The angle of elevation of an aeroplane from a point on the ground is 45°.
Question: The angle of elevation of an aeroplane from a point on the ground is 45. After a flight of 15 seconds, the elevation changes to 30. If the aeroplane is flying at a height of 3000 metres, find the speed of the aeroplane. Solution: Let angle of elevation of an aero plane is 45. After 15 second angle of elevation is change to 30. LetDEbe the height of aero plane which is 3000 meter above the ground. Let,,and. Here we have to find speed of aero plane. We have the corresponding figure as fo...
Read More →We know that the sum of the interior angles of a triangle is 180°.
Question: We know that the sum of the interior angles of a triangle is 180. Show that the sums of the interior angles of polygons with 3, 4, 5, 6, ... sides form an arithmetic progression. Find the sum of the interior angles for a 21 sided polygon. [NCERT EXEMPLAR] Solution: We know that, thesum of the interior angles of a polygon with 3 sides,a1=180, thesum of the interior angles of a polygon with 4 sides,a2=360, thesum of the interior angles of a polygon with 5 sides,a3 =540, As, $a_{2}-a_{1}=...
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Question: If $\left(\sin ^{-1} x\right)^{2}+\left(\sin ^{-1} y\right)^{2}+\left(\sin ^{-1} z\right)^{2}=\frac{3}{4} \pi^{2}$, find the value of $x^{2}+y^{2}+z^{2}$ Solution: We know that the maximum value of $\sin ^{-1} x, \sin ^{-1} y, \sin ^{-1} z$ is $\frac{\pi}{2}$ and minimum value of $\sin ^{-1} x, \sin ^{-1} y, \sin ^{-1} z$ is $-\frac{\pi}{2}$ Now, For maximum value $\mathrm{LHS}=\left(\sin ^{-1} x\right)^{2}+\left(\sin ^{-1} y\right)^{2}+\left(\sin ^{-1} z\right)^{2}$ $=\left(\frac{\pi}...
Read More →In the given figure, ABCD is a rhombus. Then,
Question: In the given figure,ABCDis a rhombus. Then, (a) $A C^{2}+B D^{2}=A B^{2}$ (b) $A C^{2}+B D^{2}=2 A B^{2}$ (c) $A C^{2}+B D^{2}=4 A B^{2}$ (d) $2\left(A C^{2}+B D^{2}\right)=3 A B^{2}$ Solution: (c) $A C^{2}+B D^{2}=4 A B^{2}$ Explanation:We know that the diagonals of a rhombus bisect each other at right angles. Here, $O A=\frac{1}{2} A C, O B=\frac{1}{2} B D$ and $\angle A O B=90^{\circ}$ Now, $A B^{2}=O A^{2}+O B^{2}=\frac{1}{4}(A C)^{2}+\frac{1}{4}(B D)^{2}$ $\therefore 4 A B^{2}=\le...
Read More →We know that the sum of the interior angles of a triangle is 180°.
Question: We know that the sum of the interior angles of a triangle is 180. Show that the sums of the interior angles of polygons with 3, 4, 5, 6, ... sides form an arithmetic progression. Find the sum of the interior angles for a 21 sided polygon. [NCERT EXEMPLAR] Solution: We know that, thesum of the interior angles of a polygon with 3 sides,a1=180, thesum of the interior angles of a polygon with 4 sides,a2=360, thesum of the interior angles of a polygon with 5 sides,a3 =540,...
Read More →A carpenter was hired to build 192 window frames.
Question: A carpenter was hired to build 192 window frames. The first day he made five frames and each day thereafter he made two more frames than he made the day before. How many days did it take him to finish the job? [NCERT EXEMPLAR] Solution: We have: $S=192, a=5, d=2$ Now, $S_{n}=192$ $\Rightarrow \frac{n}{2}[2 a+(n-1) d]=192$ $\Rightarrow \frac{n}{2}[2 \times 5+(n-1) \times 2]=192$ $\Rightarrow \frac{n}{2}[10+2 n-2]=192$ $\Rightarrow \frac{n}{2}[2 n+8]=192$ $\Rightarrow n(n+4)=192$ $\Right...
Read More →There are two temples, one on each bank of a river, just opposite to each other. One temple is 50 m high.
Question: There are two temples, one on each bank of a river, just opposite to each other. One temple is 50 m high. From the top of this temple, the angles of depression of the top and the foot of the other temple are 30 and 60 respectively. Find the width of the river and the height of the other temple. Solution: Letandare two temples each at the bank of river. The top of the templeCEmakes angle of depressions at the top and bottom of towerABare 30 and 60 Let $C E=50 \mathrm{~m}$ and $A B=H \ma...
Read More →In the given figure, ABCD is a parallelogram, M is the mid-point of BD and BD bisects ∠B as well as ∠D.
Question: In the given figure,ABCDis a parallelogram,Mis the mid-point ofBDandBDbisectsBas well as D. Then, AMB= ?(a) 45(b) 60(c) 90(d) 30 Solution: (c) 90Explanation:B = D $\Rightarrow \frac{1}{2} \angle B=\frac{1}{2} \angle D$ ⇒ ADB =ABD ∆ABD is an isosceles triangle and M is the midpoint of BD. We can also say that M is the median of ∆ABD. AM BD and, hence, AMB = 90...
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Question: If $\sin ^{-1} x+\sin ^{-1} y+\sin ^{-1} z+\sin ^{-1} t=2 \pi$, then find the value of $x^{2}+y^{2}+z^{2}+t^{2}$ Solution: We know that the maximum value of $\sin ^{-1} x, \sin ^{-1} y, \sin ^{-1} z$ and $\sin ^{-1} t$ is $\frac{\pi}{2}$ Now, $\mathrm{LHS}=\sin ^{-1} x+\sin ^{-1} y+\sin ^{-1} z+\sin ^{-1} t$ $=\frac{\pi}{2}+\frac{\pi}{2}+\frac{\pi}{2}+\frac{\pi}{2}$ $=2 \pi=\mathrm{RHS}$ Now, $\sin ^{-1} x=\frac{\pi}{2}, \sin ^{-1} y=\frac{\pi}{2}, \sin ^{-1} z=\frac{\pi}{2}$ and $\sin...
Read More →A man standing on the deck of a ship, which is 8 m above water level.
Question: A man standing on the deck of a ship, which is 8 m above water level. He observes the angle of elevation of the top of a hill as 60 and the angle of depression of the base of the hill as 30. Calculate the distance of the hill from the ship and the height of the hill. Solution: LetHbe height of hillCEand a man is standing on a ships at the height of 8meter above from the water level. LetAB =8,BC = x,AD = BC,AB = DC,DE=h., and We have to findxandH The corresponding figure is as follows I...
Read More →A man starts repaying a loan as first instalment of Rs 100 = 00.
Question: A man starts repaying a loan as first instalment of Rs 100 = 00. If he increases the instalments by Rs 5 every month, what amount he will pay in the 30th instalment? Solution: Let $a_{30}$ be the amount a man repays in the 30 th instalment. Letdbe the common increment in his instalment every month. Letabe the initial repayment. Here,a= 100,d= 5,n= 30 Amount to be repaid in the 30th instalment: $a_{30} \Rightarrow a+(n-1) d$ $=100+29 \times 5$ $=245$ Hence, the man repays Rs 245 in his ...
Read More →The income of a person is Rs 300,000 in the first year and he receives an increase of Rs 10000
Question: The income of a person is Rs 300,000 in the first year and he receives an increase of Rs 10000 to his income per year for the next 19 years. Find the total amount, he received in 20 years. Solution: Let $S_{n}$ denote the total amount the person receives in $n$ years. Letdbe the common increment in his income every year. Letadenote the initial income of the person. $S_{20}=\frac{20}{2}\{2 \times 300,000+(20-1) 10,000\}$ $=79,00,000$ Therefore, the total amount the person receives in 20...
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Question: $5 \tan ^{-1} x+3 \cot ^{-1} x=2 \pi$ Solution: $5 \tan ^{-1} x+3 \cot ^{-1} x=2 \pi$ $\Rightarrow 5 \tan ^{-1} x+3\left(\frac{\pi}{2}-\tan ^{-1} x\right)=2 \pi$ $\left[\because \cot ^{-1} x=\frac{\pi}{2}-\tan ^{-1} x\right]$ $\Rightarrow 5 \tan ^{-1} x+\frac{3 \pi}{2}-3 \tan ^{-1} x=2 \pi$ $\Rightarrow 2 \tan ^{-1} x=\frac{\pi}{2}$ $\Rightarrow \tan ^{-1} x=\frac{\pi}{4}$ $\Rightarrow x=\tan \frac{\pi}{4}=1$...
Read More →A man on the deck of a ship is 10 m above the water level. He observes that the
Question: A man on the deck of a ship is 10 m above the water level. He observes that the angle of elevation of the top of a cliff is 45 and the angle of depression of the base is 30. Calculate the distance of the cliff from the ship and the height of the cliff. Solution: LetHbe the height of the cliffCE. And a man is standing on the ships at the height of10 meter above from the water level. LetAB= 10,BC = x,AD = BC,AB = DC,DE = h.and We have to findHandx The corresponding figure is as follows I...
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Question: $\tan ^{-1} x+2 \cot ^{-1} x=\frac{2 \pi}{3}$ Solution: $\tan ^{-1} x+2 \cot ^{-1} x=\frac{2 \pi}{3}$ $\Rightarrow \tan ^{-1} x+2\left(\frac{\pi}{2}-\tan ^{-1} x\right)=\frac{2 \pi}{3}$ $\left[\because \cot ^{-1} x=\frac{\pi}{2}-\tan ^{-1} x\right]$ $\Rightarrow \tan ^{-1} x+\pi-2 \tan ^{-1} x=\frac{2 \pi}{3}$ $\Rightarrow \tan ^{-1} x=\frac{\pi}{3}$ $\Rightarrow \tan ^{-1} x=\frac{\pi}{3}$ $\Rightarrow \mathrm{x}=\tan \frac{\pi}{3}=\sqrt{3}$...
Read More →Shamshad Ali buys a scooter for Rs 22000. He pays Rs 4000 cash and agrees to pay the balance in annual instalments of Rs 1000 plus 10% interest on the unpaid amount.
Question: Shamshad Ali buys a scooter for Rs 22000. He pays Rs 4000 cash and agrees to pay the balance in annual instalments of Rs 1000 plus 10% interest on the unpaid amount. How much the scooter will cost him. Solution: Cost of the scooter = Rs 22000 Shamshad Ali pays Rs 4000 in cash. $\therefore$ Unpaid amount $=$ Rs $22000-$ Rs $4000=$ Rs 18000 Number of years taken by Shamshed Ali to pay the whole amount $=18000 \div 1000=18$ He agrees to pay the balance in annual instalments of Rs 1000 plu...
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Question: $4 \sin ^{-1} x=\pi-\cos ^{-1} x$ Solution: $4 \sin ^{-1} x=\pi-\cos ^{-1} x$ $\Rightarrow 4 \sin ^{-1} x=\pi-\left(\frac{\pi}{2}-\sin ^{-1} x\right)$ $\left[\because \cos ^{-1} x=\frac{\pi}{2}-\sin ^{-1} x\right]$ $\Rightarrow 4 \sin ^{-1} x=\frac{\pi}{2}+\sin ^{-1} x$ $\Rightarrow 3 \sin ^{-1} x=\frac{\pi}{2}$ $\Rightarrow \sin ^{-1} x=\frac{\pi}{6}$ $\Rightarrow x=\sin \frac{\pi}{6}=\frac{1}{2}$...
Read More →A fire in a building B is reported on telephone to two fire stations P and Q,
Question: A fire in a building B is reported on telephone to two fire stations P and Q, 20 km a part from each other on a straight road. P observes that the fire is at an angle of 60 to the road and Q observes that it is at an angle of 45 to the road. Which station should send its team and how much will this team have to travel? Solution: LetABbe the building of heighth.PObserves that the fire is at an angle of 60 to the road andQobserves that the fire is at an angle of 45 to the road.\ Let $Q A...
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Question: $\sin ^{-1} x=\frac{\pi}{6}+\cos ^{-1} x$ Solution: $\sin ^{-1} x=\frac{\pi}{6}+\cos ^{-1} x$ $\Rightarrow \sin ^{-1} x=\frac{\pi}{6}+\frac{\pi}{2}-\sin ^{-1} x$ $\left[\because \cos ^{-1} x=\frac{\pi}{2}-\sin ^{-1} x\right]$ $\Rightarrow 2 \sin ^{-1} x=\frac{2 \pi}{3}$ $\Rightarrow \sin ^{-1} x=\frac{\pi}{3}$ $\Rightarrow \sin ^{-1} x=\frac{\pi}{3}$ $\Rightarrow \mathrm{x}=\sin \frac{\pi}{3}=\frac{\sqrt{3}}{2}$...
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Question: $\sin \left(\sin ^{-1} \frac{1}{5}+\cos ^{-1} x\right)=1$ Solution: $\sin \left(\sin ^{-1} \frac{1}{5}+\cos ^{-1} x\right)=1$ $\Rightarrow \sin ^{-1} \frac{1}{5}+\cos ^{-1} x=\sin ^{-1} 1$ $\Rightarrow \sin ^{-1} \frac{1}{5}+\cos ^{-1} x=\frac{\pi}{2}$ $\Rightarrow \sin ^{-1} \frac{1}{5}=\frac{\pi}{2}-\cos ^{-1} x$ $\Rightarrow \sin ^{-1} \frac{1}{5}=\sin ^{-1} x$ $\left[\because \sin ^{-1} x=\frac{\pi}{2}-\cos ^{-1} x\right]$ $\Rightarrow x=\frac{1}{5}$...
Read More →A farmer buys a used tractor for Rs 12000.
Question: A farmer buys a used tractor for Rs 12000. He pays Rs 6000 cash and agrees to pay the balance in annual instalments of Rs 500 plus 12% interest on the unpaid amount. How much the tractor cost him? Solution: Cost of the tractor = Rs 12000 It is given that the farmer pays Rs 6000 in cash. Unpaid amount = Rs 6000 He has to pay Rs 6000 in annual instalments of Rs 500 plus 12% interest on the unpaid amount. $\therefore$ Number of years taken by the farmer to pay the whole amount $=6000 \div...
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Question: If $\left(\sin ^{-1} x\right)^{2}+\left(\cos ^{-1} x\right)^{2}=\frac{17 \pi^{2}}{36}$, find $x$ Solution: $\left(\sin ^{-1} x\right)^{2}+\left(\cos ^{-1} x\right)^{2}=\frac{17 \pi^{2}}{36}$ $\Rightarrow\left(\sin ^{-1} x\right)^{2}+\left(\frac{\pi}{2}-\sin ^{-1} x\right)^{2}=\frac{17 \pi^{2}}{36}$ Let $\sin ^{-1} x=y$ $\therefore(y)^{2}+\left(\frac{\pi}{2}-y\right)^{2}=\frac{17 \pi^{2}}{36}$ $\Rightarrow y^{2}+\frac{\pi^{2}}{4}+y^{2}-2 \times \frac{\pi}{2} \times y=\frac{17 \pi^{2}}{3...
Read More →Find the angle of elevation of the sum (sun's altitude) when the length of
Question: Find the angle of elevation of the sum (sun's altitude) when the length of the shadow of a vertical pole is equal to its height. Solution: Letbe the angle of elevation of sun. Letbe the vertical pole of heightandbe the shadow of equal length. Here we have to find angle of elevation of sun. We have the corresponding figure as follows So we use trigonometric ratios to find the required angle. In a triangle, $\Rightarrow \quad \tan \theta=\frac{A B}{B C}$ $\Rightarrow \quad \tan \theta=\f...
Read More →Two men on either side of the cliff 80 m high observes the angles of a
Question: Two men on either side of the cliff 80 m high observes the angles of a elevation of the top of the cliff to be 30 and 60 respectively. Find the distance between the two men. Solution: LetABandADbe the two men either side of cliff and height of cliff is 80 m. And makes an angle of elevation, 30 and 60 respectively of the top of the cliff We have given thatAC= 80 m. LetBC = xandCD = y. And, Here we have to find height of cliff. We have the corresponding figure as follows So we use trigon...
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Question: If $\cot \left(\cos ^{-1} \frac{3}{5}+\sin ^{-1} x\right)=0$, find the values of $x$. Solution: $\cot \left(\cos ^{-1} \frac{3}{5}+\sin ^{-1} x\right)=0$ $\Rightarrow \cos ^{-1} \frac{3}{5}+\sin ^{-1} x=\cot 0$ $\Rightarrow \cos ^{-1} \frac{3}{5}+\sin ^{-1} x=\frac{\pi}{2}$ $\Rightarrow \cos ^{-1} \frac{3}{5}=\frac{\pi}{2}-\sin ^{-1} x$ $\Rightarrow \cos ^{-1} \frac{3}{5}=\cos ^{-1} x \quad\left[\because \cos ^{-1} x=\frac{\pi}{2}-\sin ^{-1} x\right]$ $\Rightarrow x=\frac{3}{5}$...
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