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Word problems relating guy wire in trigonometry

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Guy Wire Word Problems in Trigonometry

A guy wire anchoring a pole to the ground forms a right triangle, with the pole as the opposite side, the ground distance as the adjacent side, and the wire as the hypotenuse. Learn to set up the triangle and solve for wire length or anchor distance using sine and tangent, with a worked example.

Setting up a guy wire word problem

A guy wire anchors a tall pole to the ground at an angle, forming a right triangle: the pole is the vertical leg (opposite side), the ground distance to the anchor is the adjacent side, and the wire itself is the hypotenuse.

Guy wire word problem A vertical pole 8 meters tall is supported by a guy wire anchored to the ground, making a 50 degree angle with the ground. Using sine, the wire length is about 10.44 meters. Using tangent, the anchor point is about 6.71 meters from the base of the pole. 50° pole = 8 m ≈ 6.71 m wire ≈ 10.44 m
An 8-meter pole braced by a guy wire at 50° needs a wire about 10.44 m long, anchored about 6.71 m away.

Worked example

A guy wire supports an 8-meter pole, making a 50° angle with the ground. Find the wire's length and the distance from the pole to the anchor point.

Wire length (SOH): sin 50° = 8 / wire, so wire = 8 / sin 50° ≈ 10.44 m.

Anchor distance (TOA): tan 50° = 8 / distance, so distance = 8 / tan 50° ≈ 6.71 m.

The same setup, different object

Notice this is structurally identical to a ladder problem — a vertical or near-vertical object, an angle to the ground, and a diagonal connector. Once you recognize the triangle, the choice of ratio depends only on which two sides (or side and angle) are already known, the same reasoning used across all trigonometry word problems.

Multiple wires

A pole often has more than one guy wire at different angles. Each wire forms its own right triangle with the same pole height, so the same three ratios apply — just recompute for each wire's own angle.

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