel sides of a trapezium are a and b respectively.
Question: The parallel sides of a trapezium areaandbrespectively. The line joining the mid-points of its non-parallel sides will be (a) $\frac{1}{2}(a-b)$ (b) $\frac{1}{2}(a+b)$ (C) $\frac{2 a b}{(a+b)}$ (d) $\sqrt{a b}$ Solution: (b) $\frac{1}{2}(a+b)$ Explanation: Suppose ABCD is a trapezium.Draw EF parallel to AB.JoinBDto cutEFatM.Now, in∆DAB,Eis the midpoint ofADandEM||AB. $\therefore M$ is the mid point of $\mathrm{BD}$ and $\mathrm{EM}=\frac{1}{2}(a)$ Similarly,Mis the mid point of BD andM...
Read More →P is any point on the side BC of a ∆ABC. P is joined to A.
Question: Pis any point on the sideBCof a ∆ABC.Pis joined toA. IfDandEare the midpoints of the sidesABandACrespectively andMandNare the midpoints ofBPandCPrespectively then quadrilateralDENMis(a) a trapezium(b) a parallelogram(c) a rectangle(d) a rhombus Solution: Given: In∆ABC,M,N,DandEare the mid-points ofBP,CP,ABandAC, respectivley.In∆ABP,∵∵DandMare the mid-points ofAB,andBP, respectively. (Given) $\therefore B M=\frac{1}{2} A P$ and $B M \| A P$ (Mid-point theorem) ...(i) Again, in∆ACP, $\be...
Read More →Evaluate:
Question: Evaluate: (i) $\tan \left\{\cos ^{-1}\left(-\frac{7}{25}\right)\right\}$ (ii) $\operatorname{cosec}\left\{\cot ^{-1}\left(-\frac{12}{5}\right)\right\}$ (iii) $\cos \left(\tan ^{-1} \frac{3}{4}\right)$ Solution: (i) $\tan \left\{\cos ^{-1}\left(-\frac{7}{25}\right)\right\}=\tan \left\{\cos ^{-1}\left(\pi-\frac{7}{25}\right)\right\}$ $=-\tan \left\{\cos ^{-1}\left(\frac{7}{25}\right)\right\}$ $=-\tan \left\{\tan ^{-1}\left[\frac{\sqrt{1-\left(\frac{7}{25}\right)^{2}}}{\frac{7}{25}}\right...
Read More →Two poles of equal heights are standing opposite to each other on either side of the
Question: Two poles of equal heights are standing opposite to each other on either side of the road which is 80 m wide. From a point between them on the road the angles of elevation of the top of the poles are 60 and 30 respectively. Find the height of the poles and the distance of the point from the poles. Solution: Letandbe the two poles of equal heighthm.be the points makes an angle of elevation from the top of poles are 60 and 30 respectively. Let $O A=80-x, O D=x .$ And $\angle B O A=30^{\c...
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