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

Question 1: In the figure, given below, it is given that $\displaystyle AB$ is perpendicular to $\displaystyle BD$ and is of length $\displaystyle X$ meters. $\displaystyle DC = 30\text{ m,} \angle ADB = 30^o$ and $\displaystyle \angle ACB = 45^o$ Without using tables, find $\displaystyle X .$

Answer:

$\displaystyle \text{From } \triangle ABC$

$\displaystyle \frac{x}{BC} = \tan 45^o \Rightarrow x = BC$

$\displaystyle \text{Similarly, } \text{From } \triangle ADB$

$\displaystyle \frac{x}{30+BC} = \tan 30^o$

Substituting from above

$\displaystyle x = 30 30^o + x \tan 30^o$

$\displaystyle x (1- \tan 30^o) = 30 \tan 30$

$\displaystyle \Rightarrow x = \frac{30 \tan 30^o}{1- \tan 30^o} = 40.98\text{ m }$

$\displaystyle \\$

Question 2: Find the height of a tree when it is found that on walking away from it $\displaystyle 20\text{ m }$ ? in a horizontal line through its base, the elevation of its top changes $\displaystyle \text{From } 60^o$ to $\displaystyle 30^o .$

Answer:

$\displaystyle \text{From } \triangle ABC$

$\displaystyle \frac{x}{BC} = \tan 60^o \Rightarrow BC = \frac{x}{\tan 60^o}$

$\displaystyle \text{Similarly, } \text{From } \triangle ADC$

$\displaystyle \frac{x}{20+BC} = \tan 30^o$

Substituting from above

$\displaystyle x = (20 + \frac{x}{\tan 60^o}) \tan 30^o$

$\displaystyle x(1- \frac{\tan 30^o}{\tan 60^o} ) = 30 \tan 30^o$

$\displaystyle \Rightarrow x = \frac{11.547}{0.6667} = 17.32 \text{ m } = 40.98\text{ m }$

$\displaystyle \\$

Question 3: Find the height of a building, when it is found that on walking towards it $\displaystyle 40\text{ m }$ in a horizontal line through its base the angular elevation of its top changes $\displaystyle \text{From } 30^o$ to $\displaystyle 45^o .$

Answer:

$\displaystyle \text{From } \triangle ABC$

$\displaystyle \frac{h}{BC} = \tan 45^o \Rightarrow h = BC$

$\displaystyle \text{Similarly, } \text{From } \triangle ADB$

$\displaystyle \frac{h}{40+BC} = \tan 30^o$

Substituting from above

$\displaystyle h = (40+ h) \tan 30^o$

$\displaystyle h (1- \tan 30^o) = 40 \tan 30$

$\displaystyle \Rightarrow h = \frac{40 \tan 30^o}{1- \tan 30^o} = 54.64\text{ m }$

$\displaystyle \\$

Question 4: From the top of a light house $\displaystyle 100\text{ m }$ high, the angles of depression of two ships are observed as $\displaystyle 48^o$ and $\displaystyle 36^o$ 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) $\displaystyle \text{From } \triangle ABC$

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

$\displaystyle \text{Similarly, } \text{From } \triangle ADB$

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

Therefore, the dis\tance between the ships is =

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

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

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

(ii) If the ships were on the opposite sides, then the distance between the ships is =

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

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

$\displaystyle \\$

Question 5: Two pillars of equal heights stand on either side of a roadway, which is $\displaystyle 150\text{ m }$ wide. At a point in the roadway between the pillars the elevations of the tops of the pillars are $\displaystyle 60^o$ and $\displaystyle 30^o$ ; find the height position of the pillar and the position of the point.

Answer:

$\displaystyle \text{From } \triangle ABC$

$\displaystyle \frac{h}{x} = \tan 60^o$

$\displaystyle \text{Similarly, } \text{From } \triangle DEC$

$\displaystyle \frac{h}{150-x} = \tan 30^o$

$\displaystyle \therefore x \tan 60^o = (150 - x) \tan 30^o$

$\displaystyle x (\tan 60^o + \tan 30^o) = 150 \tan 30^o$

$\displaystyle x = \frac{150 \tan 30^o}{\tan 60^o+\tan 30^o} = 37.5\text{ m }$

Hence $\displaystyle h = 37.5 \times \tan 60^o = 64.95\text{ m }$

$\displaystyle \\$

Question 6: From the figure, given below, calculate the length of $\displaystyle CD .$

Answer:

$\displaystyle \text{From } \triangle ABC$

$\displaystyle \frac{AB}{15} = \tan 47^o$

$\displaystyle \text{Similarly, } \text{From } \triangle AED$

$\displaystyle \frac{AB}{15} = \tan 22^o$

$\displaystyle \text{Therefore } CD = AB - AE = 15 \tan 47^o - 15 \tan 22^o = 16.085-6.060 = 10.025$

$\displaystyle \\$

Question 7: The angle of elevation of the top of a tower is observed to be $\displaystyle 50^o .$ At a Point, $\displaystyle 30\text{ m }$ vertically above the first point of observation, the elevation is found to be $\displaystyle 45^o$ Find: (i) the height of the tower, (ii) its horizontal distance from the points of Observation

Answer:

$\displaystyle \text{From } \triangle ABC$

$\displaystyle \frac{h}{BC} = \tan 60^o$

$\displaystyle \text{Similarly, } \text{From } \triangle ADE$

$\displaystyle \frac{h - 30}{BC} = \tan 45^o \Rightarrow BC = (h - 30)$

$\displaystyle \text{Therefore } h = (h - 30 ) \tan 60^o$

Given $\displaystyle (h - 30 ) \tan 60^o = -30 \tan 60^o$

$\displaystyle \Rightarrow h = \frac{-30 \tan 60^o}{1 - \tan 60^o} = 70.98\text{ m }$

$\displaystyle \text{Therefore } BC = \frac{h}{ \tan 60^o} = \frac{70.98}{ \tan 60^o} = 40.98\text{ m }$

$\displaystyle \\$

Question 8: From the top of a cliff, $\displaystyle 60\text{ m }$ high, the angles of depression of the top and bottom of a tower are observed to be $\displaystyle 30^o$ and $\displaystyle 60^o .$ Find the height of the tower.

Answer:

$\displaystyle \text{From } \triangle ABC$

$\displaystyle \frac{60}{BC} = \tan 60^o$

$\displaystyle \text{Similarly, } \text{From } \triangle AED$

$\displaystyle \frac{60- h}{BC} = \tan 30^o$

$\displaystyle \text{Therefore } 60 - h = \frac{60}{\tan 60^o} . \tan 30^o$

$\displaystyle \Rightarrow h = 60 - 60 ( \frac{\tan 30^o}{\tan 60^o}) = 40\text{ m }$

$\displaystyle \\$

Question 9: A man on a cliff observes a boat. at an angle of depression $\displaystyle 30^o$ , which is sailing towards the shore to the point immediately beneath him. Three minutes later, the angle of depression of the boat is found to be $\displaystyle 60^o .$ Assuming that the boat sails at a uniform speed, determine: (i) how much more time it will take to reach the shore. (ii) the speed of the boat in meter per second if the height of the cliff is $\displaystyle 500\text{ m } .$

Answer:

$\displaystyle \text{From } \triangle ABC$

$\displaystyle \frac{500}{BC} = \tan 60^o \Rightarrow BC = \frac{500}{\tan 60^o}$

$\displaystyle \text{Similarly, } \text{From } \triangle ABD$

$\displaystyle \frac{500}{x+BC} = \tan 30^o$

$\displaystyle \therefore 500 = (x + \frac{500}{\tan 60^o} ) \tan 30^o$

$\displaystyle \Rightarrow 500 (1 - \frac{\tan 30^o}{\tan 60^o} ) = x \tan 30^o$

$\displaystyle \Rightarrow x = 577.35\text{ m }$

Also $\displaystyle BC = \frac{500}{\tan 60^o} = 288.675\text{ m }$

Therefore the time that the boat will take to reach the shore $\displaystyle = \frac{288.675}{577.35} \times 3 = 1.5 \text{minutes } .$

Speed of the boat $\displaystyle = \frac{577.35}{3 \times 60} = 3.21 \frac{m}{s}$

$\displaystyle \\$

Question 10: A man in a boat rowing away from a lighthouse $\displaystyle 150\text{ m }$ high, takes $\displaystyle 2$ minutes to change the angle of elevation of the top of the lighthouse $\displaystyle \text{From } 60^o$ to $\displaystyle 45^o .$ Find the speed of the boat.

Answer:

$\displaystyle \text{From } \triangle ABC$

$\displaystyle \frac{150}{BC} = \tan 60^o$

$\displaystyle \text{Similarly, } \text{From } \triangle ABD$

$\displaystyle \frac{150}{x+BC} = \tan 45^o$

$\displaystyle \therefore 150 = x + \frac{150}{\tan 60^o}$

$\displaystyle \Rightarrow x = 150 (1 - \frac{1}{\tan 60^o} ) = 63.34\text{ m }$

$\displaystyle \therefore Speed of the boat = \frac{63.34}{2 \times 60} = 0.53 \frac{m}{s}$

$\displaystyle \\$

Question 11: A person standing on the bank of a river observes that the angle of elevation of the top of a tree standing on the opposite bank is $\displaystyle 60^o .$ When he moves $\displaystyle 40\text{ m }$ away from the bank, he finds the angle of elevation to be $\displaystyle 30^o .$ Find: (i) the height of the free, correct to $\displaystyle 2$ decimals places, (ii) the width of the river.

Answer:

$\displaystyle \text{From } \triangle ABC$

$\displaystyle \frac{h}{BC} = \tan 60^o$

$\displaystyle \text{Similarly, } \text{From } \triangle ABD$

$\displaystyle \frac{h}{40+BC} = \tan 30^o$

$\displaystyle \therefore h = \tan 30^o ( \frac{h}{\tan 30^o} + 40)$

$\displaystyle \Rightarrow h (1 - \frac{\tan 30^o}{\tan 60^o} ) = \tan 30^o \times 40$

$\displaystyle h = \frac{\tan 30^o \times 40}{(1 - \frac{\tan 30^o}{\tan 60^o})}$

$\displaystyle \text{Therefore } BC = \frac{34.64}{\tan 60^o} = 20\text{ m }$

$\displaystyle \\$

Question 12: The horizontal distance between two towers is $\displaystyle 5\text{ m }$ and the angular depression of the top of the first tower as seen from the top of the second. which is $\displaystyle 160\text{ m }$ high, is $\displaystyle 45^o .$ Find the height of the first tower.

Answer:

$\displaystyle \text{From } \triangle ABC$

$\displaystyle \frac{160-h}{75} = \tan 45^o$

$\displaystyle \Rightarrow 160 - h = 75$

$\displaystyle \Rightarrow h = 160 - 75 = 85\text{ m }$

$\displaystyle \\$

Question 13: The length of the shadow of a tower standing on level plane is found to be $\displaystyle 2y$ meters longer when the sun’s altitude is $\displaystyle 30^o$ then when it is $\displaystyle 45^o .$ Prove that the height of the tower is $\displaystyle y(\sqrt{3}+1)$ meters.

Answer:

$\displaystyle \text{From } \triangle ABC$

$\displaystyle \frac{h}{BC} = \tan 45^o \Rightarrow h = BC$

$\displaystyle \text{Similarly, } \text{From } \triangle ABD$

$\displaystyle \frac{h}{h + 2y} = \tan 30^o$

$\displaystyle \text{Substituting } h = 2y \frac{\tan 30^o}{1- \tan 30^o}$

$\displaystyle \Rightarrow h = 2y \frac{1}{\sqrt{3}} (\frac{\sqrt{3}}{\sqrt{3} -1})$

$\displaystyle \Rightarrow h = 2y (\frac{1}{\sqrt{3} -1})$

Multiplying both numerator and denominator by $\displaystyle (\sqrt{3}+1)$ we get

$\displaystyle \Rightarrow h = 2y ( \frac{\sqrt{3}+1}{2} ) = y (\sqrt{3}+1)$

$\displaystyle \\$

Question 14: After $\displaystyle 10$ seconds, its elevation is observed to be $\displaystyle 30^o$ ; find the uniform speed of the Aeroplane in km per hour.

Answer:

$\displaystyle \text{From } \triangle EDC$

$\displaystyle \frac{1}{BC} = \tan 60^o \Rightarrow DC = \frac{1}{\sqrt{3}}$

$\displaystyle \text{Similarly, } \text{From } \triangle ADB$

$\displaystyle \frac{1}{DC + BC} = \tan 30^o$

$\displaystyle \Rightarrow 1 = \tan 30^o ( \frac{1}{\sqrt{3}} +BC)$

$\displaystyle BC = \frac{1}{\tan 30^o} - \frac{1}{\sqrt{3}} = \sqrt{3} - \frac{1}{\sqrt{3}}$

$\displaystyle \Rightarrow BC = \frac{2}{\sqrt{3}}$

$\displaystyle \therefore Speed = \frac{\frac{2}{\sqrt{3}}}{\frac{10}{3600}} = 415.69 \frac{km}{hr}$

$\displaystyle \\$

Question 15: From the top of a hill, the angles of depression of two consecutive kilometer stones, due east, are found to be $\displaystyle 30^o$ and $\displaystyle 45^o$ respectively. Find the distances of the two stones from the foot of the hill. [2007]

Answer: