Class 10: Heights and Distances – Exercise 24(c) – ICSE / ISC / CBSE Mathematics Portal for K12 Students

Question 1: Find $\displaystyle AD$ :

Answer:

$\displaystyle \frac{AE}{BE} = \tan 32^{\circ}$

$\displaystyle \Rightarrow AE = 20 \times \tan 32^{\circ} = 12.$

$\displaystyle AD = 5 + 12.5 = 17.5 \text{ m }$

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Question 2: In the following diagram, $\displaystyle AB$ is a floor-board; $\displaystyle PQRS$ is a cubical box with each edge $\displaystyle = 1 \text{ m }$ and $\displaystyle \angle B = 60^{\circ} .$ Calculate the length of the board $\displaystyle AB .$

Answer:

$\displaystyle \frac{PS}{PB} = \sin 60^{\circ}$

$\displaystyle PB = \frac{1}{\sin 60^{\circ}} = 1.15$

$\displaystyle \frac{PQ}{AP} = \cos 60^{\circ}$

$\displaystyle AP = \frac{1}{\cos 60^{\circ}} = 2$

$\displaystyle \therefore AB = PB + AP = 1.15 + 3 = 3.15 \text{ m }$

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Question 3: Calculate $\displaystyle BC .$

Answer:

$\displaystyle \frac{CD}{20} = \tan 42^{\circ}$

$\displaystyle CD = 20 \tan 42^{\circ} = 18$

$\displaystyle \frac{20}{BC+18} = \tan 35^{\circ}$

$\displaystyle 20 = BC \tan 35^{\circ} + 18 \tan 35^{\circ}$

$\displaystyle BC = \frac{20 - 18 \tan 35^{\circ}}{\tan 35^{\circ}} = 10.56$

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Question 4: Calculate $\displaystyle AB .$

Answer:

$\displaystyle \frac{AX}{6} = \cos 30^{\circ}$

$\displaystyle AX = 6 \cos 30^{\circ} = 5.196 \text{ m }$

$\displaystyle \frac{XB}{5} = \cos 43^{\circ}$

$\displaystyle XB = 5 \cos 43^{\circ} = 3.856 \text{ m }$

$\displaystyle AB = 5.196 + 3.856 = 8.85 \text{ m }$

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Question 5: The radius of a circle is given as $\displaystyle 15 \text{ cm }$ and chord $\displaystyle AB$ subtends an angle of $\displaystyle 131^{\circ}$ at the center $\displaystyle C$ of the circle. Using trigonometry, calculate: (i) the length of $\displaystyle AB$ ; (ii) the distance of $\displaystyle AB$ from the center $\displaystyle C .$

Answer:

$\displaystyle \frac{AX}{15} = \cos 24.5^{\circ}$

$\displaystyle AX = 15 \cos 24.5^{\circ} = 13.64 \text{ cm }$

Therefore $\displaystyle AB = 2 \times 13.64 = 27.3 \text{ cm }$

$\displaystyle \frac{OX}{15} = \sin 24.5^{\circ}$

$\displaystyle OX = 15 \sin 24.5^{\circ} = 6.22 \text{ cm }$

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Question 6: At a point on level ground, the angle of elevation of a vertical tower is found to be such that its tangent is $\displaystyle \frac{5}{12}$ on walking $\displaystyle 192 \text{ m }$ eters towards the tower; the tangent of the angle is found to be $\displaystyle \frac{3}{4}$ Find the height of the tower.

Answer:

$\displaystyle \frac{h}{DB} = \frac{3}{4} \Rightarrow h = DB . \frac{3}{4}$

$\displaystyle \frac{h}{CB} = \frac{5}{12} \Rightarrow h = \frac{5}{12} (192+DB)$

Therefore

$\displaystyle \frac{5}{12} (192+DB) = DB . \frac{3}{4}$

$\displaystyle 960 + 5 DB = 9 DB \Rightarrow DB = 240 \text{ m }$

$\displaystyle \text{Hence } h = 240 \times \frac{3}{4} = 180 \text{ m }$

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Question 7: A vertical tower stands on horizontal plane and is surmounted by a vertical flagstaff of height $\displaystyle h \text{ m }$ eter. At a point on the plane, the angle of elevation of the bottom of the flagstaff is $\displaystyle \alpha$ and that of the top of flagstaff is $\displaystyle \beta .$ Prove that the height of the tower is $\displaystyle \frac{h \tan \alpha}{\tan \beta - \tan \alpha} .$

Answer:

$\displaystyle \frac{h+DB}{CB} = \tan \beta$

$\displaystyle \frac{DB}{CB} = \tan \alpha$

$\displaystyle CB = \frac{DB}{\tan \alpha}$

$\displaystyle \frac{h + DB}{\frac{DB}{\tan \alpha}} = \tan \beta$

$\displaystyle h + DB = DB \Big( \frac{ \tan \beta}{\tan \alpha} \Big)$

$\displaystyle \Rightarrow h = \Big( \frac{\tan \beta - \tan \alpha}{\tan \alpha} \Big) DB$

$\displaystyle \therefore DB = \Big( \frac{\tan \alpha}{\tan \beta - \tan \alpha} \Big) h$

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Question 8: With reference to the given figure of a man stands on the ground at point $\displaystyle A ,$ which is on the same horizontal plane as $\displaystyle B ,$ the foot of the vertical pole $\displaystyle BC .$ The height of the pole is $\displaystyle 10 m .$ The man’s eye is $\displaystyle 2 \text{ m }$ above the ground. He observes the angle of elevation of $\displaystyle C ,$ the top of the pole, as $\displaystyle x^{\circ}$ where $\displaystyle \tan x = \frac{2}{5}$ Calculate: (i) the distance $\displaystyle AB$ in meters; (ii) angle of elevation of the top of the pole when he is standing $\displaystyle 15 \text{ m }$ eters from the pole. Give your answer to the nearest degree. [1999]

Answer:

$\displaystyle \tan x = \frac{2}{5}$

$\displaystyle \frac{CE}{DE} = \frac{2}{5}$

$\displaystyle DE = \frac{5}{2} \times ( 10 - 2) = 20 \text{ m }$

i) $\displaystyle AB = DE = 20 \text{ m }$

ii) if the distance from pole $\displaystyle = 15 \text{ m }$

$\displaystyle \therefore \tan y = \frac{8}{15} = 0.5333 \Rightarrow y = 28.072^{\circ}$

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Question 9: The angles of elevation of the top of a tower from two points on the ground at distances $\displaystyle a$ and $\displaystyle b \text{ m }$ eters from the base of the tower and in the same straight line with it are complementary. Prove that the height of the tower is $\displaystyle \sqrt{ab} \text{ m } .$

Answer:

$\displaystyle \tan \alpha = \frac{h}{a}$

$\displaystyle \tan \beta = \frac{h}{b}$

$\displaystyle \alpha + \beta = 90^{\circ}$

$\displaystyle \tan (\alpha + \beta) = \tan (90^{\circ})$

$\displaystyle \frac{\tan \alpha + \tan \beta}{1 - \tan \alpha . \tan \beta} = \tan (90^{\circ})$

$\displaystyle \Rightarrow 1 - \tan \alpha . \tan \beta = 0$

$\displaystyle \frac{h}{a} . \frac{h}{b} = 1$

$\displaystyle \Rightarrow h^2 = ab$

$\displaystyle \Rightarrow h = \sqrt{ab}$

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Question 10: From a window $\displaystyle A$ , $\displaystyle 10 \text{ m }$ above the ground the angle of elevation of the top $\displaystyle C$ of a tower is $\displaystyle x^{\circ} ,$ where $\displaystyle \tan x^{\circ} = \frac{5}{2}$ and the angle of depression of the foot $\displaystyle D$ of the tower is $\displaystyle y^{\circ}$ where $\displaystyle \tan y^{\circ} = \frac{1}{4} .$ See the given figure. Calculate the height $\displaystyle CD$ of the tower in meters. [2000]

Answer:

$\displaystyle \tan x = \frac{5}{2} \Rightarrow \frac{EC}{AE} = \frac{5}{2}$

$\displaystyle \tan y = \frac{1}{4} \Rightarrow \frac{ED}{AE} = \frac{1}{4} \Rightarrow AE = 4 \times 10 = 10 \text{ m }$

$\displaystyle \therefore EC = 40 \times \frac{5}{2} = 100 \text{ m }$

$\displaystyle \therefore CD = ED + EC = 10 + 100 = 110 \text{ m }$

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Question 11: A vertical tower is $\displaystyle 20 \text{ m }$ high. A man standing at some distance from the tower knows that the cosine of the angle of elevation of the top of the tower is $\displaystyle 0.53 .$ How far is he standing from the foot of the tower?

Answer:

$\displaystyle \cos \theta = 0.53$

$\displaystyle \Rightarrow \frac{AB}{CB}$

$\displaystyle CB = \sqrt{20^2 + x^2}$

$\displaystyle \therefore x = 0.53 \sqrt{20^2 + x^2}$

$\displaystyle 3.56 x^2 = 20^2 + x^2$

$\displaystyle \Rightarrow x = \sqrt{ \frac{400}{2.56} }$

$\displaystyle \Rightarrow x = 12.5 \text{ m }$

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Question 12: A man standing on the bank of a river observes that the angle of elevation of the top of the tower is $\displaystyle 60^{\circ}.$ When he moves $\displaystyle 50 \text{ m }$ away from the bank. he finds the angle of elevation to be $\displaystyle 30^{\circ}.$ Calculate: (i) the width of the river and (ii) the height of the tree. [2003]

Answer:

Let the height of the tree $\displaystyle = h$

Let the width of the river $\displaystyle = w$

$\displaystyle \therefore \frac{h}{w} = \tan 60^{\circ} = \sqrt{3}$

$\displaystyle \frac{h}{w+50} = \tan 30^{\circ} = \frac{1}{\sqrt{3}}$

$\displaystyle \therefore h = \sqrt{3} w$

$\displaystyle \Rightarrow \sqrt{3} h = w + 50$

$\displaystyle \Rightarrow \sqrt{3} (\sqrt{3} w) = w + 50$

$\displaystyle \Rightarrow 3w = w + 50$

$\displaystyle \Rightarrow 2w = 50$

$\displaystyle \Rightarrow w = 25 \text{ m }$

$\displaystyle \therefore h = 25\sqrt{3} = 43.3 \text{ m }$

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Question 13: A $\displaystyle 20 m$ high vertical pole and a vertical tower are on the same level ground in such a way that the angle of elevation of the top of the tower, as seen from the foot of the pole, is $\displaystyle 60^{\circ}$ and angle of elevation of the top of the pole as seen from the foot of the tower is $\displaystyle 30^{\circ}.$ Find: (i) the height of the tower (ii) the horizontal distance between the pole and the tower.

Answer:

$\displaystyle \frac{20}{CB} = \tan 30^{\circ} = \frac{1}{\sqrt{3}}$

$\displaystyle CB = 20\sqrt{3} \text{ m }$

$\displaystyle \therefore \frac{AB}{CB} = \tan 60^{\circ} = \sqrt{3}$

$\displaystyle \Rightarrow AB = 20 \sqrt{3} \times \sqrt{3} = 60 \text{ m }$

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Question 14: A vertical pole and vertical tower are on the same ground level ground in such a way that from the top of the pole the angle of elevation of the top of the tower is $\displaystyle 60^{\circ}$ and the angle of depression of the bottom of the tower is $\displaystyle 30^{\circ}$ Find: (i) the height of the tower, if the height of the pole is $\displaystyle 20 m$ ;(ii)the height of the pole; if the height of the tower is $\displaystyle 75 \text{ m.}$

Answer:

i) $\displaystyle \frac{DE}{AE} = \tan 60 \Rightarrow DE = AE (\sqrt{3})$

$\displaystyle \frac{EC}{AE} = \tan 30 \Rightarrow 20 \sqrt{3}$

$\displaystyle \therefore DE = 60 \text{ m }$

Height of tower $\displaystyle = 60 + 20 = 80 \text{ m }$

$\displaystyle \text{ii) } \frac{75-h}{AE} = \tan 60 \Rightarrow 75 - h = \sqrt{3} AE$

$\displaystyle \frac{h}{AE} = \tan 30 \Rightarrow AE = \sqrt{3} h$

$\displaystyle \therefore 75 - h = \sqrt{3} (\sqrt{3} h)$

$\displaystyle 75 = 4 - h$

$\displaystyle h = 18.5 \text{ m }$

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Question 15: From a point, $\displaystyle 36 m$ above the surface of a lake, the angle of elevation of a bird is observed to be $\displaystyle 30^{\circ}$ and the angle of depression of its image in the water of the lake is observed to be $\displaystyle 60^{\circ}.$ Find the actual height of the bird above the surface of the lake.

Answer:

$\displaystyle \frac{h}{BC} = \tan 30 \Rightarrow BC = \sqrt{3} h$

$\displaystyle \frac{EF + 36}{BC} = \tan 60$

$\displaystyle \Rightarrow EF + 36 = \sqrt{3} h ( \sqrt{3}) = 3h$

$\displaystyle \therefore EF = 3h - 36$

Now, $\displaystyle AE = EF$

$\displaystyle \Rightarrow h + 36 = 3h - 36$

$\displaystyle \Rightarrow 72 = 2h$

$\displaystyle \Rightarrow h = 36 \text{ m }$

Therefore height of the bird $\displaystyle = 72 \text{ m }$

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Question 16: A man observes the angle of elevation of the top of a building to be $\displaystyle 30^{\circ}.$ He walks towards it in a horizontal line through its base. On covering $\displaystyle 60 m ,$ the angle of elevation changes to $\displaystyle 60^{\circ} .$ Find the height of the building correct to the nearest meter.

Answer:

$\displaystyle \frac{h}{CB} = \tan 60 \Rightarrow CB = \frac{h}{\sqrt{3}}$

$\displaystyle \frac{h}{DB} = \tan 30 \Rightarrow DB = \sqrt{3} h$

$\displaystyle \therefore \sqrt{3} h - \frac{h}{\sqrt{3}} = 60$

$\displaystyle \Rightarrow h \Big( \frac{3-1}{\sqrt{3}} \Big) = 60$

$\displaystyle h = 30 \sqrt{3} = 51.96 \text{ m }$ $\displaystyle \approx 52 \text{ m }$

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Question 17: As observed from the top of a, $\displaystyle 80 m$ tall lighthouse, the angles of depression of two ships, on the same side of the lighthouse in horizontal line with its base, are $\displaystyle 30^{\circ}$ and $\displaystyle 40^{\circ}$ respectively. Find the distance between the two ships. Give your answer correct to the nearest meter. [2012]

Answer:

$\displaystyle \frac{80}{BC} = \tan 40 \Rightarrow \frac{80}{0.839} = 95.34$

$\displaystyle \frac{80}{BC+CD} = \tan 30$

$\displaystyle \Rightarrow 80\sqrt{3} = 95.34 + CD$

$\displaystyle CD = 43.22 \text{ m }$ $\displaystyle \approx 43 \text{ m }$

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Question 18: In the given figure, from the top of a building $\displaystyle AB = 60 m$ high, the angles of depression of the top and bottom of a vertical lamp post $\displaystyle CD$ are observed to be $\displaystyle 30^{\circ}$ and $\displaystyle 60^{\circ}$ respectively. Find: (i) the horizontal distance between $\displaystyle AB$ and $\displaystyle CD .$ (ii) the height of the lamp Post.

Answer:

i) $\displaystyle \frac{60}{BC} = \tan 60 \Rightarrow BC = \frac{60}{\sqrt{3}} = 20 \sqrt{3} = 34.34 \text{ m }$

ii) $\displaystyle \frac{60-h}{20\sqrt{3}} = \tan 30 \Rightarrow \frac{60-h}{20\sqrt{3}} = \frac{1}{\sqrt{3}} \Rightarrow h = 40 \text{ m }$

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Question 19: An airplane, at an altitude of $\displaystyle 250 m$ observes the angles of depression of two boats on the opposite banks of a river to be $\displaystyle 45^{\circ}$ and $\displaystyle 60^{\circ}$ respectively. Find the width of the river. Write the answer correct to the nearest whole number. [2014]