asked 183k views
5 votes
In ∆PQR, p = 8, r, = 11, and < Q = 55o. Solve the triangle.?

asked
User Xuyanjun
by
8.5k points

1 Answer

1 vote

Answer:

  • q = 9.2
  • P = 45.6°
  • R = 79.4°

Explanation:

The two given sides are either side of the given angle. This provides insufficient information for using the Law of Sines, which needs an angle and a side opposite. However, it is exactly what is needed to make use of the Law of Cosines.

Thus, the Law of Cosines can be used to find side length q.

q² = p² +r² -2pr·cos(Q)

q² = 8² +11² -2·8·11·cos(55°) ≈ 64 +121 -100.949 ≈ 84.051

q ≈ √(84.051) ≈ 9.1679 ≈ 9.2

Then the Law of Sines can be used to find angle P. We choose to find the smaller angle in order to avoid the ambiguity that can come with trying to determine if the larger angle is acute or obtuse.

sin(R)/r = sin(Q)/q

P = arcsin(p/q·sin(Q)) ≈ arcsin(8/9.1679·sin(55°)) ≈ 45.627° ≈ 45.6°

R = 180° -Q -P = 180° -55° -45.6° = 79.4°

__

The solution is ...

q = 9.2

∠P = 45.6°

∠R = 79.4°

In ∆PQR, p = 8, r, = 11, and < Q = 55o. Solve the triangle.?-example-1
answered
User KappaNossi
by
8.0k points
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