This page provides comprehensive Chapter 9: Applications of Trigonometry - Standard Worksheet - SJMaths. Standard level practice worksheet for Class 10 Applications of Trigonometry. Practice for CBSE Board Exams.
Question 1: A circus artist is climbing a 20 m long rope, which is tightly stretched and tied from the top of a vertical pole to the ground. Find the height of the pole, if the angle made by the rope with the ground level is 30°.
Solution: Step 1: Let height of pole be $h$. Length of rope (hypotenuse) $= 20$ m. Step 2: $\sin 30^\circ = \frac{\text{Opposite}}{\text{Hypotenuse}} = \frac{h}{20}$. Step 3: $\frac{1}{2} = \frac{h}{20} \Rightarrow h = 10$ m. Answer: 10 m.
Question 2: A tree breaks due to storm and the broken part bends so that the top of the tree touches the ground making an angle 30° with it. The distance between the foot of the tree to the point where the top touches the ground is 8 m. Find the height of the tree.
Solution: Step 1: Let broken part be $x$ and standing part be $y$. Base $= 8$ m. Step 2: $\tan 30^\circ = \frac{y}{8} \Rightarrow y = \frac{8}{\sqrt{3}}$. Step 3: $\cos 30^\circ = \frac{8}{x} \Rightarrow \frac{\sqrt{3}}{2} = \frac{8}{x} \Rightarrow x = \frac{16}{\sqrt{3}}$. Step 4: Total height $= x+y = \frac{24}{\sqrt{3}} = 8\sqrt{3}$ m.
Question 3: The angle of elevation of the top of a tower from a point on the ground, which is 30 m away from the foot of the tower, is 30°. Find the height of the tower.
Solution: Step 1: Let height be $h$. Base $= 30$ m. Step 2: $\tan 30^\circ = \frac{h}{30}$. Step 3: $\frac{1}{\sqrt{3}} = \frac{h}{30} \Rightarrow h = \frac{30}{\sqrt{3}} = 10\sqrt{3}$ m.
Question 4: A 1.5 m tall boy is standing at some distance from a 30 m tall building. The angle of elevation from his eyes to the top of the building increases from 30° to 60° as he walks towards the building. Find the distance he walked towards the building.
Solution: Step 1: Height above eye level $= 30 - 1.5 = 28.5$ m. Step 2: Let initial distance be $x+y$ and final distance be $y$. Step 3: $\tan 60^\circ = \frac{28.5}{y} \Rightarrow y = \frac{28.5}{\sqrt{3}}$. Step 4: $\tan 30^\circ = \frac{28.5}{x+y} \Rightarrow x+y = 28.5\sqrt{3}$. Step 5: $x = 28.5\sqrt{3} - \frac{28.5}{\sqrt{3}} = 28.5(\frac{2}{\sqrt{3}}) = 19\sqrt{3}$ m.
Question 5: From a point on the ground, the angles of elevation of the bottom and the top of a transmission tower fixed at the top of a 20 m high building are 45° and 60° respectively. Find the height of the tower.
Solution: Step 1: Let building height $AB=20$, tower height $BC=h$, point $P$ on ground at distance $x$. Step 2: In $\Delta PAB$, $\tan 45^\circ = 20/x \Rightarrow x=20$. Step 3: In $\Delta PAC$, $\tan 60^\circ = (20+h)/x \Rightarrow \sqrt{3} = (20+h)/20$. Step 4: $20\sqrt{3} = 20+h \Rightarrow h = 20(\sqrt{3}-1)$ m.
Question 6: A statue, 1.6 m tall, stands on the top of a pedestal. From a point on the ground, the angle of elevation of the top of the statue is 60° and from the same point the angle of elevation of the top of the pedestal is 45°. Find the height of the pedestal.
Solution: Step 1: Let pedestal height be $h$ and distance be $x$. Step 2: $\tan 45^\circ = h/x \Rightarrow x=h$. Step 3: $\tan 60^\circ = (h+1.6)/x \Rightarrow \sqrt{3}h = h+1.6$. Step 4: $h(\sqrt{3}-1) = 1.6 \Rightarrow h = \frac{1.6}{\sqrt{3}-1} = 0.8(\sqrt{3}+1)$ m.
Question 7: The angle of elevation of the top of a building from the foot of the tower is 30° and the angle of elevation of the top of the tower from the foot of the building is 60°. If the tower is 50 m high, find the height of the building.
Solution: Step 1: Tower height $CD=50$. Building height $AB=h$. Distance $BD$. Step 2: In $\Delta CDB$, $\tan 60^\circ = 50/BD \Rightarrow BD = 50/\sqrt{3}$. Step 3: In $\Delta ABD$, $\tan 30^\circ = h/BD \Rightarrow h = BD/\sqrt{3}$. Step 4: $h = (50/\sqrt{3}) / \sqrt{3} = 50/3 = 16 \frac{2}{3}$ m.
Question 8: Two poles of equal heights are standing opposite each other on either side of the road, which is 80 m wide. From a point between them on the road, the angles of elevation of the top of the poles are 60° and 30°, respectively. Find the height of the poles and the distances of the point from the poles.
Solution: Step 1: Let height be $h$. Distances be $x$ and $80-x$. Step 2: $\tan 60^\circ = h/x \Rightarrow h = x\sqrt{3}$. Step 3: $\tan 30^\circ = h/(80-x) \Rightarrow 1/\sqrt{3} = x\sqrt{3}/(80-x)$. Step 4: $80-x = 3x \Rightarrow 4x = 80 \Rightarrow x = 20$. Step 5: Distances: 20 m, 60 m. Height: $20\sqrt{3}$ m.
Question 9: A TV tower stands vertically on a bank of a canal. From a point on the other bank directly opposite the tower, the angle of elevation of the top of the tower is 60°. From another point 20 m away from this point on the line joining this point to the foot of the tower, the angle of elevation of the top of the tower is 30°. Find the height of the tower and the width of the canal.
Solution: Step 1: Let height be $h$ and width be $x$. Step 2: $\tan 60^\circ = h/x \Rightarrow h = x\sqrt{3}$. Step 3: $\tan 30^\circ = h/(x+20) \Rightarrow 1/\sqrt{3} = x\sqrt{3}/(x+20)$. Step 4: $x+20 = 3x \Rightarrow 2x = 20 \Rightarrow x = 10$. Answer: Width = 10 m, Height = $10\sqrt{3}$ m.
Question 10: From the top of a 7 m high building, the angle of elevation of the top of a cable tower is 60° and the angle of depression of its foot is 45°. Determine the height of the tower.
Solution: Step 1: Building $AB=7$. Tower $CD$. Draw horizontal $AE$. Step 2: Angle of depression 45° $\Rightarrow \angle ADE = 45^\circ$. In $\Delta ADE$, $\tan 45^\circ = 7/AE \Rightarrow AE=7$. Step 3: In $\Delta AEC$, $\tan 60^\circ = CE/AE \Rightarrow CE = 7\sqrt{3}$. Step 4: Total height $= CE + ED = 7\sqrt{3} + 7 = 7(\sqrt{3}+1)$ m.