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Difficulty: HardRight Triangle Trigonometry (SOHCAHTOA)

In right triangle PQRPQR, the right angle is at vertex QQ. Point SS lies on side QRQR such that line segment PSPS bisects ∠QPR\angle QPR. If the length of PQPQ is 1414 units and tan⁡(∠QPS)=34\tan(\angle QPS) = \frac{3}{4}, what is the length, in units, of side PRPR?

  1. A
    17.517.5
  2. B
    2828
  3. C
    3535
  4. 5050Answer
  5. E
    5656

Answer

The length of side PRPR is 5050 units.
In right triangle PQSPQS, the tangent ratio gives QS=14⋅34=10.5QS = 14 \cdot \frac{3}{4} = 10.5, which yields sin⁡(∠QPS)=35\sin(\angle QPS) = \frac{3}{5} and cos⁡(∠QPS)=45\cos(\angle QPS) = \frac{4}{5}. Because PSPS bisects ∠QPR\angle QPR, the angle at PP for triangle PQRPQR is twice ∠QPS\angle QPS. Using the cosine double-angle relationship, cos⁡(∠QPR)=cos⁡2(∠QPS)−sin⁡2(∠QPS)=1625−925=725\cos(\angle QPR) = \cos^2(\angle QPS) - \sin^2(\angle QPS) = \frac{16}{25} - \frac{9}{25} = \frac{7}{25}. Finally, applying SOHCAHTOA to right triangle PQRPQR gives cos⁡(∠QPR)=PQPR=14PR=725\cos(\angle QPR) = \frac{PQ}{PR} = \frac{14}{PR} = \frac{7}{25}, which solves to PR=50PR = 50.

Step-by-Step Solution

1
Use right triangle PQSPQS to find QSQS and the trigonometric values of θ=∠QPS\theta = \angle QPS.
QS=10.5QS = 10.5, sin⁡(θ)=35\sin(\theta) = \frac{3}{5}, and cos⁡(θ)=45\cos(\theta) = \frac{4}{5}.
Since △PQS\triangle PQS has a right angle at QQ, tan⁡(θ)=oppositeadjacent=QS14=34\tan(\theta) = \frac{\text{opposite}}{\text{adjacent}} = \frac{QS}{14} = \frac{3}{4}, giving QS=10.5QS = 10.5. The hypotenuse PS=142+10.52=17.5PS = \sqrt{14^2 + 10.5^2} = 17.5, so sin⁡(θ)=10.517.5=35\sin(\theta) = \frac{10.5}{17.5} = \frac{3}{5} and cos⁡(θ)=1417.5=45\cos(\theta) = \frac{14}{17.5} = \frac{4}{5}.
2
Determine cos⁡(∠QPR)\cos(\angle QPR) using the double-angle identity for cosine.
cos⁡(∠QPR)=725.\cos(\angle QPR) = \frac{7}{25}.
Because PSPS bisects ∠QPR\angle QPR, ∠QPR=2θ\angle QPR = 2\theta. Using cos⁡(2θ)=cos⁡2(θ)−sin⁡2(θ)\cos(2\theta) = \cos^2(\theta) - \sin^2(\theta), we get cos⁡(2θ)=(45)2−(35)2=1625−925=725\cos(2\theta) = \left(\frac{4}{5}\right)^2 - \left(\frac{3}{5}\right)^2 = \frac{16}{25} - \frac{9}{25} = \frac{7}{25}.
3
Apply the cosine definition SOHCAHTOA in right triangle PQRPQR to solve for hypotenuse PRPR.
PR = 50.
In right triangle PQRPQR, cos⁡(∠QPR)=adjacenthypotenuse=PQPR=14PR\cos(\angle QPR) = \frac{\text{adjacent}}{\text{hypotenuse}} = \frac{PQ}{PR} = \frac{14}{PR}. Setting 14PR=725\frac{14}{PR} = \frac{7}{25} yields 7⋅PR=3507 \cdot PR = 350, so PR=50PR = 50.

Key Concept

Right Triangle Trigonometry (SOHCAHTOA) and Trigonometric Ratios of Composite Angles
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Topics

Right Triangle Trigonometry (SOHCAHTOA)

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