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Precalculus · Question of the Day

Precalculus Question of the Day

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Friday, October 23, 2026

In right triangle △PQR\triangle PQR, ∠R=90∘\angle R = 90^\circ. The hypotenuse is PQ=10PQ = 10 and the leg adjacent to ∠P\angle P is PR=8PR = 8. Using the given information, what is the value of cos⁡(∠P)\cos(\angle P)?

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Question of the Day

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In right triangle △PQR\triangle PQR, ∠R=90∘\angle R = 90^\circ. The hypotenuse is PQ=10PQ = 10 and the leg adjacent to ∠P\angle P is PR=8PR = 8. Using the given information, what is the value of cos⁡(∠P)\cos(\angle P)?

  1. 54\frac{5}{4}
  2. 45\frac{4}{5} (correct answer)
  3. 35\frac{3}{5}
  4. 53\frac{5}{3}

Explanation: This question tests the ability to solve right triangles using trigonometric ratios. Trigonometric ratios relate the angles of a right triangle to the ratios of its sides: sin(θ) = opposite/hypotenuse, cos(θ) = adjacent/hypotenuse, and tan(θ) = opposite/adjacent. From the perspective of angle P, the side PR=8 is adjacent, and PQ=10 is the hypotenuse, so we use cos(P) = adjacent/hypotenuse = 8/10 = 4/5. Choice B is correct because it uses the cosine ratio with the adjacent side and hypotenuse: cos(P) = 8/10 = 4/5. Choice C reverses the sine and cosine ratios, using the opposite side instead of adjacent, which swaps the ratios. Remember the SOH-CAH-TOA mnemonic for choosing the correct trig ratio: Sine = Opposite/Hypotenuse, Cosine = Adjacent/Hypotenuse, Tangent = Opposite/Adjacent, where opposite and adjacent are always relative to the angle in question. For complementary angles in a right triangle, sin(θ) = cos(90° - θ), which explains why the sine of one acute angle equals the cosine of the other.