Class 10: Height and Distances – ICSE Board Problems – ICSE / ISC / CBSE Mathematics Portal for K12 Students

Question 1: 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 $\displaystyle 2 m$ 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 2: From a window $\displaystyle A$ , $\displaystyle 10 m$ 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 3: 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 $\displaystyle 50 m$ 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 4: 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 5: 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]

Answer:

$\displaystyle \frac{250}{BD} = \tan 45 \Rightarrow BD = 250 \text{ m }$

$\displaystyle \frac{250}{CD} = \tan 60 \Rightarrow CD = \frac{250}{\sqrt{3}}$

Therefore the width of the river

$\displaystyle = BD + CD = 250 + \frac{250}{\sqrt{3}} = 250 + 144.34 = 394.34 \text{ m }$

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Question 6: From the top of a light house $\displaystyle 100 m$ high, the angles of depression of two ships are observed as $\displaystyle 48^{\circ}$ and $\displaystyle 36^{\circ}$ respectively. Find the distance between the two ships (in the nearest meter) if:

(i) the ships are on the same side of the light house,

(ii) the ships are on the opposite sides of the light house. [2010]

Answer:

(i) From $\displaystyle \triangle ABC$

$\displaystyle \frac{100}{BC} = \tan 48^{\circ} \Rightarrow BC = \frac{100}{\tan 48^{\circ}}$

Similarly, from $\displaystyle \triangle ADB$

$\displaystyle \frac{100}{DB} = \tan 36^{\circ} \Rightarrow DB = \frac{100}{\tan 36^{\circ}}$

Therefore, the dis\tance between the ships is =

$\displaystyle DB - BC = \frac{100}{\tan 36^{\circ}} - \frac{100}{\tan 48^{\circ}}$

$\displaystyle = 100 ( \frac{\tan 48^{\circ} - \tan 36^{\circ}}{ \tan 48^{\circ} . \tan 36^{\circ}})$

$\displaystyle = \frac{100 \times 0.3841}{0.8069} = 47.60 \approx 48 \text{ m }$

(ii) If the ships were on the opposite sides, then the dis\tance between the ships is =

$\displaystyle DB + BC = \frac{100}{\tan 36^{\circ}} + \frac{100}{tan 48^{\circ}}$

$\displaystyle = 90.040 + 136.63 = 227.68 \approx 228 \text{ m }$

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Question 7: From the top of a hill, the angles of depression of two consecutive kilometer stones, due east, are found to be $\displaystyle 30^{\circ}$ and $\displaystyle 45^{\circ}$ respectively. Find the distances of the two stones from the foot of the hill. [2007]