>> Elementary Mathematics >> Height and Distance

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Contents:

- Elementary Mathematics
- Area and perimeter
- Circles
- Clocks
- Factorisation
- Geometry
- Height and Distance
- Indices and Surd
- LCM and HCF
- Loci and concurrency
- Logarithms
- Polynomials
- Quadratic Equations
- Quadrilateral and parallelogram
- Rational expression
- Real Analysis
- Rectangular and Cartesian products
- Set theory
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- Simplification
- Statistics
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- Triangle
- Trigonometric ratio
- Trigonometry
- Volume and surface area

1). The angle of elevation of the top of a tower at a point on the ground is \( \Large 30 ^{\circ} \). If on walking 20m towards the tower the angle of elevation becomes \( \Large 60 ^{\circ} \) then the heights of the tower is:
Let the height of the tower be h. | ||||

2). When the elevation of sun changes from \( \Large 45 ^{\circ} \) to \( \Large 30 ^{\circ} \) the shadow of a tower increases by 60 m, The height of the tower is:
Let the height of the tower be h. | ||||

3). A person standing on the bank of a river observes that the angle subtended by a tree on the opposite bank is \( \Large 60 ^{\circ} \) when he retire 40 m from the bank he finds the angle to be \( \Large 30 ^{\circ} \). The breadth of the river is:
Let the height of the tree be h and breadth of the river be b. | ||||

4). A flag-staff is upon the top of a building. If at a distance of 40 m from the base of building the angles of elevation of the tapes of the flag-staff and building are \( \Large 60 ^{\circ} \) and \( \Large 30 ^{\circ} \) respectively, then the height of the flag-staff is:
| ||||

5). A house of height 100 m subtends a right angle at the window of an opposite house. If the height of the window be 64 m, the distance between the two houses is:
In \( \Large \triangle DAB, \tan \theta = \frac{64}{d} \) \( \Large d = 64 \cot \theta \) ...(i) In \( \Large \triangle CDE, \tan \left(90 ^{\circ} - \theta \right) = \frac{ \left(100 - 64\right) }{d} \) \( \Large d = 36 \tan \theta \) ...(ii) From Eqs. (i) and (ii), we get \( \Large d^{2} = 36 \times 64 => d = 48 m\) | ||||

6). From the top of a light house 60 m high with its base at the sea level the angle of depression of a boat is \( \Large 15 ^{\circ} \). The distance of the boat from the foot of light house is:
Let BC be the light house. In \( \Large \triangle ABC, \tan 15 ^{\circ} = \frac{60}{d} \) => \( \Large d = 60 \cot 15 ^{\circ} = 60 \left(\frac{\sqrt{3}+1}{\sqrt{3}-1}\right) m \) | ||||

7). Two poles of equal heights stand on either side of a 100 m wide road. At a point between the poles the angles of elevation of the tops of the poles are \( \Large 30 ^{\circ} \) and \( \Large 60 ^{\circ} \). The height of each pole is:
Let the height of pole AD = BC = h In \( \Large \triangle OBC \tan 60 ^{\circ} = \frac{h}{x} \) => \( \Large x = \frac{h}{\sqrt{3}} \) In \( \Large \triangle AOD \tan 30 ^{\circ} = \frac{h}{100-x} \) => \( \Large h = \left(100 - x\right) \frac{1}{\sqrt{3}} = \left(100 - \frac{h}{\sqrt{3}}\right) \frac{1}{\sqrt{3}} \) [from (i)] => \( \Large 3h = 100\sqrt{3} - h \) => \( \Large h = \frac{100\sqrt{3}}{4} \) => \( \Large h = 25\sqrt{3} m \) | ||||

8). At a distance 2h metre from the foot of a tower of height h meter the top of the tower and pole at the top of tower subtend equal angles. Height of the pole should be
In In \( \Large \triangle ABD, \tan \alpha = \frac{h}{2h} \) => \( \Large \tan \alpha = \frac{1}{2} \) In \( \Large \triangle ABC, \tan 2\alpha = \frac{h+p}{2h} \) => \( \Large \frac{2 \tan \alpha }{1 - \tan^{2}a} = \frac{h+p}{2h} \) => \( \Large \frac{2 \left(\frac{1}{2}\right) }{1- \left(\frac{1}{2}\right)^{2} } = \frac{h+p}{2h} \) => \( \Large \frac{4}{3} = \frac{h+p}{2h} => 8h = 3h + 3p \) => \( \Large 5h = 3p =>p = \frac{5h}{3} m \) | ||||

9). A kite is flying at an inclination of \( \Large 60 ^{\circ} \) with the horizontal. If the length of the thread is 120 m, then the height of the kite is:
Triangle ABC, \( \Large sin 60 ^{\circ} = \frac{h}{120} \) => \( \Large h = 120 \times \frac{\sqrt{3}}{2} \) = \( \Large 60\sqrt{3} m \) | ||||

10). If a flag staff of 6 m high placed on the top of a tower throws a \( \Large 2\sqrt{3} m\) along the ground, then the angle (in degrees) that the sun makes with the ground is:
Triangle DCB, \( \Large \tan \theta = \frac{h}{x} \) |