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Question 1: 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 $\frac{h(\tan \beta + \tan \alpha)}{\tan \beta - \tan \alpha}$.
Solution:
Step 1: Let cloud height be $H$. Height above observation point is $H-h$. Reflection depth is $H$. Depth below observation point is $H+h$.
Step 2: $\tan \alpha = \frac{H-h}{x}$ and $\tan \beta = \frac{H+h}{x}$.
Step 3: Divide: $\frac{\tan \alpha}{\tan \beta} = \frac{H-h}{H+h}$.
Step 4: $H \tan \alpha + h \tan \alpha = H \tan \beta - h \tan \beta \Rightarrow h(\tan \alpha + \tan \beta) = H(\tan \beta - \tan \alpha)$.
Step 5: $H = \frac{h(\tan \beta + \tan \alpha)}{\tan \beta - \tan \alpha}$. -
Question 2: A round balloon of radius $r$ subtends an angle $\alpha$ at the eye of the observer while the angle of elevation of its centre is $\beta$. Prove that the height of the centre of the balloon is $r \sin \beta \csc(\alpha/2)$.
Solution:
Step 1: Let $O$ be observer, $C$ be centre. In $\Delta OAC$ (tangent from O), $\sin(\alpha/2) = \frac{r}{OC} \Rightarrow OC = r \csc(\alpha/2)$.
Step 2: In $\Delta OCM$ (vertical triangle), $\sin \beta = \frac{h}{OC} \Rightarrow h = OC \sin \beta$.
Step 3: Substitute $OC$: $h = r \sin \beta \csc(\alpha/2)$. -
Question 3: The angle of elevation of a jet plane from a point A on the ground is 60°. After a flight of 30 seconds, the angle of elevation changes to 30°. If the jet plane is flying at a constant height of $3600\sqrt{3}$ m, find the speed of the jet plane.
Solution:
Step 1: $h = 3600\sqrt{3}$. $\tan 60^\circ = \frac{h}{x} \Rightarrow x = 3600$.
Step 2: $\tan 30^\circ = \frac{h}{x+d} \Rightarrow x+d = 3600\sqrt{3} \times \sqrt{3} = 10800$.
Step 3: $d = 10800 - 3600 = 7200$ m.
Step 4: Speed $= \frac{7200}{30} = 240$ m/s $= 864$ km/h. -
Question 4: A straight highway leads to the foot of a tower. A man standing at the top of the tower observes a car at an angle of depression of 30°, which is approaching the foot of the tower with a uniform speed. Six seconds later, the angle of depression of the car is found to be 60°. Find the time taken by the car to reach the foot of the tower from this point.
Solution:
Step 1: Let height $h$. Distances $x$ (at 60°) and $y$ (at 30°).
Step 2: $\tan 60^\circ = h/x \Rightarrow h = x\sqrt{3}$. $\tan 30^\circ = h/y \Rightarrow h = y/\sqrt{3}$.
Step 3: $x\sqrt{3} = y/\sqrt{3} \Rightarrow y = 3x$. Distance traveled $= y-x = 2x$.
Step 4: Time for $2x$ is 6s. Time for remaining distance $x$ is 3s. -
Question 5: 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 $\beta$ with the horizontal. Show that $\frac{p}{q} = \frac{\cos \beta - \cos \alpha}{\sin \alpha - \sin \beta}$.
Solution:
Step 1: Let length be $L$. Initial coords $(L \cos \alpha, L \sin \alpha)$. Final coords $(L \cos \beta, L \sin \beta)$.
Step 2: $p = x_2 - x_1 = L(\cos \beta - \cos \alpha)$.
Step 3: $q = y_1 - y_2 = L(\sin \alpha - \sin \beta)$.
Step 4: Ratio $\frac{p}{q} = \frac{L(\cos \beta - \cos \alpha)}{L(\sin \alpha - \sin \beta)} = \frac{\cos \beta - \cos \alpha}{\sin \alpha - \sin \beta}$. -
Question 6: From the top of a light house, the angles of depression of two ships on the opposite sides of it are observed to be $\alpha$ and $\beta$. If the height of the light house is $h$ meters and the line joining the ships passes through the foot of the light house, show that the distance between the ships is $\frac{h(\tan \alpha + \tan \beta)}{\tan \alpha \tan \beta}$.
Solution:
Step 1: Distances $x, y$. $\tan \alpha = h/x \Rightarrow x = h \cot \alpha$. $\tan \beta = h/y \Rightarrow y = h \cot \beta$.
Step 2: Distance $= x+y = h(\cot \alpha + \cot \beta)$.
Step 3: $h(\frac{1}{\tan \alpha} + \frac{1}{\tan \beta}) = \frac{h(\tan \alpha + \tan \beta)}{\tan \alpha \tan \beta}$. -
Question 7: A bird is sitting on the top of a 80 m high tree. From a point on the ground, the angle of elevation of the bird is 45°. The bird flies away horizontally in such a way that it remained at a constant height from the ground. After 2 seconds, the angle of elevation of the bird from the same point is 30°. Find the speed of flying of the bird.
Solution:
Step 1: $h=80$. $\tan 45^\circ = 80/x \Rightarrow x=80$.
Step 2: $\tan 30^\circ = 80/y \Rightarrow y=80\sqrt{3}$.
Step 3: Distance $= y-x = 80(\sqrt{3}-1)$.
Step 4: Speed $= \frac{80(\sqrt{3}-1)}{2} = 40(\sqrt{3}-1)$ m/s. -
Question 8: The angle of elevation of a jet fighter from a point A on the ground is 60°. After a flight of 15 seconds, the angle of elevation changes to 30°. If the jet is flying at a speed of 720 km/hr, find the constant height at which the jet is flying.
Solution:
Step 1: Speed $= 720 \times \frac{5}{18} = 200$ m/s. Distance in 15s $= 3000$ m.
Step 2: Let height $h$. $x = h \cot 60^\circ = h/\sqrt{3}$.
Step 3: $x+3000 = h \cot 30^\circ = h\sqrt{3}$.
Step 4: $h\sqrt{3} - h/\sqrt{3} = 3000 \Rightarrow \frac{2h}{\sqrt{3}} = 3000 \Rightarrow h = 1500\sqrt{3}$ m. -
Question 9: A man on the deck of a ship, which is 10 m above water level, observes the angle of elevation of the top of a hill as 60° and the angle of depression of the base of the hill as 30°. Find the distance of the hill from the ship and the height of the hill.
Solution:
Step 1: Deck height 10. $\tan 30^\circ = 10/x \Rightarrow x = 10\sqrt{3}$ (distance).
Step 2: Hill height $H = 10 + h'$. $\tan 60^\circ = h'/x \Rightarrow h' = x\sqrt{3} = 10\sqrt{3} \times \sqrt{3} = 30$.
Step 3: Total height $10+30=40$ m. -
Question 10: A vertical tower stands on a horizontal plane and is surmounted by a vertical flagstaff of height $h$. At a point on the plane, the angles of elevation of the bottom and the top of the flagstaff are $\alpha$ and $\beta$, respectively. Prove that the height of the tower is $\frac{h \tan \alpha}{\tan \beta - \tan \alpha}$.
Solution:
Step 1: Let tower height $H$, distance $x$. $\tan \alpha = H/x \Rightarrow x = H \cot \alpha$.
Step 2: $\tan \beta = \frac{H+h}{x}$.
Step 3: $H+h = x \tan \beta = H \cot \alpha \tan \beta = H \frac{\tan \beta}{\tan \alpha}$.
Step 4: $h = H(\frac{\tan \beta}{\tan \alpha} - 1) \Rightarrow H = \frac{h \tan \alpha}{\tan \beta - \tan \alpha}$.