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Difficulty: MediumLaw of Sines and Law of Cosines

A civil engineer is designing a triangular bridge support structure with vertices PP, QQ, and RR. The support beam PQPQ is 8080 feet long, the beam QRQR is 5050 feet long, and the interior angle ∠PQR\angle PQR measures 60∘60^\circ. What is the length, in feet, of the support beam PRPR?

Answer: 70 feet

Answer

The length of the support beam PRPR is 7070 feet.
Using the Law of Cosines formula c2=a2+b2−2abcos⁡(C)c^2 = a^2 + b^2 - 2ab\cos(C) with side lengths 8080 and 5050 and included angle 60∘60^\circ, we calculate PR2=802+502−2(80)(50)cos⁡(60∘)=6400+2500−4000=4900PR^2 = 80^2 + 50^2 - 2(80)(50)\cos(60^\circ) = 6400 + 2500 - 4000 = 4900. Taking the square root gives PR=70PR = 70 feet.

Step-by-Step Solution

1
Identify the given dimensions and included angle
Side PQ=80PQ = 80 ft, side QR=50QR = 50 ft, and included angle ∠PQR=60∘\angle PQR = 60^\circ.
The Law of Cosines applies directly when two side lengths and the included angle (SAS) are known.
2
Set up the Law of Cosines equation for the unknown side PRPR
PR2=PQ2+QR2−2(PQ)(QR)cos⁡(∠PQR)PR^2 = PQ^2 + QR^2 - 2(PQ)(QR)\cos(\angle PQR)
This formula generalizes the Pythagorean theorem to non-right triangles.
3
Substitute the known values into the equation and evaluate
PR2=802+502−2(80)(50)cos⁡(60∘)=6400+2500−4000=4900PR^2 = 80^2 + 50^2 - 2(80)(50)\cos(60^\circ) = 6400 + 2500 - 4000 = 4900
Evaluating squares and using cos⁡(60∘)=0.5\cos(60^\circ) = 0.5 simplifies the calculation.
4
Take the positive square root to find PRPR
PR=4900=70PR = \sqrt{4900} = 70
Side length must be a positive real number.

Key Concept

Applying the Law of Cosines to solve for an unknown side in a Side-Angle-Side (SAS) triangle.
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Topics

Law of Sines and Law of Cosines

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