RowQ
The Vault
RowQ
The Vault
CBSE Class 10 Maths · 11 questions · 26 marks
This is the chapter where trigonometry stops being abstract and starts measuring real towers, kites and ships. Every problem reduces to the same routine: draw the right triangle, mark the angle of elevation or depression at the correct vertex, and pick the ratio that links what you know to what you want. The single most common mistake is placing the angle at the object instead of at the observer's eye, so label the figure before writing any equation.
A vertical pole 15 m high casts a shadow 15√3 m long on level ground. The angle of elevation of the sun at that moment is:
Answer
The angle of elevation is 30°. If θ is the angle of elevation, tan θ = height/shadow length. tan θ = 15/(15√3) = 1/√3. Since tan 30° = 1/√3, we get θ = 30°. Notice the shadow is longer than the pole, which already tells you the angle must be less than 45°.
A kite is flying at a height of 60 m above the ground and its string is taut, making an angle of 30° with the horizontal ground. The length of the string is:
Answer
The string is 120 m long. The string is the hypotenuse and the height is the side opposite the 30° angle, so sin 30° = 60/L. Since sin 30° = 1/2, we get 1/2 = 60/L. Therefore L = 120 m.
A ladder 10 m long leans against a vertical wall and makes an angle of 60° with the ground. The height on the wall reached by the ladder is:
Answer
The ladder reaches 5√3 m up the wall. The ladder is the hypotenuse and the required height is opposite the 60° angle, so sin 60° = h/10. Since sin 60° = √3/2, h = 10 × √3/2 = 5√3 m ≈ 8.66 m. The option 5 m would be the distance of the ladder's foot from the wall, since 10 cos 60° = 5.
From the top of a 50 m high cliff, the angle of depression of a boat at sea is 30°. The distance of the boat from the foot of the cliff is:
Answer
The boat is 50√3 m from the foot of the cliff. The angle of depression from the top equals the angle of elevation of the top from the boat, so that angle is 30° at the boat. tan 30° = 50/d, and tan 30° = 1/√3. So 1/√3 = 50/d, giving d = 50√3 ≈ 86.6 m.
Assertion (A): As an observer walks towards the foot of a vertical tower on level ground, the angle of elevation of the top of the tower increases. Reason (R): For a fixed height h, the angle of elevation θ satisfies tan θ = h/d, and tan θ increases as the horizontal distance d decreases.
Answer
Both A and R are true and R is the correct explanation of A. Reason R states the correct relation tan θ = h/d. With h fixed, reducing d makes the fraction h/d larger, so tan θ increases. For acute angles tan θ increases as θ increases, so a larger value of tan θ means a larger θ. Hence the elevation angle grows as the observer approaches, which is Assertion A, and R is exactly why it happens.
A vertical tower casts a shadow 30 m long on level ground when the angle of elevation of the sun is 60°. Find the height of the tower.
Answer
Let the height of the tower be h metres. tan 60° = h/30, and tan 60° = √3. So h = 30√3 metres. Using √3 ≈ 1.732, h ≈ 51.96 metres.
A girl whose eye level is 1.6 m above the ground stands 45 m away from the foot of a factory chimney. The angle of elevation of the top of the chimney from her eyes is 30°. Find the height of the chimney. (Take √3 = 1.732)
Answer
Let the height of the chimney above her eye level be h metres. In the right triangle formed by her eye, the horizontal line at eye level and the top of the chimney: tan 30° = h/45 1/√3 = h/45, so h = 45/√3 = 45√3/3 = 15√3 metres. Using √3 = 1.732, h = 15 × 1.732 = 25.98 metres. The chimney's total height = h + eye level height = 25.98 + 1.6 = 27.58 metres.
A straight rope is tied from the top of a vertical pole 12 m high to a peg on level ground, and the rope makes an angle of 30° with the ground. Find the length of the rope.
Answer
The rope is the hypotenuse and the pole is opposite the 30° angle. sin 30° = 12/L 1/2 = 12/L, so L = 24 metres. The rope is 24 m long.
A lighthouse stands 90 m above sea level. From its top, the angles of depression of two ships on opposite sides of the lighthouse are 30° and 45°. Find the distance between the two ships. (Take √3 = 1.732)
Answer
Let L be the top of the lighthouse and F its foot, with LF = 90 m. Let the ships be at P and Q on opposite sides of F. The angle of depression of P is 30°, so the angle of elevation of L from P is also 30° (alternate angles). In right triangle LFP: tan 30° = 90/FP. 1/√3 = 90/FP, so FP = 90√3 metres. Similarly, the angle of elevation of L from Q is 45°. In right triangle LFQ: tan 45° = 90/FQ. 1 = 90/FQ, so FQ = 90 metres. Since P and Q are on opposite sides of the lighthouse, the distance between the ships is PQ = FP + FQ = 90√3 + 90 = 90(√3 + 1) metres. Using √3 = 1.732: PQ = 90(2.732) = 245.88 metres. The ships are about 245.88 m apart.
From the top of a building 24 m high, the angle of elevation of the top of a nearby transmission tower is 45° and the angle of depression of its foot is 30°. Find the height of the tower and the horizontal distance between the building and the tower. (Take √3 = 1.732)
Answer
Let AB be the building with A at the top and B at the foot, so AB = 24 m. Let CD be the tower with C at the top and D at the foot, and let the horizontal distance BD = d. Draw AE horizontal from A to meet the tower at E, so AE = d and ED = AB = 24 m. Step 1 — Use the angle of depression of the foot D, which equals the angle of elevation of A from D, i.e. 30°. tan 30° = 24/d 1/√3 = 24/d, so d = 24√3 metres ≈ 41.57 m. Step 2 — Use the angle of elevation of the top C from A, which is 45°. tan 45° = CE/AE = CE/d 1 = CE/(24√3), so CE = 24√3 metres. Step 3 — Total height of the tower: CD = CE + ED = 24√3 + 24 = 24(√3 + 1) metres. Using √3 = 1.732: CD = 24(2.732) = 65.57 metres. So the tower is about 65.57 m tall and stands about 41.57 m away from the building.
Read the following and answer the questions that follow: An adventure park is installing a straight zip-line wire that runs from the top of a vertical tower down to a landing point on level ground. The tower is 30 m high and the wire is planned to make an angle of 30° with the ground. (a) Find the length of wire needed for this design. (b) How far from the foot of the tower will the landing point be? (Take √3 = 1.732) (c) The designers consider a steeper wire making 45° with the ground instead. What length of wire would that need? (Take √2 = 1.414) (d) Which design uses less wire, and by how much?
Answer
(a) The wire is the hypotenuse and the tower is opposite the 30° angle. sin 30° = 30/L, and sin 30° = 1/2, so L = 60 metres. (b) Let the horizontal distance be d. tan 30° = 30/d, and tan 30° = 1/√3, so d = 30√3 metres. Using √3 = 1.732, d = 30 × 1.732 = 51.96 metres. (c) With a 45° wire: sin 45° = 30/L', and sin 45° = 1/√2, so L' = 30√2 metres. Using √2 = 1.414, L' = 30 × 1.414 = 42.42 metres. (d) The 45° design uses less wire. Saving = 60 - 42.42 = 17.58 metres of wire.
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