A parachutist is descending vertically and makes angles of elevation of 45° and 60°
Question: A parachutist is descending vertically and makes angles of elevation of 45 and 60 at two observing points 100 m apart from each other on the left side of himself. Find the maximum height from which he falls and the distance of the point where he falls on the ground form the just observation point. Solution: Let BCbe the heighthof the parachutist and makes an angle of elevationsandrespectively at two observing pointsapart from each other. So we use trigonometric ratios. In triangle BCD ...
Read More →Show that the quadrilateral formed by joining the midpoints of the pairs of adjacent sides of a square is a square.
Question: Show that the quadrilateral formed by joining the midpoints of the pairs of adjacent sides of a square is a square. Solution: LetABCDbe a square andP,Q, RandSbe the midpoints ofAB, BC, CDandDA,respectively.Join the diagonalsACandBD.LetBDcutSRatFandACcutRQatE. LetObe the intersection point ofACandBD.In∆ABC,we have: $\therefore P Q \| A C$ and $P Q=\frac{1}{2} A C$ [By midpoint theorem] Again, in ∆DAC,the pointsSandRare the midpoints ofADandDC,respectively. $\therefore S R \| A C$ and $S...
Read More →The shadow of a tower, when the angle of elevation of the sun is 45°,
Question: The shadow of a tower, when the angle of elevation of the sun is 45, is found to be 10 m longer than when it was 60. Find the height of the tower. Solution: Lethbe height of towerABand angle of elevation are 45 and 60 are given. In a triangleOAC, given thatAB= 10+xandBC=x Now we have to find height of tower. So we use trigonometrical ratios. In a triangleOAB, $\Rightarrow \quad \tan A=\frac{O B}{A B}$ $\Rightarrow \tan 45^{\circ}=\frac{O B}{A B}$ $\Rightarrow \quad 1=\frac{h}{10+x}$ $\...
Read More →Show that the quadrilateral formed by joining the midpoints of the pairs of adjacent sides of a rhombus is a rectangle.
Question: Show that the quadrilateral formed by joining the midpoints of the pairs of adjacent sides of a rhombus is a rectangle. Solution: LetABCDbe a rhombus andP,Q, RandSbe the midpoints ofAB, BC, CDandDA, respectively.Join the diagonals,ACandBD.In∆ABC,we have: $P Q \| A C$ and $P Q=\frac{1}{2} A C$ [By midpoint theorem] Again, in∆DAC,the pointsSandRare the midpoints ofADandDC,respectively. $\therefore S R \| A C$ and $S R=\frac{1}{2} A C$ [By midpoint theorem] Now,PQ∣∣ACandSR∣∣AC⇒PQ∣∣ SR Als...
Read More →A vertical tower stands on a horizontal plane and is surmounted by a vertical flag-staff
Question: A vertical tower stands on a horizontal plane and is surmounted by a vertical flag-staff of height 5 metres. At a point on the plane, the angles of elevation of the bottom and the top of the flag-staff are respectively 30 and 60. Find the height of the tower. Solution: LetBCbe the tower of heighthm andABbe the flag staff with distance 5m.Then angle of elevation from the top and bottom of flag staff are 60 and 30 respectively. Let $C D=x$ and $\angle A D C=60^{\circ}, \angle B D C=30^{\...
Read More →Evaluate each of the following:
Question: Evaluate each of the following: (i) $\cos ^{-1}\left\{\cos \left(-\frac{\pi}{4}\right)\right\}$ (ii) $\cos ^{-1}\left\{\cos \frac{5 \pi}{4}\right\}$ (iii) $\cos ^{-1}\left\{\cos \left(\frac{4 \pi}{3}\right)\right\}$ (iv) $\cos ^{-1}\left(\cos \frac{13 \pi}{6}\right)$ (v) $\cos ^{-1}(\cos 3)$ (vi) $\cos ^{-1}(\cos 4)$ (vii) $\cos ^{-1}(\cos 5)$ (viii) $\cos ^{-1}(\cos 12)$ Solution: We know \cos ^{-1}(\cos \theta)=\theta \text { if } 0 \leq \theta \leq \pi (i) We have $\cos ^{-1}\left\{...
Read More →A person observed the angle of elevation of the top of a tower as 30°.
Question: A person observed the angle of elevation of the top of a tower as 30. He walked 50 m towards the foot of the tower along level ground and found the angle of elevation of the top of the tower as 60. Find the height of the tower. Solution: Letbe the tower of height. And person makes an angle of elevation of top of tower is 30, he walksm towards the foot of tower then makes an angle of elevation 60 Let $B C=x, C D=50$, and $\angle A C B=60^{\circ}, \angle A D B=30^{\circ}$ Now we have to ...
Read More →A vertically straight tree, 15 m high,
Question: A vertically straight tree, 15 m high, is broken by the wind in such a way that it top just touches the ground and makes an angle of 60 with the ground. At what height from the ground did the tree break? Solution: Letbe the tree of desired heightxm and tree is broken by wind then tree makes an angle. LetAC=15 x Here we have to find heightx So we use trigonometric ratios. In a triangle, $\Rightarrow \quad \sin C=\frac{A B}{A C}$ $\Rightarrow \quad \sin 60^{\circ}=\frac{x}{15-x}$ $\Right...
Read More →A vertical tower stands on a horizontal plane and is surmounted by a vertical flag-staff.
Question: A vertical tower stands on a horizontal plane and is surmounted by a vertical flag-staff. At a point on the plane 70 metres away from the tower, an observer notices that the angles of elevation of the top and the bottom of the flagstaff are respectively 60 and 45. Find the height of the flag-staff and that of the tower. Solution: LetBCbe the tower of heightxm andABbe the flag staff of heighty, 70 m away from the tower, makes an angle of elevation are 60 and 45 respectively from top and...
Read More →The length of a string between a kite and a point on the ground is 90 metres.
Question: The length of a string between a kite and a point on the ground is 90 metres. If the string makes an angle with the ground level such that tan = 15/8, how high is the kite? Assume that there is no slack in the string. Solution: Letbe the kite of heightm and the length of string between kite and a point on the ground ismeters. Someters and Now we have to find the height of kite. We have the corresponding figure as follows Given that $\tan \theta=\frac{15}{8}$ Since $\sec ^{2} \t...
Read More →Solve this
Question: (i) $\sin ^{-1}\left(\sin \frac{\pi}{6}\right)$ (ii) $\sin ^{-1}\left(\sin \frac{7 \pi}{6}\right)$ (iii) $\sin ^{-1}\left(\sin \frac{5 \pi}{6}\right)$ (iv) $\sin ^{-1}\left(\sin \frac{13 \pi}{7}\right)$ (v) $\sin ^{-1}\left(\sin \frac{17 \pi}{8}\right)$ (vi) $\sin ^{-1}\left\{\left(\sin -\frac{17 \pi}{8}\right)\right\}$ (vii) $\sin ^{-1}(\sin 3)$ (viii) $\sin ^{-1}(\sin 4)$ (ix) $\sin ^{-1}(\sin 12)$ (x) $\sin ^{-1}(\sin 2)$ Solution: We know $\sin \left(\sin ^{-1} \theta\right)=\theta...
Read More →Solve this
Question: (i) $\sin ^{-1}\left(\sin \frac{\pi}{6}\right)$ (ii) $\sin ^{-1}\left(\sin \frac{7 \pi}{6}\right)$ (iii) $\sin ^{-1}\left(\sin \frac{5 \pi}{6}\right)$ (iv) $\sin ^{-1}\left(\sin \frac{13 \pi}{7}\right)$ (v) $\sin ^{-1}\left(\sin \frac{17 \pi}{8}\right)$ (vi) $\sin ^{-1}\left\{\left(\sin -\frac{17 \pi}{8}\right)\right\}$ (vii) $\sin ^{-1}(\sin 3)$ (viii) $\sin ^{-1}(\sin 4)$ (ix) $\sin ^{-1}(\sin 12)$ (x) $\sin ^{-1}(\sin 2)$ Solution: We know $\sin \left(\sin ^{-1} \theta\right)=\theta...
Read More →A kit is flying at a height of 75 metres from the ground level,
Question: A kit is flying at a height of 75 metres from the ground level, attached to a string inclined at 60 to the horizontal. Find the length of the string to the nearest metre. Solution: LetACbe the string of lengthhm andCbe the point, makes an angle of 60 and the kite is flying at the height of 75 m from the ground level. In a triangle, given that height of kite isAB= 75 m and angle C = 60 Now we have to find the length of string. So we use trigonometric ratios. In a triangle, $\Rightarrow ...
Read More →An electric pole is 10 m high. A steel wire tied to top of the pole is affixed at a point on the ground
Question: An electric pole is 10 m high. A steel wire tied to top of the pole is affixed at a point on the ground to keep the pole up right. If the wire makes an angle of 45 with the horizontal through the foot of the pole, find the length of the wire. Solution: LetACbe the wire of lengthm andCbe the point, makes an angle of 45 In a triangleABC, given that height of electric pole isBC= 2m and angleC= 45 Now we have to find the length of wire. So we use trigonometrically ratios. In a triangleABC,...
Read More →Evaluate each of the following:
Question: Evaluate each of the following: (i) $\cot ^{-1} \frac{1}{\sqrt{3}}-\operatorname{cosec}^{-1}(-2)+\sec ^{-1}\left(\frac{2}{\sqrt{3}}\right)$ (ii) $\cot ^{-1}\left\{2 \cos \left(\sin ^{-1} \frac{\sqrt{3}}{2}\right)\right\}$ (iii) $\operatorname{cosec}^{-1}\left(-\frac{2}{\sqrt{3}}\right)+2 \cot ^{-1}(-1)$ (iv) $\tan ^{-1}\left(-\frac{1}{\sqrt{3}}\right)+\cot ^{-1}\left(\frac{1}{\sqrt{3}}\right)+\tan ^{-1}\left(\sin \left(-\frac{\pi}{2}\right)\right)$ Solution: (i) $\cot ^{-1} \frac{1}{\s...
Read More →A ladder is placed along a wall of a house such that its upper end is touching the top of the wall.
Question: A ladder is placed along a wall of a house such that its upper end is touching the top of the wall. The foot of the ladder is 2 m away from the wall and the ladder is making an angle of 60 with the level of the ground. Determine the height of the wall. Solution: LetABbe the wall of heighthm andCbe the points, makes an angle 60 and foot of the ladder is 2m away from the wall. We have to find height of wall In a triangleABC, given thatBC= 2m and angleC= 60 Now we have to find the height ...
Read More →The angle of elevation of a ladder leaning against a wall is 60° and the foot
Question: The angle of elevation of a ladder leaning against a wall is 60 and the foot of the ladder is 9.5 m away from the wall. Find the length of the ladder. Solution: Letbe the ladder of lengthm andCbe the points, makes an angle of elevation 60 with the wall and foot of the ladder is 9.5 meter away from wall. In a triangleABC, given thatBC= 9.5 m and angle C = 60 Now we have to find length of ladder. So we use trigonometrically ratios. In a triangleABC, $\Rightarrow \quad \cos C=\frac{B C}{A...
Read More →A tower stand vertically on the ground.
Question: A tower stand vertically on the ground. From a point on the ground 20 m away from the foot of the tower, the angle of elevation of the top of the tower is 60. What is the height of the tower? Solution: Letbe the tower of heightm andCbe the point on the ground, makes an angle of elevationwith the top of tower. In a triangle, given thatBC= 20 m and Now we have to find height of tower, so we use trigonometrical ratios. In the triangle, $\Rightarrow \tan C=\frac{A B}{B C}$ $\Rightarrow \sq...
Read More →Find the domain
Question: Find the domain of $f(x)=\cot x+\cot ^{-1} x$ Solution: Let $f(x)=g(x)+h(x)$, where $g(x)=\cot x$ and $h(x)=\cot ^{-1} x$ Therefore, the domain of $f(x)$ is given by the intersection of the domain of $g(x)$ and $h(x)$ The domain of $g(x)$ is $\mathrm{R}-\{n \pi, n \dot{E} Z\}$ The domain of $h(x)$ is $(0, \pi)$ Therfore, the intersection of $g(x)$ and $h(x)$ is $\mathrm{R}-\{n \pi, n \dot{E} Z\}$...
Read More →If a, b, c are in A.P., prove that:
Question: Ifa,b,care in A.P., prove that: (i) (ac)2= 4 (ab) (bc) (ii)a2+c2+ 4ac= 2 (ab+bc +ca) (iii)a3+c3+ 6abc= 8b3. Solution: Sincea, b, care in A.P., we have: 2b= a+c $\Rightarrow b=\frac{a+c}{2}$ (i) Consider RHS: 4 (ab) (bc) Substituting $b=\frac{a+c}{2}:$ $\Rightarrow 4\left\{a-\frac{a+c}{2}\right\}\left\{\frac{a+c}{2}-c\right\}$ $\Rightarrow 4\left\{\frac{2 a-a-c}{2}\right\}\left\{\frac{a+c-2 c}{2}\right\}$ $\Rightarrow(a-c)(a-c)$ $\Rightarrow(a-c)^{2}$ Hence, proved. (ii) Consider RHS: 2...
Read More →Find the principal values of each of the following:
Question: Find the principal values of each of the following: (i) $\cot ^{-1}(-\sqrt{3})$ (ii) $\cot ^{-1}(\sqrt{3})$ (iii) $\cot ^{-1}\left(-\frac{1}{\sqrt{3}}\right)$ (iv) $\cot ^{-1}\left(\tan \frac{3 \pi}{4}\right)$ Solution: (i) Let $\cot ^{-1}(-\sqrt{3})=y$ Then, $\cot y=-\sqrt{3}$ We know that the range of the principal value branch is $(0, \pi)$. Thus, $\cot y=-\sqrt{3}=\cot \left(\frac{5 \pi}{6}\right)$ $\Rightarrow y=\frac{5 \pi}{6} \in(0, \pi)$ Hence, the principal value of $\cot ^{-1...
Read More →If a, b, c are in A.P., prove that:
Question: Ifa,b,care in A.P., prove that: (i) (ac)2= 4 (ab) (bc) (ii)a2+c2+ 4ac= 2 (ab+bc +ca) (iii)a3+c3+ 6abc= 8b3. Solution: Sincea, b, care in A.P., we have: 2b= a+c $\Rightarrow b=\frac{a+c}{2}$ (i) Consider RHS: 4 (ab) (bc) Substituting $b=\frac{a+c}{2}:$ $\Rightarrow 4\left\{a-\frac{a+c}{2}\right\}\left\{\frac{a+c}{2}-c\right\}$ $\Rightarrow 4\left\{\frac{2 a-a-c}{2}\right\}\left\{\frac{a+c-2 c}{2}\right\}$ $\Rightarrow(a-c)(a-c)$ $\Rightarrow(a-c)^{2}$ Hence, proved. (ii) Consider RHS: 2...
Read More →Draw a right triangle in which sides (other than the hypotenuse) are of lengths 8 cm and 6 cm.
Question: Draw a right triangle in which sides (other than the hypotenuse) are of lengths $8 \mathrm{~cm}$ and $6 \mathrm{~cm}$. Then construct another triangle whose sides are $\frac{3}{4}$ times the corresponding sides of the first triangle. Solution: Given that Construct a right triangle of sides let $A B=8 \mathrm{~cm}, A C=6 \mathrm{~cm}$, and $\angle A=90^{\circ}$ and then a triangle similar to it whose sides are $(3 / 4)^{\text {th }}$ of the corresponding sides of $\triangle A B C$. We f...
Read More →Solve the following
Question: If $\frac{b+c}{a}, \frac{c+a}{b}, \frac{a+b}{c}$ are in A.P., prove that: (i) $\frac{1}{a}, \frac{1}{b}, \frac{1}{c}$ are in A.P. (ii)bc,ca,abare in A.P. Solution: (i) Since $\frac{b+c}{a}, \frac{c+a}{b}, \frac{a+b}{c}$ are in A.P., we have: $\frac{c+a}{b}-\frac{b+c}{a}=\frac{a+b}{c}-\frac{c+a}{b}$ $\Rightarrow \frac{a c+a^{2}-b^{2}-b c}{a b}=\frac{a b+b^{2}-c^{2}-a c}{b c}$ $\Rightarrow \frac{(a+b)(a-b)+c(a-b)}{a b}=\frac{(b+c)(b-c)+a(b-c)}{b c}$ $\Rightarrow \frac{(a-b)(a+b+c)}{a b}=...
Read More →Construct a triangle similar to a given ΔXYZ with its sides equal to (3/4)th of the corresponding
Question: Construct a triangle similar to a given ΔXYZwith its sides equal to (3/4)thof the corresponding sides of ΔXYZ. Write the steps of construction. Solution: Given that Construct a $\triangle X Y Z$ of given data, Let $X Y=5 \mathrm{~cm}, Y Z=6 \mathrm{~cm}$ and $\angle Y=60^{\prime}$ and then a triangle similar to it whose sides are $(3 / 4)$ of the corresponding sides of $\triangle X Y Z$. We follow the following steps to construct the given Step of construction Step: I- First of all we ...
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