Some Applications of Trigonometry · Class 10 Mathematics: 46 flashcards

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1. What is the line of sight?

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The straight line from the observer's eye to the object; it is the hypotenuse of the right triangle.

2. Define the angle of elevation.

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The angle between the horizontal through the eye and the line of sight, when the object is above the observer.

3. Define the angle of depression.

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The angle between the horizontal through the eye and the line of sight, when the object is below the observer.

4. From what line are both angles measured?

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The horizontal through the observer's eye — never the vertical.

5. Why does the angle of depression equal the angle of elevation?

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The two horizontals are parallel and the line of sight is a transversal, so the angles are alternate interior angles.

6. Which ratio links the height and the ground distance?

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Tangent: $\tan\theta = \frac{\text{height}}{\text{distance}}$.

7. Which ratio involves the line of sight and the height?

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Sine, because the line of sight is the hypotenuse.

8. Which ratio involves the line of sight and the ground distance?

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Cosine.

9. What is special about an elevation of 45°?

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The height equals the horizontal distance.

10. At 30°, how does the line of sight compare with the height?

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It is exactly twice the height, since $\sin 30° = \frac12$.

11. A tower is 30 m from an observer, elevation 30°. Find its height.

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$h = 30\tan 30° = \frac{30}{\sqrt3} = 10\sqrt3 \approx 17.3$ m.

12. A tower is 20 m tall. At what distance is its elevation 45°?

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20 m — at 45° the two legs are equal.

13. A kite is 60 m high and its string makes 60° with the ground. How long is the string?

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$\ell = \frac{60}{\sin 60°} = \frac{120}{\sqrt3} = 40\sqrt3$ m.

14. Why must you draw a dashed horizontal for an angle of depression?

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Because there is no ground at eye level for the angle to be measured from; without it the angle is taken from the vertical by mistake.

15. An angle is taken at an eye height of 1.5 m. What must you do at the end?

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Add 1.5 m, because the triangle only gives the height above eye level.

16. Is $\frac{40}{\sqrt3}$ a finished answer?

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No. Rationalise it to $\frac{40\sqrt3}{3}$.

17. Which angles does this chapter allow?

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30°, 45° and 60° only.

18. How many right triangles may one problem use?

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At most two.

19. A 40 m lighthouse has a boat at a depression of 45°. How far out is the boat?

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40 m, since the elevation at the boat is also 45° and $\tan 45° = 1$.

20. What is the first thing to write in a heights-and-distances answer?

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A labelled diagram.

21. If the sight line is not mentioned, which ratio is almost always needed?

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Tangent.

22. A ladder at 60° to the ground reaches 6 m up a wall. How long is the ladder?

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$\frac{6}{\sin 60°} = \frac{12}{\sqrt3} = 4\sqrt3$ m.

23. Height 15 m, elevation 60°. Find the ground distance.

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$\frac{15}{\sqrt3} = 5\sqrt3$ m.

24. Does the angle of elevation halve when you double your distance?

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No — angles do not scale like lengths.

25. Why can a single angle of elevation not give the height of a tower?

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It gives one equation with two unknowns — the height and the distance.

26. What do the two right triangles in these problems always share?

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The vertical side, which is the height.

27. Which point gets the larger angle of elevation?

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The nearer one.

28. Two observers on the same side, gap $g$. How are the distances related?

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They are $d$ and $d + g$ — they differ by the gap.

29. A tower between two points $g$ apart. How are the distances related?

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They are $x$ and $g - x$ — they add to the gap.

30. How do you find the height of a flagstaff on a tower?

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Find the tower-plus-flagstaff height and the tower height from the same point, then subtract.

31. Angles 60° and 30° from two points 20 m apart on the same side. Find the height.

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$d\sqrt3 = \frac{d+20}{\sqrt3}$ gives $d = 10$, so $h = 10\sqrt3 \approx 17.3$ m.

32. A tower between two points 80 m apart, angles 60° and 30°. Find the height.

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$x\sqrt3 = \frac{80-x}{\sqrt3}$ gives $x = 20$, so $h = 20\sqrt3 \approx 34.6$ m.

33. From 20 m away, the tower is at 45° and the flagstaff top at 60°. How tall is the flagstaff?

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$20\sqrt3 - 20 = 20(\sqrt3 - 1) \approx 14.6$ m.

34. A man walks 40 m towards a tower and the elevation rises from 30° to 45°. Find the height.

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$h = \frac{40}{\sqrt3 - 1} = 20(\sqrt3 + 1) \approx 54.6$ m.

35. How do you rationalise $\frac{40}{\sqrt3 - 1}$?

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Multiply above and below by the conjugate $\sqrt3 + 1$, giving $\frac{40(\sqrt3+1)}{2} = 20(\sqrt3+1)$.

36. What happens if both given angles are equal?

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The two equations become identical, so nothing can be found — the problem is unsolvable.

37. Should you substitute the standard values before or after eliminating the height?

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After. Arranging the algebra first keeps the working clean.

38. Why draw one figure rather than two?

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Because a single figure shows how the two distances are related through the gap.

39. What check confirms you assigned the angles correctly?

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The larger angle must have come out with the smaller distance.

40. In a flagstaff problem, what do the two triangles share?

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The same observer and the same horizontal distance.

41. How many right triangles may a syllabus problem use?

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At most two.

42. Which angles are permitted?

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30°, 45° and 60° only.

43. Angles 45° and 30° from two points 40 m apart on the same side. Find the height.

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$d = \frac{40}{\sqrt3-1} = 20(\sqrt3+1)$, and $h = d$ since $\tan 45° = 1$, so $h = 20\sqrt3 + 20 \approx 54.6$ m.

44. Two observers on opposite sides of a 60 m road see a tower between them at 45° each. Where is the tower?

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Exactly halfway, 30 m from each, and the height is 30 m.

45. What is the last thing to do in one of these solutions?

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Rationalise, state the unit, and check the nearer distance is the smaller one.

46. A moving observer problem is really which type?

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Two points on the same side — the two positions are the two points.

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