EPFO APFC 2023 General Studies · Question 39 of 120

Correct — C, (c) 13·65. Set up the two right triangles and name the unknown you are actually asked for. Let the tower stand at a horizontal distance d from the building, let the part of the tower above the level of the building's top be x, and remember that the building is 10 metres tall, so the whole tower is x + 10. From the bottom of the building the angle of elevation of the tower's top is 45°, and the horizontal leg of that triangle is d while the vertical leg is the full height of the tower. So tan 45° = (x + 10)/d, and since tan 45° = 1, the distance d equals x + 10. That is the geometric gift in this question: a 45° elevation makes the horizontal distance equal to the height, so one unknown disappears immediately. From the top of the building the angle of elevation is 30°, measured along a horizontal line at the level of the roof. The horizontal leg is still d and the vertical leg is now only x, the part of the tower above the roof. So tan 30° = x/d, that is x/d = 1/√3, giving d = x√3. Equate the two expressions for d: x + 10 = x√3, so x(√3 − 1) = 10 and x = 10/(√3 − 1). Rationalise before substituting — this is the step that decides the mark. Multiply numerator and denominator by (√3 + 1): x = 10(√3 + 1)/((√3)² − 1) = 10(√3 + 1)/2 = 5(√3 + 1). Now put in the value the paper supplies, √3 = 1·73: x = 5 × 2·73 = 13·65 metres exactly. That is option (c), and it is exact rather than approximate under the given value of the surd. If instead you substitute first and divide 10 by 0·73, you get 13·698…, which is not one of the printed values and leaves you choosing by eye between 13·65 and 13·41. Rationalising turns a division by an awkward decimal into a single multiplication, and on this question it is the difference between an exact match and a guess. As a check, the whole tower is x + 10 = 23·65 metres and the horizontal distance is the same 23·65 metres, which is consistent with the 45° sighting from the ground.

• (a)10·65 — Since the answer is exactly 5(√3 + 1), each wrong option corresponds to a wrong value of the surd, and 10·65 would require √3 to be about 1·13 — nowhere near the 1·73 the question itself supplies. The option can also be killed by working backwards through the geometry, which is a good habit on any height-and-distance item: if the part of the tower above the roof were 10·65 metres, the whole tower would be 20·65 metres, the horizontal distance would be 20·65 metres too because of the 45° sighting, and the elevation from the roof would satisfy tan θ = 10·65/20·65 = 0·516, giving an angle of about 27°, not the 30° the question states.
• (b)13·41 — 13·41 is the nearest miss in the set and the one that punishes an approximate method. It corresponds to taking √3 as roughly 1·68, and it sits close enough to the true value that a candidate who divides 10 by 0·73 in a hurry, or who rounds an intermediate step, may talk himself into it. Check it against the geometry and it fails cleanly: a part-height of 13·41 metres gives a whole tower of 23·41 metres, an equal horizontal distance, and an elevation from the roof of tan θ = 13·41/23·41 = 0·573, which is about 29·8° rather than 30°. The lesson is that on a question offering two values within a quarter of a metre of each other, the calculation must be carried exactly and the approximation applied only at the last step.
• (d)15·41 — 15·41 would require √3 to be about 2·08, which is larger than 2 and therefore impossible, since 2² = 4 exceeds 3. It is the largest value on offer and appeals to a candidate who has muddled the two tangents — using tan 30° = √3 rather than 1/√3, or measuring the 30° elevation against the full height of the tower instead of the part above the roof. The check works here too: 15·41 above the roof means a tower of 25·41 metres, an equal horizontal distance, and tan θ = 15·41/25·41 = 0·607, an elevation of about 31·2°. Any of the three wrong options can be eliminated in fifteen seconds by this reverse test if the forward calculation goes astray.

Height-and-distance problems are built from right triangles that share a horizontal leg, and almost all of them are solved by writing the tangent of each given angle and eliminating that shared leg. Three standard values do nearly all the work: tan 30° = 1/√3, tan 45° = 1 and tan 60° = √3. The 45° case deserves special attention because it collapses a triangle — the height equals the horizontal distance — and an examiner who prints 45° is usually handing you that simplification. When an observer stands on top of a building and sights the top of a taller object, the angle is measured from the horizontal at the observer's own level, so the vertical leg of that triangle is the difference of the two heights, not the full height of the object; getting this right is the single most important modelling decision in the whole family of problems. Two further habits matter for accuracy. First, rationalise any surd in a denominator before substituting a decimal approximation: 10/(√3 − 1) becomes 5(√3 + 1), which turns a division by 0·73 into a multiplication by 2·73. Second, verify by reversing — take your answer, rebuild the triangle and compute the tangent, which should reproduce the angle given in the question. That reverse test is quick enough to run on every numerical geometry item.

The quantitative block of this paper mixes proof-style statement items with straightforward computation, and this is one of the computations. What makes it worth more than its 2·5 marks is that it is entirely self-contained: no recalled fact is needed beyond three tangent values, and the paper even supplies the value of √3 to be used. The options are deliberately close together — three of the four lie within five metres of one another and two within a quarter of a metre — which means that estimation is worthless and the answer must be produced exactly. Note the printed convention as well: this booklet prints its decimal points as a raised middle dot, so the question's given value appears as 1·73 and the options as 10·65, 13·41, 13·65 and 15·41. That is the Commission's typography, not an error, and it is reproduced here as printed. The item carries no diagram, so drawing the two triangles yourself, marking which vertical leg belongs to which angle, is part of the work rather than an optional extra.

• With the tower at horizontal distance d, a 45° elevation from the foot of the building gives tan 45° = 1, so the horizontal distance equals the full height of the tower — the simplification that removes one unknown from this problem.
• A 30° elevation from the top of the building gives tan 30° = 1/√3 measured against only the part of the tower above the roof, so that part x satisfies d = x√3.
• Equating the two expressions for d gives x(√3 − 1) = 10, so x = 10/(√3 − 1) = 5(√3 + 1), which with the given value √3 = 1·73 is exactly 13·65 metres.
• The whole tower is therefore 23·65 metres tall and stands 23·65 metres away; the general result for a building of height h with angles of 30° from the top and 45° from the bottom is x = h(√3 + 1)/2 and total height h(3 + √3)/2.
• Rationalising a surd denominator before substituting the decimal approximation avoids the rounding drift that comes from dividing by 0·73, and a reverse check — rebuild the triangle from your answer and confirm that the tangent gives 30° — settles any doubt between two close options.

• Measuring the 30° elevation against the whole height of the tower rather than against the part above the building's roof; the vertical leg of that triangle is the difference of the two heights
• Answering the wrong quantity: the question asks for the height of the tower above the top of the building, not the total height of the tower, and 23·65 metres is the total
• Substituting √3 = 1·73 before rationalising, which turns the calculation into a division by 0·73 and produces a value that matches none of the printed options exactly
• Estimating rather than computing when two options differ by a quarter of a metre; the reverse check — rebuild the triangle and take the tangent — is the fastest way to separate them

Heights and distances appear in this paper as a single computation with the required surd value supplied in the stem, and the configurations are drawn from a small standard set. Expect one angle of 45° so that a distance can be eliminated at once, expect the second observation point to be raised above the ground so that the difference of heights is in play, and expect the options to be clustered so that only an exact calculation discriminates. Practise the two moves this item rewards — rationalise before substituting, and verify by rebuilding the triangle — and the whole family becomes routine.

No directly related past PYQ was found.

• practice — not a real PYQ

  The angle of elevation of the top of a tower from a point on the ground is 30°. On walking 40 metres towards the tower along level ground the angle of elevation becomes 60°. What is the height of the tower in metres (given that √3 = 1·73)?

  • (a)20·00
  • (b)27·68
  • (c)34·60
  • (d)40·00

  Answer(c) 34·60 — if the height is h, the first point is at h√3 from the foot because tan 30° = 1/√3, and the second is at h/√3 because tan 60° = √3. The difference of the two distances is the 40 metres walked, so h√3 − h/√3 = 40, which gives 2h/√3 = 40 and h = 20√3 = 20 × 1·73 = 34·60 metres. As always, eliminate the shared horizontal leg first and substitute the decimal value only at the end.

• practice — not a real PYQ

  The top of a tower is seen at an angle of elevation of 45° from the foot of a building and at 30° from its top. The horizontal distance between the building and the tower is equal to

  • (a)the height of the building
  • (b)the full height of the tower
  • (c)half the height of the tower
  • (d)the difference between the heights of the tower and the building

  Answer(b) the full height of the tower — the sighting from the foot of the building is at 45°, and tan 45° = 1, so the vertical leg of that triangle equals its horizontal leg. The vertical leg there is the whole height of the tower measured from the ground, so the horizontal distance equals the tower's full height. The 30° sighting from the roof uses the same horizontal distance but a shorter vertical leg, namely the part of the tower above the roof.