Chapter 28

A Complete Resource Book in Mathematics for JEE Main · 51 exercises

Problem 2

Two straight roads \(O A\) and \(O B\) intersect at \(O\). A tower is situated within the angle formed by them and subtends angles of \(45^{\circ}\) and \(30^{\circ}\) at the points \(A\) and \(B\) where the roads are nearest to it. If \(O A=a\) and \(O B=b\), then the height of the tower is (A) \(\sqrt{\frac{a^{2}+b^{2}}{2}}\) (B) \(\sqrt{a^{2}+b^{2}}\) (C) \(\sqrt{\frac{a^{2}-b^{2}}{2}}\) (D) \(\sqrt{a^{2}-b^{2}}\)

5 step solution

Problem 3

\(A\) and \(B\) are two points in the horizontal plane through \(O\), the foot of pillar \(O P\) of height \(h\), such that \(\triangle A O B=\theta\). If the elevation of the top of the pilar from \(A\) and \(B\) are also equal to \(\theta\), then \(A B\) is equal to (A) \(h \cot \theta\) (B) \(h \cos \theta \sec \frac{\theta}{2}\) (C) \(h \cot \theta \sin \frac{\theta}{2}\) (D) \(h \cos \theta \operatorname{cosec} \frac{\theta}{2}\)

6 step solution

Problem 5

An observer finds that the angular elevation of a tower is \(A\). On advancing \(3 \mathrm{~m}\) towards the tower the elevation is \(45^{\circ}\) and on advancing \(2 \mathrm{~m}\) nearer, the elevation is \(90^{\circ}\) \(-A\). The height of the tower is (A) \(2 \mathrm{~m}\) (B) \(4 \mathrm{~m}\) (C) \(6 \mathrm{~m}\) (D) \(8 \mathrm{~m}\)

8 step solution

Problem 6

A lamp post standing at a point \(A\) on a circular path of radius \(\mathrm{r}\) subtends an angle \(\alpha\) at some point \(B\) on the path, and \(A B\) subtends an angle of \(45^{\circ}\) at any other point on the path, then height of the lampost is (A) \(\sqrt{2} r \cot \alpha\) (B) \((r / \sqrt{2}) \tan \alpha\) (C) \(\sqrt{2} r \tan \alpha\) (D) \((r / \sqrt{2}) \cot \alpha\)

6 step solution

Problem 7

\(P Q\) is a vertical tower, \(P\) is the foot, \(Q\) the top of the tower, \(A, B, C\) are three points in the horizontal plane through \(P\). The angles of elevation of \(Q\) from \(A, B, C\) are equal and each is equal to \(\theta\). The sides of the triangle \(A B C\) are \(a, b, c\) and the area of the triangle \(A B C\) is \(\Delta\). The height of the tower is (A) \((a b c) \tan \theta / 4 \Delta\) (B) \((a b c) \cot \theta / 4 \Delta\) (C) \((a b c) \tan \theta / 4 \Delta\) (D) none of these

6 step solution

Problem 9

A person standing at the foot of a tower walks a distance \(3 a\) away from the tower and observes that the angle of elevation of the top of the tower is \(\alpha\). He then walks a distance \(4 a\) perpendicular to the previous direction and observes the angle of elevation to be \(\beta\). Then height of the tower is (A) \(3 a \tan \alpha\) or \(5 a \tan \beta\) (B) \(5 a \tan \alpha\) or \(3 a \tan \beta\) (C) \(4 a \tan \beta\) (D) \(7 a \tan \alpha\)

6 step solution

Problem 10

The angle of elevation of the top of a tree at point \(B\) due south of it is \(60^{\circ}\) and at a point \(C\) due north of it is \(30^{\circ} . D\) is a point due north of \(C\) where the angle of elevation is \(15^{\circ}\). If \(\sqrt{3}=1 \frac{8}{11}\) and \(B C \times C D=2^{3} \times 3^{2}\) \(\times 19 \times 11\), the height of the tree is (A) 33 (B) 38 (C) 57 (D) 88

4 step solution

Problem 12

The angular elevation of tower \(C D\) at a point \(A\) due south of it is \(60^{\circ}\) and at a point \(B\) due west of \(A\), the elevation is \(30^{\circ}\). If \(A B=3 \mathrm{~km}\), the height of the tower is (A) \(2 \sqrt{3} \mathrm{~km}\) (B) \(2 \sqrt{6} \mathrm{~km}\) (C) \(\frac{3 \sqrt{3} \mathrm{~km}}{2}\) (D) \(\frac{3 \sqrt{6} \mathrm{~km}}{4}\)

4 step solution

Problem 13

An isosceles triangle of wood of base \(2 a\) and height \(h\) is placed with its base on the ground and vertex directly above. The triangle faces the sun whose altitude is \(30^{\circ}\). Then the tangent of the angle at the apex of the shadow is (A) \(\frac{2 h a}{\sqrt{3}}\) (B) \(\frac{2 h a \sqrt{3}}{3 h^{2}-a^{2}}\) (C) \(\frac{a^{2}+h^{2}}{2 \sqrt{3}}\) (D) \(\frac{2 a h \sqrt{3}}{3 h^{2}+a^{2}}\)

5 step solution

Problem 14

The length of the shadow of a rod inclined at \(10^{\circ}\) to the vertical towards the sun is \(2.05\) metre when the elevation of the sun is \(38^{\circ} .\) The length of the rod is (A) \(\frac{2.05 \sin 38^{\prime \prime}}{\sin 42^{\prime \prime}}\) (B) \(\frac{2.05 \cos 38^{\prime \prime}}{\sin 42^{\prime \prime}}\) (C) \(\frac{2.05 \sin 42^{\prime \prime}}{\sin 38^{\prime \prime}}\) (D) \(\frac{2.05 \cos 42 "}{\sin 38^{n}}\)

5 step solution

Problem 15

\(A B C D\) is a rectangular field. A vertical lamp post of height \(12 \mathrm{~m}\) stands at the corner \(A\). If the angle of elevation of its top from \(B\) is \(60^{\circ}\) and from \(C\) is \(45^{\circ}\), then the area of the field is (A) \(48 \sqrt{2} s q-m\) (B) \(48 \sqrt{3} s q . m\) (C) \(48 s q . m\) (D) \(48 \sqrt{3} s q . m\)

7 step solution

Problem 17

The angle of elevation of a stationary cloud from a point \(2,500 \mathrm{~m}\) above a lake is \(15^{\circ}\) and the angle of depression of its reflection in the lake is \(45^{\circ}\). The height of the cloud above the lake level is (A) \(2500 \sqrt{3} \mathrm{~m}\) (B) \(2500 \mathrm{~m}\) (C) \(500 \sqrt{3} \mathrm{~m}\) (D) none of these

7 step solution

Problem 18

From the top of a tower \(100 \mathrm{~m}\) height, the angles of depression of two objects \(200 \mathrm{~m}\) apart on the horizontal plane and in a line passing through the foot of the tower and on the same side of the tower are \(45^{\circ}-A\) and \(45^{\circ}+A\). The angle \(\mathrm{A}\) is equal to (A) \(15^{\circ}\) (B) \(35^{\circ}\) (C) \(22 \frac{1}{2}^{\circ}\) (D) \(45^{\circ}\)

7 step solution

Problem 19

\(A B C\) is triangular park with \(A B=A C=100 \mathrm{~m}\). A clock tower is situated at the mid-point of \(B C\). The angles of elevation of the top of the tower at \(A\) and \(B\) are \(\cot ^{-1} 3.2\) and \(\operatorname{cosec}^{-1} 2.6\) respectively. The height of the tower is (A) \(50 \mathrm{~m}\) (B) \(25 m\) (C) \(40 \mathrm{~m}\) (D) none of these

5 step solution

Problem 22

A flag staff \(5 \mathrm{~m}\) high is placed on a building \(25 \mathrm{~m}\) high. If flag and building both subtend equal angles on the observer at a height \(30 \mathrm{~m}\), the distance between the observer and the top of the flag is (A) \(\frac{5 \sqrt{3}}{2}\) (B) \(5 \sqrt{\frac{3}{2}}\) (C) \(5 \sqrt{\frac{2}{3}}\) (D) \(\frac{5 \sqrt{2}}{3}\)

7 step solution

Problem 23

The angle of elevation of the top of a vertical pole when observed from each vertex of a regular hexagon is \(\frac{\pi}{3}\). If the area of the circle circumscribing the hexagon be \(A\) metre \(^{2}\) then the height of the tower is (A) \(\frac{2 A}{\sqrt{3 \pi}}\) metre (B) \(\frac{A}{\sqrt{3 \pi}}\) metre (C) \(2 \sqrt{\frac{A}{3 \pi}}\) metre (D) \(\sqrt{\frac{A}{3 \pi}}\) metre

6 step solution

Problem 25

From a point on a hill-side of constant inclination, the angle of elevation of the top of a flagstaff on its summit is observed to be \(\alpha\) and \(a\) metre nearer the top of the hill, it is \(\beta .\) If \(h\) is the height of the flagstaff, the inclination of the hill to the horizontal is (A) \(\sin ^{-1}\left(\frac{a \sin \alpha \sin \beta}{h \sin (\beta-\alpha)}\right)\) (B) \(\cos ^{-1}\left(\frac{a \sin \alpha \sin \beta}{h \sin (\beta-\alpha)}\right)\) (C) \(\tan ^{-1}\left(\frac{a \sin \alpha \sin \beta}{h \sin (\beta-\alpha)}\right)\) (D) none of these

7 step solution

Problem 26

A flagstaff stands verticality on a pillar, the height of the flagstaff being double the height of the pillar. A man on the ground at a distance finds that both the pillar and the flagstaff subtend equal angles at his eyes. The ratio of the height of the pillar and the distance of the man from the pillar is (A) \(\sqrt{3}: 1\) (B) \(1: \sqrt{3}\) (C) \(2: \sqrt{3}\) (D) none of these

5 step solution

Problem 28

ABC is an equilateral triangular plot. An electric pole stands at the vertex and makes an angle of \(60^{\circ}\) at either of the other two vertices. If the height of the triangle is \(100 \mathrm{~m}\), the height of the pole is (A) \(200 \mathrm{~m}\) (B) \(100 \mathrm{~m}\) (C) \(150 \mathrm{~m}\) (D) none of these

5 step solution

Problem 29

A ladder rests against a wall at an angle \(\alpha\) to the horizontal. If the foot is pulled away through a distance \(a\), it slides a distance \(\mathrm{b}\) down the wall, finally making an angle \(\beta\) with the horizontal. Then, \(\tan \left(\frac{\alpha+\beta}{2}\right)\) equal to (A) \(\frac{a}{b}\) (B) \(\frac{b}{a}\) (C) \(a b\) (D) none of these

6 step solution

Problem 31

A tower \(A B\) leans towards west making an angle \(\alpha\) with the vertical. The angular elevation of \(B\), the top most point of the tower is \(\beta\), as observed from a point \(C\) due east of \(A\) at a distnace \(d\) from \(A\). If the angular elevation of \(B\) from a point due east of \(C\) at a distance \(2 d\) from \(C\) is \(\gamma\), then (A) \(2 \tan \alpha=2 \cot \beta-\cot \gamma\) (B) \(2 \tan \alpha=3 \cot \beta-\cot \gamma\) (C) \(\tan \alpha=\cot \beta-\cot \gamma\) (D) none of these

5 step solution

Problem 32

\(A B\) is a vertical pole with \(B\) at the ground level and \(A\) at the top. A man finds that the angle of elevation of the point \(A\) from a certain point \(C\) on the ground is \(60^{\circ} .\) He moves away from the pole along the line \(B C\) to a point \(D\) such that \(C D=7 \mathrm{~m} .\) From \(D\) the angle of elevation of the point \(A\) is \(45^{\circ}\). Then the height of the pole is (A) \(\frac{7 \sqrt{3}}{2} \cdot \frac{1}{\sqrt{3}-1} \mathrm{~m}\) (B) \(\frac{7 \sqrt{3}}{2} \cdot(\sqrt{3}+1) \mathrm{m}\) (C) \(\frac{7 \sqrt{3}}{2} \cdot(\sqrt{3}-1) \mathrm{m}\) (D) \(\frac{7 \sqrt{3}}{2} \cdot \frac{1}{\sqrt{3}+1}\)

7 step solution

Problem 34

The longer side of a parallelogram is \(10 \mathrm{~cm}\) and the shorter is \(6 \mathrm{~cm}\). If the longer diagonal makes an angle \(30^{\circ}\) with the longer side, the length of the longer diag. onal (in \(\mathrm{cm}\) ) is (A) \(5 \sqrt{3}+\sqrt{11}\) (B) \(4 \sqrt{3}+\sqrt{11}\) (C) \(5 \sqrt{3}+\sqrt{13}\) (D) none of these

6 step solution

Problem 35

The angle of elevation of a tower from a point \(\mathrm{A}\) due south of it is \(\mathrm{x}\) and from a point \(\mathrm{B}\) due east of \(\mathrm{A}\) is \(\mathrm{y}\). If \(\mathrm{AB}=1\), then the height \(\mathrm{h}\) of the tower is given by (A) \(\frac{l}{\sqrt{\cot ^{2} y-\cot ^{2} x}}\) (B) \(\frac{l}{\sqrt{\tan ^{2} y-\tan ^{2} x}}\) (C) \(\frac{2 l}{\sqrt{\cot ^{2} y-\cot ^{2} x}}\) (D) none of these

6 step solution

Problem 37

If from the top of a tower, 60 metre high, the angles of depression of the top and floor of a house are \(\alpha\) and \(\beta\) respectively and if the height of the house is \(\frac{60 \sin (\beta-\alpha)}{x}\), then \(x=\) (A) \(\sin \alpha \sin \beta\) (B) \(\cos \alpha \cos \beta\) (C) \(\sin \alpha \cos \beta\) (D) \(\cos \alpha \sin \beta\)

7 step solution

Problem 38

Due south of a tower which is leaning towards north there are two stations at distances \(x\) and \(y\) respectively from its foot. If \(\alpha, \beta\) respectively be the angles of elevation of the top of the tower at these stations, then the inclination \(\theta\) of the tower to the horizontal is given by \(\cot \theta=\) (A) \(\frac{y \cot \alpha-x \cot \beta}{y-x}\) (B) \(\frac{y \cot \alpha+x \cot \beta}{y-x}\) (C) \(\frac{y \cot \alpha-x \cot \beta}{y+x}\) (D) \(\frac{y \tan \alpha-x \sin \beta}{y-x}\)

6 step solution

Problem 39

Two straight roads OA and OB intersect at O. A tower is situated within the angle formed by them and subtends angles of \(45^{\circ}\) and \(30^{\circ}\) at the points \(\mathrm{A}\) and \(\mathrm{B}\) where the roads are nearest to it. If \(\mathrm{OA}=\mathrm{a}\) and \(\mathrm{OB}=\mathrm{b}\), then the height of the tower is (A) \(\sqrt{\frac{a^{2}+b^{2}}{2}}\) (B) \(\sqrt{a^{2}+b^{2}}\) (C) \(\sqrt{\frac{a^{2}-b^{2}}{2}}\) (D) \(\sqrt{a^{2}-b^{2}}\)

5 step solution

Problem 40

A and B are two points in the horizontal plane through O, the foot of pillar OP of height \(h\), such that \(\angle A O B=\) \(\theta\). If the elevation of the top of the pillar from \(\mathrm{A}\) and \(\mathrm{B}\) are also equal to \(\theta\), then \(\mathrm{AB}\) is equal to (A) \(\operatorname{hcot} \theta\) (B) \(h \cos \theta \sec \frac{\theta}{2}\) (C) \(h \cot \theta \sin \frac{\theta}{2}\) (D) \(h \cos \theta \operatorname{cosec} \frac{\theta}{2}\)

4 step solution

Problem 41

A flag is mounted on the semicircular dome of radius \(\mathrm{r}\). The elevation of the top of the flag at any point on the ground is \(30^{\circ}\). Moving d distance towards the dome, when the flag is just visible, the angle of elevation is \(45^{\circ}\). The relation between \(\mathrm{r}\) and \(\mathrm{d}\) is (A) \(\mathrm{r}=\frac{d}{\sqrt{2}(\sqrt{3}-1)}\) (B) \(\mathrm{r}=d \frac{2 \sqrt{2}}{\sqrt{3}+1}\) (C) \(\mathrm{r}=\frac{d}{\sqrt{2}(\sqrt{3}+1)}\) (D) \(\mathrm{r}=d \frac{2 \sqrt{2}}{\sqrt{3}-1}\)

5 step solution

Problem 42

A lamp post standing at a point \(\mathrm{A}\) on a circular path of radius \(\mathrm{r}\) subtends an angle \(\alpha\) at some point \(\mathrm{B}\) on the path, and AB subtends an angle of \(45^{\circ}\) at any other point on the path, then height of the lampost is (A) \(\sqrt{2} r \cot \alpha\) (B) \((r / \sqrt{2}) \tan \alpha\) (C) \(\sqrt{2} r \tan \alpha\) (D) \((r / \sqrt{2}) \cot \alpha\)

5 step solution

Problem 43

PQ is a vertical tower, \(\mathrm{P}\) is the foot, \(\mathrm{Q}\) the top of the tower, \(\mathrm{A}, \mathrm{B}, \mathrm{C}\) are three points in the horizontal plane through \(\mathrm{P}\). The angles of elevation of \(\mathrm{Q}\) from \(\mathrm{A}, \mathrm{B}, \mathrm{C}\) are equal and each is equal to \(\theta\). The sides of the triangle \(\mathrm{ABC}\) are \(\mathrm{a}, \mathrm{b}, \mathrm{c}\) and the area of the triangle \(\mathrm{ABC}\) is \(\Delta\). The height of the tower is (A) (abc) \(\tan \theta / 4 \Delta\) (B) (abc) \(\cot \theta / 4 \Delta\) (C) (abc) \(\tan \theta / 4 \Delta\) (D) none of these

6 step solution

Problem 44

The angle of elevation of a cloud from a point \(\mathrm{h}\) metres above the surface of a lake is \(\theta\) and the angles of depression of its reflection is \(\phi .\) Then the height of the cloud is (A) \(\frac{h \sin (\phi+\theta)}{\sin (\phi-\theta)}\) (B) \(\frac{h}{\sin (\phi-\theta)}\) (C) \(h \tan (\phi-\theta)\) (D) \(\frac{h \sin (\phi-\theta)}{\sin (\phi+\theta)}\)

7 step solution

Problem 45

A person standing at the foot of a tower walks a distance \(3 \mathrm{a}\) away from the tower and observes that the angle of elevation of the top of the tower is \(\alpha\). He then walks a distance \(4 \mathrm{a}\) perpendicular to the previous direction and observes the angle of elevation to be \(\beta\). Then height of the tower is (A) \(3 \mathrm{a} \tan \alpha\) or \(5 \mathrm{a} \tan \beta\) (B) \(5 \mathrm{a} \tan \alpha\) or \(3 \mathrm{a} \tan \beta\) (C) \(4 \mathrm{a} \tan \beta^{\prime}\) (D) \(7 \mathrm{a} \tan \alpha\)

5 step solution

Problem 46

A stationary balloon is observed from three points \(\mathrm{A}\), \(\mathrm{B}\) and \(\mathrm{C}\) on the plane ground and is found that its angle of elevation from each point is \(\alpha\). If \(\angle \mathrm{ABC}=\beta\) and \(\mathrm{AC}=\mathrm{b}\), the height of the balloon is (A) \(\frac{b}{2} \tan \alpha \operatorname{cosec} \beta\) (B) \(\frac{b}{2} \tan \alpha \sin \beta\) (C) \(\frac{b}{2} \cot \alpha \operatorname{cosec} \beta\) (D) \(\frac{b}{2} \cot \alpha \sin \beta\)

5 step solution

Problem 47

An isosceles triangle of wood of base \(2 \mathrm{a}\) and height \(\mathrm{h}\) is placed with its base on the ground and vertex directly above. The triangle faces the sun whose altitude is \(30^{\circ}\). Then the tangent of the angle at the apex of the shadow is (A) \(\frac{2 h a}{\sqrt{3}}\) (B) \(\frac{2 h a \sqrt{3}}{3 h^{2}-a^{2}}\) (C) \(\frac{a^{2}+h^{2}}{2 \sqrt{3}}\) (D) \(\frac{2 a h \sqrt{3}}{3 h^{2}+a^{2}}\)

6 step solution

Problem 48

OAB is a triangle in the horizontal plane through the foot \(\mathrm{P}\) of the tower at the middle point of the side \(\mathrm{OB}\) of he triangle. If \(\mathrm{OA}=2 \mathrm{~m}, \mathrm{OB}=6 \mathrm{~m}, \mathrm{AB}=5 \mathrm{~m}\) and \(\angle \mathrm{AOB}\) is equal to the angle subtended by the tower at A then the height of the tower is (A) \(\sqrt{\frac{11 \times 39}{25 \times 3}}\) (B) \(\sqrt{\frac{11 \times 39}{25 \times 2}}\) (C) \(\sqrt{\frac{11 \times 25}{39 \times 2}}\) (D) none of these

6 step solution

Problem 50

The angle of elevation of the top of a vertical pole when observed from each vertex of a regular hexagon is \(\frac{\pi}{3}\). If the area of the circle circumscribing the hexagon be \(\mathrm{A} \mathrm{m}^{2}\) then the height of the tower is (A) \(\frac{2 A}{\sqrt{3 \pi}} \mathrm{m}\) (B) \(\frac{A}{\sqrt{3 \pi}} \mathrm{m}\) (C) \(2 \sqrt{\frac{A}{3 \pi}} \mathrm{m}\) (D) \(\sqrt{\frac{A}{3 \pi}} \mathrm{m}\)

7 step solution

Problem 51

From a point on a hill-side of constant inclination, the angle of elevation of the top of a flagstaff on its summit is observed to be \(\alpha\) and a metre nearer the top of the hill, it is \(\beta\). If \(\mathrm{h}\) is the height of the flagstaff, the inclination of the hill to the horizontal is (A) \(\sin ^{-1}\left(\frac{a \sin \alpha \sin \beta}{h \sin (\beta-\alpha)}\right)\) (B) \(\cos ^{-1}\left(\frac{a \sin \alpha \sin \beta}{h \sin (\beta-\alpha)}\right)\) (C) \(\tan ^{-1}\left(\frac{a \sin \alpha \sin \beta}{h \sin (\beta-\alpha)}\right)\) (D) none of these

7 step solution

Problem 52

\(\mathrm{ABC}\) is an equilateral triangular plot. An electric pole stands at the vertex and makes an angle of \(60^{\circ}\) at either of the other two vertices. If the height of the triangle is \(100 \mathrm{~m}\), the height of the pole is (A) \(200 \mathrm{~m}\) (B) \(100 \mathrm{~m}\) (C) \(150 \mathrm{~m}\) (D) none of these

4 step solution

Problem 53

A tower \(\mathrm{AB}\) leans towards west making an angle \(\alpha\) with the vertical. The angular elevation of \(B\), the top most point of the tower is \(\beta\), as observed from a point C due east of \(A\) at a distnace \(d\) from \(A\). If the angular elevation of B from a point due east of \(\mathrm{C}\) at a distance \(2 \mathrm{~d}\) from \(\mathrm{C}\) is \(\gamma\), then (A) \(2 \tan \alpha=2 \cot \beta-\cot \gamma\) (B) \(2 \tan \alpha=3 \cot \beta-\cot \gamma\) (C) \(\tan \alpha=\cot \beta\) - cot \(\gamma\) (D) none of these

6 step solution

Problem 54

Two flagstaffs stand on a horizontal plane. A and B are two points on the line joining their feet and between them. The angles of elevation of the tops of the flagstaffs as seen from \(\mathrm{A}\) are \(30^{\circ}\) and \(60^{\circ}\) and as seen from B are \(60^{\circ}\) and \(45^{\circ}\). If \(\mathrm{AB}\) is \(30 \mathrm{~m}\), then the distance between the flagstaffs in metres is (A) \(30+15 \sqrt{3}\) (B) \(45+15 \sqrt{3}\) (C) \(60-15 \sqrt{3}\) (D) \(60+15 \sqrt{3}\)

6 step solution

Problem 55

In a cubical hall abcdpqrs with each side \(10 \mathrm{~m}, \mathrm{G}\) is the centre of the wall berq and \(\mathrm{T}\) is the mid point of the side \(\mathrm{AB}\). The angle of elevation of \(\mathrm{G}\) at the point \(\mathrm{T}\) is (A) \(\sin ^{-1} \frac{1}{\sqrt{3}}\) (B) \(\cos ^{-1} \frac{1}{\sqrt{3}}\) (C) \(\tan ^{-1} \frac{1}{\sqrt{3}}\) (D) \(\cot ^{-1} \frac{1}{\sqrt{3}}\)

5 step solution

Problem 56

Two objects \(\mathrm{P}\) and \(\mathrm{Q}\) subtend an angle of \(30^{\circ}\) at \(\mathrm{A}\). Length of \(20 \mathrm{~m}\) and \(10 \mathrm{~m}\) are measured from \(\mathrm{A}\) at right angles to \(\mathrm{AP}\) and \(\mathrm{AQ}\) respectively to points \(\mathrm{R}\) and \(\mathrm{S}\) at each of which PQ subtends angles of \(30^{\circ}\), the length of \(\mathrm{PQ}\) is (A) \(\sqrt{300-200 \sqrt{3}}\) (B) \(\sqrt{500-200 \sqrt{3}}\) (C) \(\sqrt{500 \sqrt{3}-200}\) (D) \(\sqrt{300}\)

5 step solution

Problem 58

From the top of a building of height \(\mathrm{h}\), a tower standing on the ground is observed to make an angle \(\theta .\) If the horizontal distance between the building and the tower is \(\mathrm{h}\), then height of the tower is (A) \(\frac{2 h \sin \theta}{\sin \theta+\cos \theta}\) (B) \(\frac{2 h \tan \theta}{1+\tan \theta}\) (C) \(\frac{2 h}{1+\cot \theta}\) (D) \(\frac{2 h \cos \theta}{\sin \theta+\cos \theta}\)

7 step solution

Problem 63

Assertion: A pole of length \(h\) stands inside a triangular plot \(A B C\) and subtends equal angles \(\alpha\) at its vertices, then \(2 h \cos \alpha \sin A=a \sin \alpha\). Reason: For circumscribed radius \(R\) of a \(\triangle A B C\), \(\frac{a}{\sin A}=\frac{b}{\sin B}=\frac{c}{\sin C}=2 R .\)

6 step solution

Problem 64

Assertion: A tower leans towards west making an angle \(\alpha\) with the vertical. The angular elevation of \(B\), the top most point of the tower, is \(\beta\) as observed from a point \(C\) due east of \(A\) at a distance \(d\) from \(A\). If the angular elevation of \(B\) from a point due east of \(C\) at a distance \(2 d\) from \(C\) is \(\gamma\), then \(2 \tan \alpha=3 \cot \beta-\cot \gamma\) Reason: In any \(\triangle A B C\), if \(B D: D C=m: n\), \(\angle B A D=\alpha, \angle C A D=\beta\) and \(\angle A D C=\theta\), then \((m+n) \cot \theta=m \cot \alpha-n \cot \beta\)

4 step solution

Problem 65

A person standing on the bank of a river observes that the angle of elevation of the top of a tree on the opposite bank of the river is \(60^{\circ}\) and when he retires 40 meter away from the tree the angle of elevation becomes \(30^{\circ} .\) The breadth of the river is (A) \(20 \mathrm{~m}\) (B) \(30 \mathrm{~m}\) (C) \(40 \mathrm{~m}\) (D) \(60 \mathrm{~m}\)

5 step solution

Problem 66

A tower stands at the centre of a circular park. A and \(\mathrm{B}\) are two points on the boundary of the park such that \(\mathrm{AB}\) ( \(=\) a) subtends an angle of \(60^{\circ}\) at the foot of the tower, and the angle of elevation of the top of the tower from \(\mathrm{A}\) or \(\mathrm{B}\) is \(30^{\circ} .\) The height of the tower is \(\left.\mathbf{[ 2 0 0 7}\right]\) (A) \(\frac{2 a}{\sqrt{3}}\) (B) \(2 a \sqrt{3}\) (C) \(\frac{a}{\sqrt{3}}\) (D) \(a \sqrt{3}\)

5 step solution

Problem 67

\(\mathrm{AB}\) is a vertical pole with \(\mathrm{B}\) at the ground level and A at the top. A man finds that the angle of elevation of the point A from a certain point \(\mathrm{C}\) on the ground is \(60^{\circ} .\) He moves away from the pole along the line BC to a point \(\mathrm{D}\) such that \(\mathrm{CD}=7 \mathrm{~m}\). From D the angle of elevation of the point \(\mathrm{A}\) is \(45^{\circ}\). Then the height of the pole is \(\quad\) (A) \(\frac{7 \sqrt{3}}{2} \cdot \frac{1}{\sqrt{3}-1} m\) (B) \(\frac{7 \sqrt{3}}{2} \cdot(\sqrt{3}+1) m\) (C) \(\frac{7 \sqrt{3}}{2} \cdot(\sqrt{3}-1) m\) (D) \(\frac{7 \sqrt{3}}{2} \cdot \frac{1}{\sqrt{3}+1}\)

7 step solution

Problem 68

A bird is sitting on the top of a vertical pole \(20 \mathrm{~m}\) high which makes an angle of elevation \(45^{\circ}\) from a point \(O\) on the ground. It flies off horizontally straight away from the point \(O\). After one second, the elevation of the bird from \(O\) is reduced to \(30^{\circ}\). Then the speed (in \(\mathrm{m} / \mathrm{s}\) ) of the bird is (A) \(40(\sqrt{2}-1)\) (B) \(40(\sqrt{3}-\sqrt{2})\) (C) \(20 \sqrt{2}\) (D) \(20(\sqrt{3}-1)\)

4 step solution

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