Ambiguous Cable Layout for a Sign Gantry
A sign gantry is to be anchored to a horizontal beam at points A and B, 9.0 m apart. A single cable runs from A to the top P of a vertical mast; its length is fixed at AP = 7.0 m. The angle between BA and BP at B is measured as 38.0°. The mast must provide at least 5.0 m clearance above the beam for delivery trucks; the fabricator also needs the length of the second cable BP and the area of triangle ABP to order material. Because the known data are two sides and a non-included angle, the shop drawing is not unique. The site engineer must decide which of the two possible configurations satisfies the clearance, confirm it with a law-of-sines calculation, and report the corresponding angle at A and the length BP. A second fabricator claims the 38.0° measurement forces a unique triangle, so no check is needed.
1. Show that the given data allow two triangles and determine which one satisfies the 5.0 m clearance.
Read the worked answer
In triangle ABP, the side opposite angle B is AP = 7.0 m, and the side opposite angle P is AB = 9.0 m. By the law of sines,
sin P / 9.0 = sin 38.0° / 7.0, so sin P = 9.0 sin 38.0° / 7.0 ≈ 0.7916.
Thus P ≈ 52.3° or P ≈ 180° − 52.3° = 127.7°. Both are admissible because each leaves a positive angle at A:
- If P ≈ 52.3°, then A = 180° − 38.0° − 52.3° ≈ 89.7°.
- If P ≈ 127.7°, then A ≈ 14.3°.
The mast height above the beam is h = AP sin A. The first case gives h = 7.0 sin 89.7° ≈ 7.0 m; the second gives h = 7.0 sin 14.3° ≈ 1.73 m. Only the first configuration clears the required 5.0 m, so the second is rejected even though it is geometrically valid.
2. For the selected configuration, compute BP and the area of triangle ABP.
Read the worked answer
Using A ≈ 89.7° and the law of sines with BP opposite angle A:
BP / sin A = AP / sin B = 7.0 / sin 38.0°, BP = 7.0 sin 89.7° / sin 38.0° ≈ 7.0(0.99999)/0.61566 ≈ 11.4 m.
The area is
Area = ½(AB)(AP) sin A = ½(9.0)(7.0) sin 89.7° ≈ 31.5 m².
As a check, using sides AB and BP with included angle B gives ½(9.0)(11.37) sin 38.0° ≈ 31.5 m², which agrees. The fabricator should order about 11.4 m of second cable and allow for roughly 31.5 m² of triangular covering.
3. The angle at B is measured with a tolerance of ±1.0°. Does the clearance decision remain safe over that tolerance?
Read the worked answer
Check the extreme values. At B = 37.0°, sin P = 9.0 sin 37.0°/7.0 ≈ 0.7738, so P ≈ 50.7° or 129.3°. The corresponding angles at A are about 92.3° and 13.7°, giving heights 7.0 sin 92.3° ≈ 6.99 m and 7.0 sin 13.7° ≈ 1.66 m. At B = 39.0°, sin P ≈ 0.8091, so P ≈ 54.0° or 126.0°, giving A ≈ 87.0° or 15.0° and heights about 6.99 m and 1.81 m. Throughout the tolerance interval, one configuration remains near 7.0 m and the other stays below 2.0 m. The 5.0 m clearance decision is therefore robust, but the shop should still label the selected triangle explicitly because the drawing is not unique from the raw data.