/20 1234567891011121314151617181920 Test Code: MC9L2 1 / 20 1. Tower seen at 30° (initially), then at 60° (after 6s). Find time car will take to reach tower. A) 15 sec B) 12 sec C) 10 sec D) 6 + 6√3 sec 📝 Solution: Use tan(angles) to find both distances Time = distance / speed ⇒ add both for final time 2 / 20 2. For a slide of height 1.5 m and angle 30°, what is its length? A) √3 m B) 2 m C) 3 m D) 1.5 m 📝 Solution: In triangle ABC ∆ (AB)/(AC) = sin 30⁰ 1.5/(AC) = 1/2 AC = 3m 3 / 20 3. A kite is flying at 60 m height, the string makes 60° with ground. Find string length. A) 120 m B) 120 / √3 m C) 60 m D) 80 m 📝 Solution: sin(60°) = 60 / string ⇒ √3/2 = 60 / x ⇒ x = 120 / √3 4 / 20 4. Two poles stand 80 m apart. Angles from a point are 60° and 30°. Find pole height and distances. A) 40 m and 40 m B) 20 m and 60 m C) 30 m and 50 m D) 25 m and 55 m 📝 Solution: Let AB and CD be the poles and O is the point from where the elevation angles are measured. In ΔABO, DO = BD – BO = (80 – 20) m = 60 m Therefore, the height of poles is and the point is 20 m and 60 m far from these poles. 5 / 20 5. A tree breaks and its top touches the ground making a 30° angle. The horizontal distance is 8 m. What is the height of the tree? A) 10 m B) 12 m C) 8 m D) 14 m 📝 Solution: Let broken part = AB, ground = BC, original height = AB + AC Use cos(30°) = 8 / AB ⇒ AB = 8 / cos(30°) = 8 / (√3/2) = 16 / √3 Vertical part = AB × sin(30°) = (16 / √3) × 1/2 = 8 / √3 Total height = 8 / √3 + 8 ≈ 12 m 6 / 20 6. A tower is 30 m away from a point and appears at 30° elevation. Find its height. A) 10√3 m B) 20√3 m C) 15 m D) 30 m 📝 Solution: 7 / 20 7. From 7 m tall building, angle of elevation to tower top = 60°, depression to base = 45°. Find tower height. A) 10 B) 7 + 7√3 C) 7√3 m D) 14 m 📝 Solution: Lower triangle gives base = 7 Upper triangle gives height = 7 + x ⇒ x = 7√3 ⇒ total height = 7 + 7√3 8 / 20 8. A 1.6 m statue is on pedestal. Angles of elevation: 45° (pedestal), 60° (statue). Find pedestal height. A) 0.8 m B) 1.2 m C) 0.6 m D) 1.6 m 📝 Solution: Let pedestal height = x tan(45°) = x / d ⇒ x = d tan(60°) = (x + 1.6) / d (x + 1.6) = √3x ⇒ x = 0.8 m 9 / 20 9. A tower is seen from 4 m and 9 m away at complementary angles. Prove height is 6 m. A) Proven B) 4 m C) 9 m D) Not possible 📝 Solution: Use tan(θ) = h/4, tan(90–θ) = h/9 ⇒ tan(θ)tan(90–θ) = h²/36 = 1 ⇒ h² = 36 ⇒ h = 6 m 10 / 20 10. A bridge across a river makes an angle of 45° with the river bank. if the length of the bridge across the river is 200m, then the breadth of the river is A) 200 m B) 200√2 m C) 100√2 m D) 100 m 📝 Solution: 11 / 20 11. A 1.2 m tall girl spots a balloon moving with the wind in a horizontal line at a height of 88.2 m from the ground. The angle of elevation of the balloon from the eyes of the girl at any instant is 60°. After some time, the angle of elevation reduces to 30°. Find the distance travelled by the balloon during the interval. A) 55 m B) 57 C) 60 m D) 58√3 m 📝 Solution: 12 / 20 12. A 20 m building has a tower on top. Angles of elevation from same point are 45° (building base), 60° (tower top). Find tower height. A) 10 m B) 20(√3 – 1) m C) 20 m D) 15 m 📝 Solution: Let tower height = h From building base: tan(45°) = 20 / x ⇒ x = 20 From tower top: tan(60°) = (20 + h) / 20 ⇒ h = 20(√3 – 1) 13 / 20 13. 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. A) 7 m B) 7 (3 + 1) m C) 7 (√3 + 1) m D) 8 m 📝 Solution: 14 / 20 14. A 75 m lighthouse sees two ships at depression angles 30° and 45°. Find ship distance. A) 90 m B) 75 m C) 75(√3 – 1) m D) 50 m 📝 Solution: x = 75 / tan(30°), y = 75 / tan(45°) ⇒ x – y = 75(√3 – 1) 15 / 20 15. Tower is 50 m. Angles from building and tower foot are 30° and 60°. Find building height. A) 50(1 – 1/√3) m B) 50 / 3 m C) 30 m D) 25 m 📝 Solution: Let building height = h tan(60°) = 50 / x ⇒ x = 50 / √3 tan(30°) = h / x ⇒ h = x / √3 ⇒ h = 50 / √3 × √3 = 50 / 3 16 / 20 16. TV tower on a river bank seen at 60° from one point, and 30° from 20 m behind. Find height and width.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. A) height of the tower 13 m and width is 10 B) height of the tower 10/√3 m and width is 10 C) height of the tower 10 m and width is 20 D) height of the tower 15 m and width is 30 📝 Solution: Therefore, the height of the tower is 10√3 m and the width of the canal is 10 m. 17 / 20 17. A boy 1.5 m tall sees top of 30 m building: elevation angle changes from 30° to 60°. How far did he walk? A) 10 m B) 15 m C) 28.5(√3 – 1) m D) 28.5(1 – √3) m 📝 Solution: Height from eye = 28.5 m Use tan in both triangles and subtract distances: x = 28.5 / tan(30°), y = 28.5 / tan(60°) distance = x – y = 28.5(√3 – 1) 18 / 20 18. As observed from the top of a 75 m high lighthouse from the sea-level, the angles of depression of two ships are 30° and 45°. If one ship is exactly behind the other on the same side of the lighthouse, find the distance between the two ships. A) 75 B) 75 (√3 – 1) m C) 76 D) 75 (√2) m 📝 Solution: 19 / 20 19. A circus artist climbs a 20 m long rope inclined at 30° to the ground. What is the height of the vertical pole? A) 5 m B) 17.3 m C) 15 m D) 10 m 📝 Solution: Let the pole be vertical (AB) and the rope is hypotenuse AC = 20 m.Using trigonometry:sin(30°) = AB / AC ⇒ 1/2 = AB / 20 ⇒ AB = 10 m 20 / 20 20. A 60° inclined slide is 3 m high. Find its length. A) 3.5 m B) 4 m C) √6 m D) 6 m 📝 Solution: sin(60°) = 3 / hyp ⇒ √3/2 = 3 / hyp ⇒ hyp = 6 / √3 = 6 m Send feedback
