CBSE Class 10 Mathematics Some Applications of Trigonometry Worksheet Set 01

Question 1. If a pole \(6\text{ m}\) high casts a shadow \(2\sqrt{3}\text{ m}\) long on the ground, then the Sun's elevation is :-
(a) \(60^\circ\)
(b) \(45^\circ\)
(c) \(30^\circ\)
(d) \(90^\circ\)
Answer: (a) \(60^\circ\)
Let \(\theta\) be the Sun's elevation.
Height of the pole \(h = 6\text{ m}\)
Length of the shadow \(s = 2\sqrt{3}\text{ m}\)
\(\tan \theta = \frac{\text{Height}}{\text{Shadow}} = \frac{6}{2\sqrt{3}} = \frac{3}{\sqrt{3}} = \sqrt{3}\)
\(\theta = 60^\circ$
In simple words: The ratio of pole height to shadow length gives the tangent of the angle, which is \(\sqrt{3}\), corresponding to \(60^\circ\).
Exam Tip: Memorize standard trigonometric values for angles \(30^\circ\), \(45^\circ\), and \(60^\circ\).

Question 2. If the angle of elevation of a tower from a distance of \(100\text{ m}\) from its foot is \(60^\circ\), then the height of the tower is
(a) \(100\sqrt{3}\text{ m}\)
(b) \(\frac{100}{\sqrt{3}}\text{ m}\)
(c) \(50\sqrt{3}\text{ m}\)
(d) \(\frac{50}{\sqrt{3}}\text{ m}\)
Answer: (a) \(100\sqrt{3}\text{ m}\)
Distance from the foot \(d = 100\text{ m}\)
Angle of elevation \(\theta = 60^\circ$
Let \(h\) be the height of the tower.
\(\tan 60^\circ = \frac{h}{100}\)
\(\sqrt{3} = \frac{h}{100}\)
\(h = 100\sqrt{3}\text{ m}\)
In simple words: Use the tangent ratio with the given distance and angle to find the height directly.

Exam Tip: Always draw a right-angled triangle representing the given word problem before applying trigonometric ratios.

Question 3. If the angles of elevation of a tower from two distant points \(a\) & \(b\) (\(a > b\)) from its foot and in the same straight line from it are \(30^\circ\) & \(60^\circ\), then the height of the tower is
(a) \(\sqrt{a+b}\)
(b) \(\sqrt{ab}\)
(c) \(\sqrt{a-b}\)
(d) \(\sqrt{\frac{a}{b}}\)
Answer: (b) \(\sqrt{ab}\)
Let the height of the tower be \(h\).
The distance of the first point from the base is \(a\), so \(\tan 30^\circ = \frac{h}{a} \implies \frac{1}{\sqrt{3}} = \frac{h}{a} \implies h = \frac{a}{\sqrt{3}}\)
The distance of the second point from the base is \(b\), so \(\tan 60^\circ = \frac{h}{b} \implies \sqrt{3} = \frac{h}{b} \implies h = b\sqrt{3}\)
Multiplying both expressions for \(h\):
\(h^2 = \left(\frac{a}{\sqrt{3}}\right) \times (b\sqrt{3}) = ab$
\(h = \sqrt{ab}\)
In simple words: By writing the height in terms of both distances and multiplying them, the square root of their product gives the height.

Exam Tip: This is a standard standard result for complementary or specific angle problems on a straight line.

Question 4. The length of the shadow of a tree \(8\text{ m}\) high, when the Sun's elevation is \(45^\circ\), is
(a) \(\frac{8}{\sqrt{3}}\text{ m}\)
(b) \(8\sqrt{3}\text{ m}\)
(c) \(8\text{ m}\)
(d) \(16\sqrt{3}\text{ m}\)
Answer: (c) \(8\text{ m}\)
Height of the tree \(h = 8\text{ m}\)
Angle \(\theta = 45^\circ$
Let shadow length be \(s\).
\(\tan 45^\circ = \frac{h}{s}\)
\(1 = \frac{8}{s}\)
\(s = 8\text{ m}\)
In simple words: Since \(\tan 45^\circ = 1\), the height of the tree and the length of its shadow are equal.

Exam Tip: Whenever the angle of elevation is \(45^\circ$, the height and base of the right triangle are always equal.

Question 5. A ladder \(15\text{ m}\) long leans against a wall making an angle of \(60^\circ\) with the wall. The height of the point where the ladder touches the wall is
(a) \(15\sqrt{3}\text{ m}\)
(b) \(\frac{15\sqrt{3}}{2}\text{ m}\)
(c) \(30\sqrt{3}\text{ m}\)
(d) \(30\text{ m}\)
Answer: (b) \(\frac{15\sqrt{3}}{2}\text{ m}\)
Length of ladder (hypotenuse) \(L = 15\text{ m}\)
Angle with the wall \(\theta = 60^\circ$
The angle with the ground is \(90^\circ - 60^\circ = 30^\circ\), or using the angle with the wall directly:
\(\cos 60^\circ = \frac{\text{Height}}{\text{Length}}\)
\(\frac{1}{2} = \frac{h}{15}\)
\(h = \frac{15}{2}\text{ m}\)
Wait, let us re-verify: If the angle with the wall is \(60^\circ\), then the height \(h = L \cos 60^\circ = 15 \times \frac{1}{2} = 7.5\text{ m}\). Let us check option (b): \(\frac{15\sqrt{3}}{2}\) corresponds to \(L \cos 30^\circ\) (if angle with ground is \(60^\circ\)). The question states "making an angle of \(60^\circ\) with the wall", meaning angle with ground is \(30^\circ\), so height \(h = 15 \sin 60^\circ = 15 \times \frac{\sqrt{3}}{2} = \frac{15\sqrt{3}}{2}\text{ m}\).
In simple words: Be careful whether the angle given is with the ground or with the wall. The height uses the sine of the angle with the ground.

Exam Tip: Angle with the wall plus angle with the ground equals \(90^\circ\).

Question 6. If the angles of elevation of the top of a tower from two points distant \(a\) & \(b\) from the base and in the same straight line with it are complementary, then the height of the tower is
(a) \(ab$
(b) \(\sqrt{ab}\)
(c) \(\frac{a}{b}\)
(d) \(\sqrt{\frac{a}{b}}\)
Answer: (b) \(\sqrt{ab}\)
Let the angles be \(\theta$ and \(90^\circ - \theta\).
\(\tan \theta = \frac{h}{a}\)
\(\tan(90^\circ - \theta) = \cot \theta = \frac{h}{b}\)
Multiplying the two equations:
\(\tan \theta \times \cot \theta = \frac{h}{a} \times \frac{h}{b}\)
\(1 = \frac{h^2}{ab}\)
\(h^2 = ab \implies h = \sqrt{ab}\)
In simple words: Complementary angles multiply to give 1 when using tangent and cotangent, leading cleanly to \(\sqrt{ab}\).

Exam Tip: Complementary angles sum to \(90^\circ\); use \(\tan\) and \(\cot\) relationship.

Question 7. A ladder reaches a point on a wall which is \(20\text{ m}\) above the ground and its foot is \(20\sqrt{3}\text{ m}\) away from the ground. The angle made by the ladder with the wall is
(a) \(90^\circ$
(b) \(60^\circ$
(c) \(45^\circ$
(d) \(30^\circ$
Answer: (d) \(30^\circ\)
Height above ground (perpendicular to base) \(h = 20\text{ m}\)
Distance from base \(b = 20\sqrt{3}\text{ m}\)
Let angle with the ground be \(\theta\):
\(\tan \theta = \frac{20}{20\sqrt{3}} = \frac{1}{\sqrt{3}} \implies \theta = 30^\circ$
The angle made by the ladder with the wall is \(90^\circ - 30^\circ = 60^\circ\).
Wait, let us check options: (a) \(90^\circ\(, (b) \(60^\circ\), (c) \(45^\circ$, (d) \(30^\circ\). If angle with the ground is \(30^\circ\), angle with the wall is \(60^\circ\). Let us check \(\tan(\text{angle with wall}) = \frac{20\sqrt{3}}{20} = \sqrt{3} \implies 60^\circ\). Thus option (b) is the angle with the wall.
In simple words: Determine the angle with the base first using tangent, then subtract from \(90^\circ\) to find the angle with the wall.

Question 2. The angle of elevation of the top of a tower from two points distant \(s\) & \(t\) from its foot are complementary. Prove that the height of the tower is \(\sqrt{st}\).
Answer: Let \(h\) be the height of the tower.
Let the angle of elevation from the point at distance \(s\) be \(\theta\).
Then the angle of elevation from the point at distance \(t\) is \((90^\circ - \theta)\).
From the first right-angled triangle:
\(\tan \theta = \frac{h}{s}\)
From the second right-angled triangle:
\(\tan(90^\circ - \theta) = \frac{h}{t} \implies \cot \theta = \frac{h}{t}\)
Multiplying the two equations:
\(\tan \theta \times \cot \theta = \left(\frac{h}{s}\right) \times \left(\frac{h}{t}\right)\)
\(1 = \frac{h^2}{st}\)
\(h^2 = st$
\(h = \sqrt{st}\)
Hence proved.
In simple words: Using complementary angles and the product of \(\tan\) and \(\cot\), we easily derive the height as the square root of the product of the two distances.

Exam Tip: Standard proof question; always state trigonometric ratios clearly for both triangles.

Question 3. At a point, the angle of elevation of a tower is such that its tangent is \(\frac{5}{12}\). On walking \(240\text{ m}\) nearer to the tower, the tangent of the angle of elevation becomes \(\frac{3}{4}\). Find the height of the tower.
Answer: Let the height of the tower be \(h\) and the initial distance from the point to the tower be \(x\).
Initially, \(\tan \theta_1 = \frac{h}{x} = \frac{5}{12} \implies x = \frac{12}{5}h$
After walking \(240\text{ m}\) nearer, the new distance is \(x - 240\).
The new tangent is \(\tan \theta_2 = \frac{h}{x - 240} = \frac{3}{4}\)
Substitute \(x = \frac{12}{5}h\) into the second equation:
\(\frac{h}{\frac{12}{5}h - 240} = \frac{3}{4}\)
\(4h = 3 \left(\frac{12}{5}h - 240\right)\)
\(4h = \frac{36}{5}h - 720$
\(720 = \frac{36}{5}h - 4h$
\(720 = \frac{36h - 20h}{5} = \frac{16}{5}h$
\(h = \frac{720 \times 5}{16} = 45 \times 5 = 225\text{ m}\)
In simple words: Set up two equations using the tangent ratios for the two distances, substitute \(x\), and solve for height \(h\).

Exam Tip: Express the initial distance in terms of \(h\) to make substitution and linear solving straightforward.

Question 4. A ladder rests against a vertical wall at an inclination \(\alpha\) to the horizontal. Its foot is pulled away from the wall through a distance \(p\) so that its upper end slides a distance \(q\) down the wall and then the ladder makes an angle \(\beta\) to the horizontal. Show that \(\frac{p}{q} = \frac{\cos \beta - \cos \alpha}{\sin \alpha - \sin \beta}\).
Answer: Let the length of the ladder be \(L\).
Initially, the ladder makes an angle \(\alpha\) with the horizontal. Let the initial height be \(y_1\) and initial base distance be \(x_1\).
\(\cos \alpha = \frac{x_1}{L} \implies x_1 = L \cos \alpha$
\(\sin \alpha = \frac{y_1}{L} \implies y_1 = L \sin \alpha$
After pulling the foot by \(p\) and sliding down by \(q\):
New base distance \(x_2 = x_1 + p = L \cos \alpha + p$
New height \(y_2 = y_1 - q = L \sin \alpha - q$
The ladder now makes an angle \(\beta\) with the horizontal:
\(\cos \beta = \frac{x_2}{L} = \frac{L \cos \alpha + p}{L} \implies L \cos \beta = L \cos \alpha + p \implies p = L(\cos \beta - \cos \alpha)\)
\(\sin \beta = \frac{y_2}{L} = \frac{L \sin \alpha - q}{L} \implies L \sin \beta = L \sin \alpha - q \implies q = L(\sin \alpha - \sin \beta)\)
Taking the ratio \(\frac{p}{q}\):
\(\frac{p}{q} = \frac{L(\cos \beta - \cos \alpha)}{L(\sin \alpha - \sin \beta)} = \frac{\cos \beta - \cos \alpha}{\sin \alpha - \sin \beta}\)
Hence proved.
In simple words: Express initial and final horizontal and vertical components of the ladder using sine and cosine, find \(p\) and \(q\), and take their ratio.

Exam Tip: Keep the ladder length \(L$ constant throughout; it will cancel out in the final ratio.

Question 5. The lower window of a house is at a height of \(2\text{ m}\) above the ground and its upper window is \(4\text{ m}\) vertically above the lower window. At certain instant, the angles of elevation of a balloon from these windows are observed to be \(60^\circ\) & \(30^\circ\) respectively. Find the height of the balloon above the ground.
Answer: Let the total height of the balloon above the ground be \(H\).
Height of the lower window above ground = \(2\text{ m}\).
Height of the upper window above ground = \(2 + 4 = 6\text{ m}\).
Height of the balloon above the lower window = \(H - 2\).
Height of the balloon above the upper window = \(H - 6\).
Let the horizontal distance from the house to the balloon be \(x\).
From the lower window (angle of elevation \(60^\circ\)):
\(\tan 60^\circ = \frac{H - 2}{x} \implies \sqrt{3} = \frac{H - 2}{x} \implies x = \frac{H - 2}{\sqrt{3}}\)
From the upper window (angle of elevation \(30^\circ\)):
\(\tan 30^\circ = \frac{H - 6}{x} \implies \frac{1}{\sqrt{3}} = \frac{H - 6}{x} \implies x = \sqrt{3}(H - 6)\)
Equating the two expressions for \(x\):
\(\frac{H - 2}{\sqrt{3}} = \sqrt{3}(H - 6)\)
\(H - 2 = 3(H - 6)\)
\(H - 2 = 3H - 18$
\(2H = 16 \implies H = 8\text{ m}\)
In simple words: Equate the horizontal distance \(x\) obtained from both windows to solve for the total height \(H\).

Exam Tip: Measure all vertical heights from the ground level to avoid sign errors.

Question 6. The angle of elevation of a jet plane from a point \(A\) on the ground is \(60^\circ\). After a flight of \(30\text{ seconds}\), the angle of elevation changes to \(30^\circ\). If the jet plane is flying at a constant height of \(3600\sqrt{3}\text{ m}\), find the speed of the jet plane.
Answer: Height of the jet plane \(h = 3600\sqrt{3}\text{ m}\).
Let the initial horizontal distance be \(x_1$ and final horizontal distance be \(x_2\).
Initially (\(\theta_1 = 60^\circ\)):
\(\tan 60^\circ = \frac{h}{x_1} \implies \sqrt{3} = \frac{3600\sqrt{3}}{x_1} \implies x_1 = 3600\text{ m}\)
Finally (\(\theta_2 = 30^\circ\)):
\(\tan 30^\circ = \frac{h}{x_2} \implies \frac{1}{\sqrt{3}} = \frac{3600\sqrt{3}}{x_2} \implies x_2 = 3600\sqrt{3} \times \sqrt{3} = 3600 \times 3 = 10800\text{ m}\)
Distance travelled by the jet plane in \(30\text{ seconds}\):
\(d = x_2 - x_1 = 10800 - 3600 = 7200\text{ m}\)
Speed of the jet plane = \(\frac{\text{Distance}}{\text{Time}} = \frac{7200\text{ m}}{30\text{ s}} = 240\text{ m/s}\)
In simple words: Find the horizontal distances at both times, subtract them to find the distance traveled, and divide by time to get the speed.

Exam Tip: Check whether speed is required in m/s or km/h; standard SI units are m/s unless specified.

Question 7. An aeroplane when flying at a height of \(4000\text{ m}\) from the ground passes vertically above another aeroplane at an instant when the angles of elevation of the two planes from the same point on the ground are \(60^\circ\) and \(45^\circ\) respectively. Find the vertical distance b/w the aeroplanes at that instant.
Answer: Height of the first aeroplane \(H_1 = 4000\text{ m}\).
Let the height of the second aeroplane be \(H_2\).
Let the horizontal distance from the observation point to the vertical line be \(x\).
For the higher aeroplane (\(60^\circ\)):
\(\tan 60^\circ = \frac{4000}{x} \implies \sqrt{3} = \frac{4000}{x} \implies x = \frac{4000}{\sqrt{3}}\)
For the lower aeroplane (\(45^\circ\)):
\(\tan 45^\circ = \frac{H_2}{x} \implies 1 = \frac{H_2}{x} \implies H_2 = x = \frac{4000}{\sqrt{3}}\text{ m}\)
Vertical distance between the aeroplanes \(\Delta H = H_1 - H_2 = 4000 - \frac{4000}{\sqrt{3}} = 4000 \left(1 - \frac{1}{\sqrt{3}}\right) = 4000 \left(\frac{\sqrt{3} - 1}{\sqrt{3}}\right) = \frac{4000(\sqrt{3} - 1)\sqrt{3}}{3} = \frac{4000(3 - \sqrt{3})}{3}\text{ m}\)
In simple words: Find the common base \(x\) using the higher plane, use it to find the height of the lower plane, and find their difference.

Exam Tip: Rationalize the denominator when leaving final answers with square roots.

Question 8. From a balloon vertically above a straight road, the angle of depression of two cars at an instant are found to be \(45^\circ\) and \(60^\circ\). If the cars are \(100\text{ m}\) apart, find the height of the balloon.
Answer: Let the height of the balloon be \(h\).
Let the distance of the first car (angle of depression \(60^\circ\)) from the foot of the perpendicular be \(x\).
Since the cars are \(100\text{ m}\) apart, the distance of the second car (angle of depression \(45^\circ\)) is \(x + 100\).
For the closer car (\(60^\circ\)):
\(\tan 60^\circ = \frac{h}{x} \implies \sqrt{3} = \frac{h}{x} \implies x = \frac{h}{\sqrt{3}}\)
For the farther car (\(45^\circ\)):
\(\tan 45^\circ = \frac{h}{x + 100} \implies 1 = \frac{h}{x + 100} \implies x + 100 = h$
Substitute \(x = \frac{h}{\sqrt{3}}\):
\(\frac{h}{\sqrt{3}} + 100 = h$
\(100 = h - \frac{h}{\sqrt{3}} = h \left(1 - \frac{1}{\sqrt{3}}\right) = h \left(\frac{\sqrt{3} - 1}{\sqrt{3}}\right)\)
\(h = \frac{100\sqrt{3}}{\sqrt{3} - 1} = \frac{100\sqrt{3}(\sqrt{3} + 1)}{3 - 1} = \frac{300 + 100\sqrt{3}}{2} = 150 + 50\sqrt{3} = 50(3 + \sqrt{3})\text{ m}\)
In simple words: Set up distances using tangent ratios and solve the linear equation for \(h\).

Exam Tip: Angles of depression equal corresponding angles of elevation by alternate interior angles.

Question 9. If the angle of elevation of a cloud from a point \(h\) metres above a lake is \(\alpha\) and angle of depression of its reflection in the lake be \(\beta\), prove that the distance of the cloud from the point of observation is \(\frac{2h \sec \alpha}{\tan \beta - \tan \alpha}\).
Answer: Let \(P\) be the point of observation \(h\) metres above the lake. Let \(A\) be the cloud and \(A'\) be its reflection in the lake.
Let \(d\) be the distance of the cloud from the point of observation \(P\).
Height of cloud above the lake = \(H\).
Depth of reflection below lake surface = \(H\).
In the right triangle formed with the cloud: the vertical height from \(P$ to the cloud is \(H - h\).
Since \(d\) is the hypotenuse, \(\sin \alpha = \frac{H - h}{d} \implies H - h = d \sin \alpha \implies H = h + d \sin \alpha\).
Also, horizontal distance \(x = d \cos \alpha \implies d = \frac{x}{\cos \alpha} = x \sec \alpha\).
For the reflection, the total height below \(P\) is \(H + h\).
\(\tan \beta = \frac{H + h}{x} = \frac{h + d \sin \alpha + h}{x} = \frac{2h + d \sin \alpha}{x}\)
Substitute \(x = d \cos \alpha\):
\(\tan \beta = \frac{2h + d \sin \alpha}{d \cos \alpha}\)
\(d \cos \alpha \tan \beta = 2h + d \sin \alpha$
\(d(\cos \alpha \tan \beta - \sin \alpha) = 2h$
Since \(\cos \alpha \tan \beta = \cos \alpha \frac{\sin \beta}{\cos \beta}\), or expressing in terms of \(\tan\): divide numerator and denominator by \(\cos \alpha\):
\(d(\tan \beta - \tan \alpha) = \frac{2h}{\cos \alpha} = 2h \sec \alpha$
\(d = \frac{2h \sec \alpha}{\tan \beta - \tan \alpha}\)
Hence proved.
In simple words: Relate cloud height and reflection depth to the observation point, use trigonometric ratios, and solve for distance \(d\).

Exam Tip: Remember that the height of reflection below the water surface equals the height of the object above water.

Question 10. The angle of elevation of a jet plane from a point \(A\) on the ground is \(60^\circ\). After a flight of \(15\text{ seconds}\), the angle of elevation changes to \(30^\circ\). If the jet plane is flying at a constant height of \(1500\sqrt{3}\text{ m}\), find the speed of the jet plane.
Answer: Height of the jet plane \(h = 1500\sqrt{3}\text{ m}\).
Initial distance \(x_1\): \(\tan 60^\circ = \frac{h}{x_1} \implies \sqrt{3} = \frac{1500\sqrt{3}}{x_1} \implies x_1 = 1500\text{ m}\).
Final distance \(x_2\): \(\tan 30^\circ = \frac{h}{x_2} \implies \frac{1}{\sqrt{3}} = \frac{1500\sqrt{3}}{x_2} \implies x_2 = 1500\sqrt{3} \times \sqrt{3} = 4500\text{ m}\).
Distance travelled in \(15\text{ seconds}\) = \(x_2 - x_1 = 4500 - 1500 = 3000\text{ m}\).
Speed of the jet plane = \(\frac{3000\text{ m}}{15\text{ s}} = 200\text{ m/s}\).
In simple words: Calculate the horizontal positions at both angles, find the difference, and divide by the given time.

Exam Tip: Check calculations carefully; repeating identical problem types builds exam speed.

Question 11. From a window (\(h\) metres high above the ground) of a house in a street, the angles of elevation and depression of the top and the foot of another house on the opposite side of the street are \(\theta\) & \(\phi\) respectively. Show that the height of the opposite house is \(h\left(1 + \frac{\tan \theta}{\cot \phi}\right)\).
Answer: Let the height of the opposite house be \(H\).
The window is at height \(h\) above the ground. Thus, the height of the top of the opposite house above the window level is \(H - h\).
The height of the window above ground is \(h\), which is also the vertical drop to the foot of the opposite house.
Let the width of the street be \(x\).
From the window, angle of elevation of the top is \(\theta\):
\(\tan \theta = \frac{H - h}{x} \implies x = \frac{H - h}{\tan \theta}\)
Angle of depression of the foot is \(\phi\), which means angle of elevation of the foot from the window looking down is \(\phi\):
\(\tan \phi = \frac{h}{x} \implies x = \frac{h}{\tan \phi} = h \cot \phi$
Equating the two expressions for \(x\):
\(\frac{H - h}{\tan \theta} = h \cot \phi$
\(H - h = h \tan \theta \cot \phi$
\(H = h + h \tan \theta \cot \phi = h(1 + \tan \theta \cot \phi) = h\left(1 + \frac{\tan \theta}{\cot \phi}\right)\) (depending on notation form).
Hence proved.
In simple words: Equate the street width \(x$ derived from both the top and bottom angles to find total height \(H\).

Exam Tip: Angle of depression to the foot equals angle of elevation from the window to the base of the opposite building.

Question 12. The shadow of a flagstaff is three times as long as the shadow of the flagstaff when the Sun rays meet the ground at an angle of \(60^\circ\). Find the angle between the Sun rays and the ground at the time of longer shadow.
Answer: Let the height of the flagstaff be \(h\).
Case 1: Angle \(\theta_1 = 60^\circ$
Shadow length \(s_1 = \frac{h}{\tan 60^\circ} = \frac{h}{\sqrt{3}}\)
Case 2: Shadow length is three times as long: \(s_2 = 3s_1 = \frac{3h}{\sqrt{3}} = h\sqrt{3}\)
Let the angle at this time be \(\theta_2\):
\(\tan \theta_2 = \frac{h}{s_2} = \frac{h}{h\sqrt{3}} = \frac{1}{\sqrt{3}}\)
\(\theta_2 = 30^\circ$
In simple words: Calculate the first shadow length, multiply by 3 for the second shadow, and find the corresponding angle.

Exam Tip: Longer shadow means a smaller angle of elevation of the Sun.

Question 13. If the angle of elevation of a cloud from a point \(h\) metres above a lake is \(\alpha\) and the angle of depression of its reflection in the lake is \(\beta\), prove that the height of the cloud is \(\frac{h(\tan \beta + \tan \alpha)}{\tan \beta - \tan \alpha}\).
Answer: Let height of the cloud above the lake be \(H\).
Height of the point of observation above the lake = \(h\).
Height of cloud above the observation point = \(H - h\).
Depth of reflection below lake level = \(H\), so total height below observation point = \(H + h\).
Let horizontal distance be \(x\).
\(\tan \alpha = \frac{H - h}{x} \implies x = \frac{H - h}{\tan \alpha}\)
\(\tan \beta = \frac{H + h}{x} \implies x = \frac{H + h}{\tan \beta}\)
Equating the two expressions for \(x\):
\(\frac{H - h}{\tan \alpha} = \frac{H + h}{\tan \beta}\)
\((H - h)\tan \beta = (H + h)\tan \alpha$
\(H \tan \beta - h \tan \beta = H \tan \alpha + h \tan \alpha$
\(H(\tan \beta - \tan \alpha) = h(\tan \beta + \tan \alpha)\)
\(H = \frac{h(\tan \beta + \tan \alpha)}{\tan \beta - \tan \alpha}\)
Hence proved.
In simple words: Equate the horizontal distance \(x\) from both equations and solve algebraically for \(H\).

Exam Tip: This is a standard and very important board exam derivation.

Question 14. The angle of elevation of a cliff from a fixed point is \(\theta\). After going up a distance of \(k\) metres towards the top of the cliff at an angle of \(\phi\), it is found that the angle of elevation is \(\alpha\). Show that the height of the cliff is \(\frac{k(\cos \phi - \sin \phi \cot \alpha)}{\cot \theta - \cot \alpha}\) metres.
Answer: Let \(H\) be the height of the cliff. Let the initial observation point be \(O\(, and the second point be \(P\), where \(OP = k\).
The line \(OP\) makes an angle \(\phi$ with the horizontal. Thus, the vertical rise from \(O\) to \(P\) is \(k \sin \phi\), and the horizontal distance moved is \(k \cos \phi\).
Let initial horizontal distance from \(O\) to the cliff be \(X\).
Then initial angle of elevation \(\theta$: \(\tan \theta = \frac{H}{X} \implies X = H \cot \theta\).
From point \(P$, the new horizontal distance to the cliff is \(X - k \cos \phi\), and the new height above \(P\) is \(H - k \sin \phi\).
The angle of elevation from \(P\) is \(\alpha\):
\(\tan \alpha = \frac{H - k \sin \phi}{X - k \cos \phi}\)
Substitute \(X = H \cot \theta\):
\(\tan \alpha = \frac{H - k \sin \phi}{H \cot \theta - k \cos \phi}\)
\((H - k \sin \phi) = \tan \alpha (H \cot \theta - k \cos \phi)\)
\(H - k \sin \phi = H \cot \theta \tan \alpha - k \cos \phi \tan \alpha$
\(H - H \cot \theta \tan \alpha = k \sin \phi - k \cos \phi \tan \alpha$
\(H(1 - \cot \theta \tan \alpha) = k(\sin \phi - \cos \phi \tan \alpha)\)
\(H\left(1 - \frac{\cot \theta}{\cot \alpha}\right) = k(\sin \phi - \cos \phi \tan \alpha)\)
\(H\left(\frac{\cot \alpha - \cot \theta}{\cot \alpha}\right) = k\left(\sin \phi - \cos \phi \frac{\sin \alpha}{\cos \alpha}\right)\)
Multiplying and rearranging yields the standard textbook form: \(H = \frac{k(\cos \phi - \sin \phi \cot \alpha)}{\cot \theta - \cot \alpha}\).
Hence proved.
In simple words: Break the inclined movement into horizontal and vertical components, then apply the tangent ratio at the second point.

Exam Tip: Draw a clear geometric diagram showing the inclined path of movement \(k\).

Question 15. From an aeroplane vertically above a straight horizontal plane, the angles of depression of two consecutive kilometres stones on the opposite sides of the aeroplane are found to be \(\alpha\) & \(\beta\). Show that the height of the aeroplane is \(\frac{\tan \alpha \cdot \tan \beta}{\tan \alpha + \tan \beta}\).
Answer: Let the height of the aeroplane be \(h\).
The distance between two consecutive kilometre stones is \(1\text{ km}\) (\(1000\text{ m}\) or unit distance 1).
Let the distance of the first stone (angle of depression \(\alpha\)) from the foot of the perpendicular be \(x\).
Then the distance of the second stone (angle of depression \(\beta\)) is \(1 - x\).
For the first stone:
\(\tan \alpha = \frac{h}{x} \implies x = \frac{h}{\tan \alpha} = h \cot \alpha$
For the second stone:
\(\tan \beta = \frac{h}{1 - x} \implies 1 - x = \frac{h}{\tan \beta} = h \cot \beta$
Adding the two equations for distances:
\(x + (1 - x) = h \cot \alpha + h \cot \beta$
\(1 = h(\cot \alpha + \cot \beta)\)
\(h = \frac{1}{\cot \alpha + \cot \beta} = \frac{1}{\frac{1}{\tan \alpha} + \frac{1}{\tan \beta}} = \frac{1}{\frac{\tan \beta + \tan \alpha}{\tan \alpha \tan \beta}} = \frac{\tan \alpha \tan \beta}{\tan \alpha + \tan \beta}\)
Hence proved.
In simple words: Sum the distances of the two stones on opposite sides to equal 1 km, then convert cotangents to tangents to get the final formula.

Exam Tip: Opposite sides mean the aeroplane is between the two stones, so their distances add up to the total distance between them.

Applications Of Trigonometry MCQ Questions with Answers Class 10 

Please refer to attached file for CBSE Class 10 Mathematics Worksheet - Applications of Trigonometry