Examkin

Question

Difficulty: MediumRight Triangle Trigonometry (SOHCAHTOA)

A sailboat travels due east from Port PP for 1515 nautical miles to Point QQ, then turns due north and travels 3636 nautical miles to Point RR. The straight line of sight from Port PP to Point RR forms an angle θ\theta with segment PQPQ. What is the value of cos⁡θ−sin⁡θ\cos \theta - \sin \theta?

  1. −713-\frac{7}{13}Answer
  2. B
    713\frac{7}{13}
  3. C
    1713\frac{17}{13}
  4. D
    −717-\frac{7}{17}
  5. E
    −125-\frac{12}{5}

Answer

The correct value of cos⁡θ−sin⁡θ\cos \theta - \sin \theta is −713-\frac{7}{13}.
The direct distance between Port PP and Point RR forms the hypotenuse of right triangle PQRPQR. By applying the Pythagorean theorem, the hypotenuse length is 152+362=39\sqrt{15^2 + 36^2} = 39. Relative to angle θ\theta at vertex PP, the adjacent leg is 1515 and the opposite leg is 3636. Therefore, cos⁡θ=1539=513\cos \theta = \frac{15}{39} = \frac{5}{13} and sin⁡θ=3639=1213\sin \theta = \frac{36}{39} = \frac{12}{13}. Subtracting these yields 513−1213=−713\frac{5}{13} - \frac{12}{13} = -\frac{7}{13}.

Step-by-Step Solution

1
Determine the length of the hypotenuse PRPR using the Pythagorean theorem.
PR=152+362=225+1296=1521=39PR = \sqrt{15^2 + 36^2} = \sqrt{225 + 1296} = \sqrt{1521} = 39 nautical miles.
Triangle PQRPQR is a right triangle with right angle at vertex QQ because the path turns from due east to due north.
2
Identify the opposite side, adjacent side, and hypotenuse relative to angle θ=∠QPR\theta = \angle QPR.
Adjacent side PQ=15PQ = 15, Opposite side QR=36QR = 36, Hypotenuse PR=39PR = 39.
Angle θ\theta is formed at vertex PP between the base path PQPQ and the direct distance PRPR.
3
Calculate cos⁡θ\cos \theta and sin⁡θ\sin \theta using SOHCAHTOA ratios.
cos⁡θ=AdjacentHypotenuse=1539=513\cos \theta = \frac{\text{Adjacent}}{\text{Hypotenuse}} = \frac{15}{39} = \frac{5}{13} and sin⁡θ=OppositeHypotenuse=3639=1213\sin \theta = \frac{\text{Opposite}}{\text{Hypotenuse}} = \frac{36}{39} = \frac{12}{13}.
Cosine is adjacent over hypotenuse and sine is opposite over hypotenuse.
4
Evaluate the expression cos⁡θ−sin⁡θ\cos \theta - \sin \theta.
cos⁡θ−sin⁡θ=513−1213=−713\cos \theta - \sin \theta = \frac{5}{13} - \frac{12}{13} = -\frac{7}{13}.
Subtracting the sine value from the cosine value gives a negative result since the opposite leg is longer than the adjacent leg.

Key Concept

Right Triangle Trigonometry (SOHCAHTOA)
Estimated Time:1m 30s
Rate this question

Topics

Right Triangle Trigonometry (SOHCAHTOA)

More questions from this topic