Bihar BSEB Matric Mathematics study materials
Bihar BSEB Matric Mathematics study resources

Checking Two Claims Before Walking to the Tower

Bihar BSEB Matric Mathematics · Put your knowledge to work in context.

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Checking Two Claims Before Walking to the Tower

A 45 m mobile tower stands on level ground in a field beside a highway. A surveyor at point P on the ground measures the angle of elevation of the top of the tower as 30°. To take a close reading at the antenna base he must move straight towards the tower to a point Q where the angle of elevation of the top becomes 60°. He carries a 50 m measuring tape and plans to cover the distance from P to Q in one stretch, measuring it with that tape. A helper claims that walking 60 m from P will land them exactly at Q, while another worker says that at a point 45 m from the foot of the tower the elevation would be exactly 45°. Before setting out, the surveyor wants both claims tested and wants to know how far Q lies from the foot of the tower.

1. Find the distance PQ and decide whether the 50 m tape is long enough.

Read the worked answer

Let A be the foot of the tower, height AB = 45 m, and let AP and AQ be the horizontal distances.

At P: tan 30° = 45/AP, so AP = 45/tan 30° = 45√3 ≈ 45 × 1.732 = 77.94 m. At Q: tan 60° = 45/AQ, so AQ = 45/√3 = 15√3 ≈ 25.98 m.

PQ = AP − AQ = 45√3 − 15√3 = 30√3 ≈ 51.96 m.

The tape is 50 m, which is about 2 m short. The surveyor cannot measure PQ in one stretch with it; he should either measure in two parts (e.g. 50 m plus the remainder) or use the computed value 30√3 m directly.

Sanity check: AQ < AP and PQ + AQ = 77.94 ✔.

2. Test the helper's claim that walking 60 m from P places them exactly at Q.

Read the worked answer

Q is about 51.96 m from P, so 60 m overshoots by about 8 m.

After walking 60 m, the walker is 77.94 − 60 = 17.94 m from the foot of the tower. That is less than AQ ≈ 25.98 m, so the walker is closer to the tower than Q.

Closer to the tower means a larger elevation angle: tan θ = 45/17.94 ≈ 2.51, so θ ≈ 68°, well above 60°.

Therefore the claim is false. Only a walk of about 52 m (exactly 30√3 m) puts the observer at Q; anything beyond that gives an angle greater than 60°.

3. Test the second claim: at a point 45 m from the foot of the tower, is the elevation exactly 45°? Where is that point relative to P and Q?

Read the worked answer

Let R be 45 m from the foot of the tower. Then tan θ = height/distance = 45/45 = 1, so θ = 45° exactly. The claim is correct.

Position check: AR = 45 m lies between AQ ≈ 25.98 m and AP ≈ 77.94 m, so R is between Q and P.

Distance from P: PR = AP − AR = 77.94 − 45 = 32.94 m, which is less than PQ ≈ 51.96 m, as expected since R lies on the way to Q.

Note the pattern: as the observer moves closer, the elevation angle increases (30° at P, 45° at R, 60° at Q), which is a quick way to catch impossible claims.