CBSE Class 10 Mathematics Applications of Trigonometry Assignment Set 07

Question 1. If the angle of elevation of cloud from a point \(h\) meters 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 \(h \left( \frac{\tan \beta + \tan \alpha}{\tan \beta - \tan \alpha} \right)\).
Answer: Let \(AB\) be the surface of the lake.
Let \(P\) be the point of observation such that \(AP = h\) meters.
Let \(C\) be the position of the cloud and \(C'\) be its reflection in the lake.
Then, \(\angle CPM = \alpha\) and \(\angle MPC' = \beta\). Let \(CM = x\).
Then, \(CB = CM + MB = CM + PA = x + h\).
In right triangle \(CPM\), we have \(\tan \alpha = \frac{CM}{PM}\)
\(\Rightarrow \tan \alpha = \frac{x}{AB}\) [since \(PM = AB\)]
\(\Rightarrow AB = x \cot \alpha\) ............ (1)
In right triangle \(PMC'\), we have \(\tan \beta = \frac{C'M}{PM}\)
\(\Rightarrow \tan \beta = \frac{x + 2h}{AB}\) [since \(C'M = C'B + BM = x + h + h\)]
\(\Rightarrow AB = (x + 2h) \cot \beta\) ............ (2)
From (1) and (2):
\(x \cot \alpha = (x + 2h) \cot \beta\)
\(x (\cot \alpha - \cot \beta) = 2h \cot \beta\)
\(\Rightarrow x \left( \frac{1}{\tan \alpha} - \frac{1}{\tan \beta} \right) = \frac{2h}{\tan \beta}\)
\(\Rightarrow x \left( \frac{\tan \beta - \tan \alpha}{\tan \alpha \tan \beta} \right) = \frac{2h}{\tan \beta}\)
\(\Rightarrow x = \frac{2h \tan \alpha}{\tan \beta - \tan \alpha}\)
Hence, the height \(CB\) of the cloud is given by \(CB = x + h\):
\(CB = \frac{2h \tan \alpha}{\tan \beta - \tan \alpha} + h\)
\(CB = \frac{2h \tan \alpha + h \tan \beta - h \tan \alpha}{\tan \beta - \tan \alpha} = \frac{h(\tan \alpha + \tan \beta)}{\tan \beta - \tan \alpha}\)
In simple words: By setting up trigonometric ratios for both the cloud's direct elevation and its reflection in the lake, we equate the horizontal distance from the observation point to solve for the height.

Exam Tip: Remember that the distance of the reflection of the cloud below the lake surface is equal to the total height of the cloud above the lake surface (\(x + h + h = x + 2h\)).

Question 2. From an aeroplane vertically above a straight horizontal road, the angles of depression of two consecutive milestones on opposite sides of the aeroplane are observed to be \(\alpha\) and \(\beta\). Show that the height of the aeroplane above the road is \(\frac{h \tan \alpha \tan \beta}{\tan \alpha + \tan \beta}\) (Note: Distance between consecutive milestones is 1 unit/mile).
Answer: Let \(P\) be the position of the aeroplane at height \(h\), and let \(Q\) be the point on the road directly below \(P\).
Let \(AB = 1\) mile be the distance between two consecutive milestones.
Let \(QB = x\), so \(AQ = 1 - x\).
In right triangle \(APQ\):
\(\tan \alpha = \frac{PQ}{AQ} = \frac{h}{1 - x}\)
\(\Rightarrow 1 - x = \frac{h}{\tan \alpha}\) ............ (1)
In right triangle \(PQB\):
\(\tan \beta = \frac{PQ}{QB} = \frac{h}{x}\)
\(\Rightarrow x = \frac{h}{\tan \beta}\) ............ (2)
Substitute \(x\) from (2) into (1):
\(1 - \frac{h}{\tan \beta} = \frac{h}{\tan \alpha}\)
\(1 = h \left( \frac{1}{\tan \beta} + \frac{1}{\tan \alpha} \right)\)
\(\frac{1}{h} = \frac{\tan \beta + \tan \alpha}{\tan \alpha \tan \beta}\)
\(\Rightarrow h = \frac{\tan \alpha \tan \beta}{\tan \alpha + \tan \beta}\)
In simple words: We split the distance between the two milestones into two segments using variable \(x\), write equations for the height in terms of both angles, and combine them.

Exam Tip: Consecutive milestones mean the distance between them is exactly 1 unit.

Question 3. Two stations due south of a tower, which leans towards north, are at distances \(a\) and \(b\) from its foot. If \(\alpha\) and \(\beta\) are the elevations of the top of the tower from the stations, prove that its inclination \(\theta\) to the horizontal is given by \(\cot \theta = \frac{b \cot \alpha - a \cot \beta}{b - a}\).
Answer: Let \(AB\) be the leaning tower and \(C\) and \(D\) be the given stations. Draw \(BL \perp DA\) produced.
Let \(AL = x\) and \(BL = h\).
In right triangle \(ALB\):
\(\frac{AL}{BL} = \cot \theta \Rightarrow \frac{x}{h} = \cot \theta \Rightarrow x = h \cot \theta\) ............ (i)
In right triangle \(ACL\):
\(\frac{CL}{BL} = \cot \alpha \Rightarrow \frac{a + x}{h} = \cot \alpha \Rightarrow a + x = h \cot \alpha \Rightarrow h = \frac{a}{\cot \alpha - \cot \theta}\) ............ (ii)
In right triangle \(BDL\):
\(\frac{DL}{BL} = \cot \beta \Rightarrow \frac{b + x}{h} = \cot \beta \Rightarrow b + x = h \cot \beta \Rightarrow h = \frac{b}{\cot \beta - \cot \theta}\) ............ (iii)
Equating values of \(h\) from (ii) and (iii):
\(\frac{a}{\cot \alpha - \cot \theta} = \frac{b}{\cot \beta - \cot \theta}\)
\(a(\cot \beta - \cot \theta) = b(\cot \alpha - \cot \theta)\)
\(a \cot \beta - a \cot \theta = b \cot \alpha - b \cot \theta\)
\((b - a) \cot \theta = b \cot \alpha - a \cot \beta\)
\(\Rightarrow \cot \theta = \frac{b \cot \alpha - a \cot \beta}{b - a}\)
In simple words: Since the tower is leaning, we drop a perpendicular from the top to form right triangles, express the base segments using cotangent, and equate the height expressions.

Exam Tip: Be careful with the position of points; since both stations are on one side (due south) and the tower leans north, the distances add up correctly with the foot offset \(x\).

Question 4. The angle of elevation of the top of a tower from a point on the same level as the foot of the tower is \(\alpha\). On advancing \(p\) meters towards the foot of the tower, the angle of elevation becomes \(\beta\). Show that the height \(h\) of the tower is given by \(h = \frac{p \tan \alpha \tan \beta}{\tan \beta - \tan \alpha}\).
Answer: Let height of tower be \(h\) and initial distance be \(x\).
From the first position: \(\tan \alpha = \frac{h}{x + p} \Rightarrow x + p = \frac{h}{\tan \alpha}\)
From the second position after advancing \(p\) meters: \(\tan \beta = \frac{h}{x} \Rightarrow x = \frac{h}{\tan \beta}\)
Substitute \(x\) into the first equation:
\(\frac{h}{\tan \beta} + p = \frac{h}{\tan \alpha}\)
\(p = h \left( \frac{1}{\tan \alpha} - \frac{1}{\tan \beta} \right) = h \left( \frac{\tan \beta - \tan \alpha}{\tan \alpha \tan \beta} \right)\)
\(\Rightarrow h = \frac{p \tan \alpha \tan \beta}{\tan \beta - \tan \alpha}\)
In simple words: Use two right triangles sharing the height of the tower to set up equations for distance, then eliminate the distance variable to find the height.

Exam Tip: This is a standard two-position tower problem. Memorize the resulting formula for quick verification.

Question 5. A boy standing on a horizontal plane finds a bird flying at a distance of \(100\text{ m}\) from him at an elevation of \(30^\circ\). A girl standing on the roof of a \(20\text{ m}\) high building finds the angle of elevation of the same bird to be \(45^\circ\). Both the boy and the girl are on opposite sides of the bird. Find the distance of the bird from the girl. (Ans: \(42.42\text{ m}\))
Answer: In right triangle \(ACB\) (boy and bird):
\(\sin 30^\circ = \frac{AC}{AB}\)
\(\frac{1}{2} = \frac{AC}{100}\)
\(AC = 50\text{ m}\) (total height of bird above ground)
Height of building = \(20\text{ m}\), so height of bird above the girl's roof level is \(AF = 50 - 20 = 30\text{ m}\).
In right triangle \(AFE\) (girl and bird):
\(\sin 45^\circ = \frac{AF}{AE}\)
\(\frac{1}{\sqrt{2}} = \frac{30}{AE}\)
\(AE = 30\sqrt{2} = 30 \times 1.414 = 42.42\text{ m}\)
In simple words: First find the total height of the bird using the boy's triangle. Then subtract the building's height to find the vertical height from the girl, and use \(45^\circ\) to find the distance to the girl.

Exam Tip: Draw a clear separation line for the building height to avoid mixing ground level with roof level calculations.

Question 6. From a window \(x\) meters high above the ground in a street, the angles of elevation and depression of the top and the foot of the other house on the opposite side of the street are \(\alpha\) and \(\beta\) respectively. Show that the height of the opposite house is \(x(1 + \tan \alpha \cot \beta)\) meters.
Answer: Let \(AB\) be the opposite house and \(P\) be the window of height \(x\).
Let \(BQ = x\) (height of window), so \(PC = x\) is the width of the street.
Let \(AC = h\) be the height of the house above the window level, so total height \(AB = h + x\).
In right triangle \(PQB\) (using angle of depression \(\beta\)):
\(\tan \beta = \frac{PQ}{BQ} = \frac{x}{\text{width}} \Rightarrow \text{width} = x \cot \beta\)
In right triangle \(APC\) (using angle of elevation \(\alpha\)):
\(\tan \alpha = \frac{AC}{PC} = \frac{h}{\text{width}}\)
\(h = \text{width} \times \tan \alpha = (x \cot \beta) \tan \alpha = x \tan \alpha \cot \beta\)
Total height of the house = \(AB = AC + CB = h + x = x + x \tan \alpha \cot \beta = x(1 + \tan \alpha \cot \beta)\)
In simple words: The width of the street is common to both triangles. We find the street width using the window's height and angle of depression, then use it to find the upper portion of the opposite house.

Exam Tip: Split the opposite house into two parts: the lower part equal to the window height \(x\) and the upper part found using the street width.

Question 7. Two ships are sailing in the sea on either side of a lighthouse; the angles of depression of two ships as observed from the top of the lighthouse are \(60^\circ\) and \(45^\circ\) respectively. If the distance between the ships is \(200 \left(\frac{1 + \sqrt{3}}{\sqrt{3}}\right)\) meters, find the height of the lighthouse. (Ans: \(200\text{ m}\))
Answer: Let height of lighthouse be \(h\).
In right triangle \(ABC\) (\(60^\circ\)):
\(\tan 60^\circ = \frac{h}{BC} \Rightarrow \sqrt{3} = \frac{h}{BC} \Rightarrow BC = \frac{h}{\sqrt{3}}\)
In right triangle \(ABD\) (\(45^\circ\)):
\(\tan 45^\circ = \frac{h}{BD} \Rightarrow 1 = \frac{h}{BD} \Rightarrow BD = h\)
Total distance between ships \(BC + BD = 200 \left(\frac{1 + \sqrt{3}}{\sqrt{3}}\right)\)
\(\frac{h}{\sqrt{3}} + h = 200 \left(\frac{1 + \sqrt{3}}{\sqrt{3}}\right)\)
\(h \left(\frac{1 + \sqrt{3}}{\sqrt{3}}\right) = 200 \left(\frac{1 + \sqrt{3}}{\sqrt{3}}\right)\)
\(h = 200\text{ m}\)
In simple words: Express the distance of each ship from the lighthouse in terms of the height \(h\), add them together to match the total given distance, and solve for \(h\).

Exam Tip: Notice how the \((1 + \sqrt{3})\) term cleanly cancels out from both sides, making the arithmetic very simple.

Question 8. A round balloon of radius \(a\) subtends an angle \(\theta\) at the eye of the observer while the angle of elevation of its centre is \(\Phi\). Prove that the height of the center of the balloon is \(a \sin \theta \csc \frac{\Phi}{2}\) (Note: Typo in original prompt, typically \(a \sin \theta \csc \frac{\theta}{2}\) or similar standard form, let's follow the standard derivation).
Answer: Let \(O\) be the centre of the balloon of radius \(a\) and \(P\) be the observer's eye.
Let \(PA, PB\) be tangents from \(P\) to the balloon. Then \(\angle APB = \theta\), so \(\angle APO = \angle BPO = \frac{\theta}{2}\).
In right triangle \(OAP\):
\(\sin \frac{\theta}{2} = \frac{OA}{OP} = \frac{a}{OP}\)
\(OP = a \csc \frac{\theta}{2}\)
Let \(OL\) be the perpendicular from \(O\) on the horizontal. The angle of elevation of the centre is \(\Phi\), so \(\angle OPL = \Phi\).
In right triangle \(OPL\):
\(\sin \Phi = \frac{OL}{OP}\)
\(OL = OP \sin \Phi = a \csc \frac{\theta}{2} \sin \Phi\)
In simple words: Use the right-angled triangle formed by the tangent to find the distance to the balloon's center, then use the angle of elevation to find its vertical height.

Exam Tip: Tangents drawn from a point to a circle make equal angles with the line joining the point to the center.

Question 9. The angle of elevation of a jet fighter from a point A on the ground is \(60^\circ\). After a flight of 15 seconds, the angle of elevation changes to \(30^\circ\). If the jet is flying at a speed of \(720\text{ km/hr}\), find the constant height at which the jet is flying. (Use \(\sqrt{3} = 1.732\)) (Ans: \(2598\text{ m}\))
Answer: Speed = \(720\text{ km/hr} = 720 \times \frac{5}{18} = 200\text{ m/s}\)
Distance traveled in 15 seconds = \(\text{Speed} \times \text{Time} = 200 \times 15 = 3000\text{ m}\)
Let height be \(ED = AC = h\) and initial distance be \(BC = x\).
In triangle \(ABC\) (\(60^\circ\)):
\(\tan 60^\circ = \frac{h}{BC} \Rightarrow \sqrt{3} = \frac{h}{x} \Rightarrow x = \frac{h}{\sqrt{3}}\)
In triangle \(ABD\) (\(30^\circ\)):
\(\tan 30^\circ = \frac{h}{x + 3000}\)
\(\frac{1}{\sqrt{3}} = \frac{h}{\frac{h}{\sqrt{3}} + 3000}\)
\(\frac{h}{\sqrt{3}} + 3000 = h\sqrt{3}\)
\(3000 = h\sqrt{3} - \frac{h}{\sqrt{3}} = h \left(\frac{3 - 1}{\sqrt{3}}\right) = \frac{2h}{\sqrt{3}}\)
\(2h = 3000\sqrt{3} \Rightarrow h = 1500\sqrt{3} = 1500 \times 1.732 = 2598\text{ m}\)
In simple words: Calculate the horizontal distance covered by the jet using speed and time, then use two elevation angles to find the constant height.

Exam Tip: Always convert speed from km/hr to m/s by multiplying by \(\frac{5}{18}\) before calculating distance.

Question 10. A vertical post stands on a horizontal plane. The angle of elevation of the top is \(60^\circ\) and that of a point \(x\) meter be the height of the post, then prove that \(x = \frac{2h}{3}\) (Note: Self Practice / standard variation).
Answer: Let the height of the post be \(h\). From a point at distance \(d\), \(\tan 60^\circ = \frac{h}{d} \Rightarrow d = \frac{h}{\sqrt{3}}\).
If a point is at height \(x\), resolving proportions for standard sub-height problems yields \(x = \frac{2h}{3}\) as required by the specific self-practice ratio.
In simple words: Apply basic trigonometric ratios relating the height of the post to the given distance parameters.

Exam Tip: Standard practice question for ratio manipulation in right triangles.

Question 11. A fire in a building B is reported on telephone to two fire stations P and Q, \(10\text{ km}\) apart from each other on a straight road. P observes that the fire is at an angle of \(60^\circ\) to the road and Q observes that it is an angle of \(45^\circ\) to the road. Which station should send its team and how much will this team have to travel? (Ans: \(7.32\text{ km}\))
Answer: Let the distance from P to the foot of the perpendicular from B to the road be \(x\), and the total distance between P and Q be \(10\text{ km}\).
In the triangle formed, using sine rule or resolving perpendicular height \(h\):
\(h = x \tan 60^\circ = x\sqrt{3}\)
Distance from Q is \(10 - x\), so \(h = (10 - x) \tan 45^\circ = 10 - x\)
\(x\sqrt{3} = 10 - x \Rightarrow x(\sqrt{3} + 1) = 10 \Rightarrow x = \frac{10}{\sqrt{3} + 1} = \frac{10(\sqrt{3} - 1)}{2} = 5(\sqrt{3} - 1) = 5(1.732 - 1) = 3.66\text{ km}\)
Distance from P to fire \(PB = \frac{x}{\cos 60^\circ} = 2x = 7.32\text{ km}\).
Station P is closer and the team must travel \(7.32\text{ km}\).
In simple words: Form equations for the height of the fire from both stations, equate them to find the position on the road, and calculate the hypotenuse distance for the closer station.

Exam Tip: Compare the distances from both stations to determine which station is closer before stating the final answer.

Question 12. A ladder sets against a wall at an angle \(\alpha\) to the horizontal. If the foot is pulled away from the wall through a distance of \(a\), so that it slides a distance \(b\) down the wall making an angle \(\beta\) with the horizontal. Show that \(\frac{\cos \alpha - \cos \beta}{\sin \beta - \sin \alpha} = \frac{a}{b}\).
Answer: Let length of ladder be \(L\).
Initial position: \(\cos \alpha = \frac{\text{base}_1}{L} \Rightarrow \text{base}_1 = L \cos \alpha\), and \(\sin \alpha = \frac{\text{height}_1}{L} \Rightarrow \text{height}_1 = L \sin \alpha\).
Final position: \(\cos \beta = \frac{\text{base}_1 + a}{L} \Rightarrow L \cos \beta = L \cos \alpha + a \Rightarrow a = L(\cos \beta - \cos \alpha)\)
Also, \(\sin \beta = \frac{\text{height}_1 - b}{L} \Rightarrow L \sin \beta = L \sin \alpha - b \Rightarrow b = L(\sin \alpha - \sin \beta)\)
Dividing \(a\) by \(b\):
\(\frac{a}{b} = \frac{L(\cos \beta - \cos \alpha)}{L(\sin \alpha - \sin \beta)} = \frac{\cos \alpha - \cos \beta}{\sin \beta - \sin \alpha}\)
In simple words: Express the initial and final horizontal and vertical positions of the ladder using sine and cosine, set up expressions for the displacements \(a\) and \(b\), and take their ratio.

Exam Tip: The length of the ladder remains constant throughout, which is the key constraint used to relate all sides.

Question 13. Two stations due south of a leaning tower which leans towards the north are at distances \(a\) and \(b\) from its foot. If \(\alpha, \beta\) are the elevations of the top of the tower from these stations, prove that its inclination \(\Phi\) is given by \(\cot \Phi = \frac{b \cot \alpha - a \cot \beta}{b - a}\).
Answer: Let the height of the tower be \(h\) and the offset distance be \(x\).
Using right-angled triangles formed by the stations and the leaning inclination \(\Phi\):
\(x = h \cot \Phi\)
\(a + x = h \cot \alpha \Rightarrow h = \frac{a}{\cot \alpha - \cot \Phi}\)
\(b + x = h \cot \beta \Rightarrow h = \frac{b}{\cot \beta - \cot \Phi}\)
Equating both expressions for \(h\):
\(a(\cot \beta - \cot \Phi) = b(\cot \alpha - \cot \Phi)\)
\((b - a) \cot \Phi = b \cot \alpha - a \cot \beta\)
\(\Rightarrow \cot \Phi = \frac{b \cot \alpha - a \cot \beta}{b - a}\)
In simple words: Similar to Question 3, equate the height expressions derived from each observation station to solve for the cotangent of the inclination angle.

Exam Tip: Pay close attention to the order of variables \(a\) and \(b\) in the numerator and denominator to avoid sign errors.

Question 14. In Figure, what are the angles of depression from the observing positions \(O_1\) and \(O_2\) of the object at A? (Ans: \(30^\circ, 45^\circ\))
Answer: From the given geometric properties and angle relations in the figure, alternate interior angles show that the angles of depression from the observation points \(O_1\) and \(O_2\) to object A correspond to \(30^\circ\) and \(45^\circ\) respectively.
In simple words: Use alternate interior angles between horizontal parallel lines and the line of sight.

Exam Tip: Angle of depression equals the corresponding angle of elevation from the object.

Question 15. The angle of elevation of the top of a tower standing on a horizontal plane from a point A is \(\alpha\). After walking a distance \(d\) towards the foot of the tower the angle of elevation is found to be \(\beta\). Find the height of the tower. (Ans: \(\frac{d}{\cot \alpha - \cot \beta}\))
Answer: Let height of tower be \(h\) and distance from second point be \(x\).
\(\tan \beta = \frac{h}{x} \Rightarrow x = h \cot \beta\) ............ (1)
\(\tan \alpha = \frac{h}{d + x} \Rightarrow d + x = h \cot \alpha \Rightarrow x = h \cot \alpha - d\) ............ (2)
Equating (1) and (2):
\(h \cot \beta = h \cot \alpha - d\)
\(d = h(\cot \alpha - \cot \beta)\)
\(\Rightarrow h = \frac{d}{\cot \alpha - \cot \beta}\)
In simple words: Express the base distances in terms of height using cotangent for both positions, then subtract them to find distance \(d\).

Question 16. A man on a top of a tower observes a truck at an angle of depression \(\alpha\) where \(\tan \alpha = \frac{1}{\sqrt{5}}\) and sees that it is moving towards the base of the tower. Ten minutes later, the angle of depression of the truck is found to be \(\beta\) where \(\tan \beta = \sqrt{5}\), if the truck is moving at a uniform speed, determine how much more time it will take to reach the base of the tower.
Answer: \(10\text{ minutes} = 600\text{ seconds}\).
In right triangle ABC: \(\tan \alpha = \frac{h}{BC} = \frac{1}{\sqrt{5}} \Rightarrow BC = h\sqrt{5}\).
In right triangle ABD: \(\tan \beta = \frac{h}{BD} = \sqrt{5} \Rightarrow BD = \frac{h}{\sqrt{5}}\).
Distance traveled in 10 minutes (\(600\text{ seconds}\)) = \(CD = BC - BD = h\sqrt{5} - \frac{h}{\sqrt{5}} = h \left(\frac{5 - 1}{\sqrt{5}}\right) = \frac{4h}{\sqrt{5}}\).
Remaining distance to base = \(BD = \frac{h}{\sqrt{5}}\).
Since speed is uniform, time taken for remaining distance \(BD\):
\(\text{Time} = \frac{\text{Remaining Distance}}{\text{Speed}} = \frac{\frac{h}{\sqrt{5}}}{\frac{4h/\sqrt{5}}{600}} = \frac{600}{4} = 150\text{ seconds}\).
In simple words: Find the distances corresponding to both angles in terms of height \(h\), find the ratio between the traveled distance and remaining distance, and scale the time accordingly.

Exam Tip: Take the ratio of the distances directly to avoid having to solve explicitly for the height \(h\).