Two elevation readings and a flagpole measured from the wrong point
A school in Coimbatore is installing a vertical flagpole on level ground. From a point A, 12 m from the foot F of the pole, the angle of elevation of the top T is 45°. From a point B on the same straight line beyond A, the angle of elevation of T is 30°. The contractor uses the 45° reading to conclude that the pole is 12 m tall. The head teacher points out that the two readings must describe the same pole and asks whether the measured distance AB is consistent with that height. A student suggests using the 30° triangle directly with AB=12 m, which gives a different height. The school also wants the distance from B to the top of the pole. All ground is level and the pole is vertical.
1. Using the 45° reading at A, find the height h of the pole. Then use the 30° reading at B to find BF and AB. Is AB=12 m consistent?
Read the worked answer
In right triangle AFT, tan 45° = h/AF = h/12, so h = 12 m. In right triangle BFT, tan 30° = h/BF, so 1/√3 = 12/BF, giving BF = 12√3 m. Since B lies beyond A on the same line, BF = BA + AF, so AB = BF - AF = 12√3 - 12 = 12(√3 - 1) ≈ 8.78 m. This is not 12 m. Therefore the readings h=12 m, angle at B=30°, and AB=12 m are not all consistent; at least one measurement is wrong.
2. If h=12 m and AB=12 m are both correct, what should the angle of elevation at B be? Compare with the measured 30°.
Read the worked answer
With h=12 m and AF=12 m, the point B is 12 m beyond A, so BF = AF + AB = 12 + 12 = 24 m. In right triangle BFT, tan(angle at B) = h/BF = 12/24 = 1/2. Thus the angle at B should be tan^-1(1/2) ≈ 26.57°, not 30°. The measured 30° is too large by about 3.43°. So if the height and the distance AB are trusted, the 30° reading is the inconsistent one.
3. Suppose instead that AB=12 m and the 30° reading are correct, while the 45° reading is wrong. Using the fact that the pole is vertical, find h and then BT.
Read the worked answer
Let AF = x. Since the 45° reading is now not trusted, h is not equal to AF automatically. From the 30° triangle, tan 30° = h/BF = h/(x+12), so h = (x+12)/√3. We need one more relation. If the 45° reading were correct we would have x=h; but we are told it is wrong, so that relation cannot be used. Instead, the problem as stated has two unknowns h and x with only one equation, so it is underdetermined. The student's earlier suggestion of putting AB=12 directly as the adjacent side in the 30° triangle is the real error: the adjacent side is BF, not AB. Without a reliable second observation or distance, h cannot be determined uniquely. If the 45° reading is retained as h=AF, then h=6(√3+1)≈16.39 m and BT=2h≈32.78 m; but that uses both readings, not only the 30° one.