Q) Two poles, 30 feet and 50 feet tall, are 40 feet apart and perpendicular to the ground. The poles are supported by wires attached from the top of each pole to the bottom of the other, as in the figure. A coupling is placed at C where the two wires cross. 

i.   What is the horizontal distance from C to the taller pole?
ii.  How high above the ground is the coupling?
iii. How far down the wire from the smaller pole is the coupling?
iv. Find the length of line joining the top of the two poles.

Ans:

i. Horizontal distance from C to the taller pole:

Step 1: Let’s start with placing the markings in the given diagram for the question:

Here, AB is the smaller tower of 30 ft height and PQ is the taller tower of 50 ft height. Distance BQ is given as 40 ft and we need to find the distance CD i.e. horizontal distance of coupling C from PQ

Step 2: Next, let’s compare Δ ABQ and Δ CEQ:

∠ ABQ = ∠ CEQ         ( ∵ both angles are 90⁰)

∠ AQB = ∠ CQE         (common angle)

Therefore, by AA similarity rule,

Δ ABQ ~ Δ CEQ

Step 3: ∴

by substituting the given values, we get:

∴ 30 X = 40 a

∴ 3 X = 4 a …………. (i)

Step 4: Similarly, let’s compare Δ PBQ and Δ CEB:

∠ PQB = ∠ CEB

( ∵ both angles are 90⁰)

∠ ABQ = ∠ CBE                (common angle)

Therefore, by AA similarity rule,

Δ PBQ ~ Δ CEB

Step 5: ∴

∴

by substituting the given values, we get:

∴ 50 (40 – X) = 40 a

∴ 5 (40 – X) = 4 a ………….. (ii)

Step 6: By comparing equations (i) and (ii) we get:

3 X = 5 (40 – X)

∴ 3 X = 200 – 5 X

∴ 8 X = 200

∴ X = = 25

Therefore, horizontal distance from C to the taller pole is 25 ft

ii. Height of the coupling from the ground:

Here we need to find the length a

∴ by putting value of X from above step 6 in equation (i), we get:

∵ 3 X = 4 a

∴ 3 (25) = 4 a             (∵ X = 25 from step 6)

∴ a = = 18.75

Therefore, the height of the coupling from the ground is 18.75 feet.

iii. Distance of coupling along the wire of smaller pole:

Step 7: Let’s look in the Δ AQB:

since Δ AQB is a right angle triangle,

∴ by Pythagoras theorem:

AQ ² = AB ² + BQ ²

∴ AQ ² = (30) ² + (40) ²  = 900 + 1600

∴ AQ ² = 2500 = (50) ²

∴ AQ = 50 ft

Step 8: Let’s mark the distance CA as Y

∴ QC = QA – CA = 50 -Y

Step 9: ∵ Δ ABQ ~ Δ CEQ     (from step 2)

∴

∴

( ∵ QE = CD = X)

∴ 50 X =  40 (50 – Y)

∴ 5 X =  4 (50 – Y)

By putting values of X from step 6 above, we get:

∴ 5 (25) = 4 (50 – Y)

∴ 125  = 200 – 4 Y

∴ 4 Y = 200 – 125 = 75

∴  Y = = 18.75

Therefore, the distance of coupling along the wire of smaller pole is 18.75 ft. 

iv. Length of line joining the top of the two poles:

Step 10: Let’s connect the tops of two poles A and P and draw a line AF parallel to ground

Now, Since Δ APF is a right angle triangle,

∴ by Pythagoras theorem:

AP ² = AF ² + PF ²

∴ AP ² = AF ² + (PQ – FQ) ²    (∵ PF = PQ – FQ)

∴ AP ² = BQ ² + (PQ – AB) ²    (∵ FQ = AB and AF = BQ)

∴ AP ² = (40) ² + (50 – 30) ²  = 1600 + 400

∴ AP ² = 2000 = 20√5  or 44.72

Therefore, the length of line joining the top of the two poles is 44.72 ft.

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