Question

Difficulty: HardTrigonometric Ratios and Identities

In right triangle RSTRST, the measure of angle TT is 9090^\circ and the length of the hypotenuse RSRS is 1010. If tan(R)+tan(S)=52\tan(R) + \tan(S) = \frac{5}{2}, what is the area of triangle RSTRST?

  1. A
    10
  2. 20Answer
  3. C
    40
  4. D
    50

Answer

20
Let the lengths of the legs of right triangle RSTRST be RT=xRT = x and ST=yST = y. The tangent ratios are tan(R)=yx\tan(R) = \frac{y}{x} and tan(S)=xy\tan(S) = \frac{x}{y}. Substituting these into the given equation gives yx+xy=52\frac{y}{x} + \frac{x}{y} = \frac{5}{2}, which simplifies to x2+y2xy=52\frac{x^2 + y^2}{xy} = \frac{5}{2}. By the Pythagorean theorem, x2+y2=102=100x^2 + y^2 = 10^2 = 100. Thus, 100xy=52\frac{100}{xy} = \frac{5}{2}, which solves to xy=40xy = 40. The area of the triangle is 12xy=12(40)=20\frac{1}{2}xy = \frac{1}{2}(40) = 20.

Step-by-Step Solution

1
Express the tangent of angles RR and SS in terms of the leg lengths RTRT and STST.
tan(R)=STRT\tan(R) = \frac{ST}{RT} and tan(S)=RTST\tan(S) = \frac{RT}{ST}.
By definition, the tangent of an acute angle in a right triangle is the ratio of the opposite leg to the adjacent leg.
2
Substitute these expressions into the given equation and simplify using the Pythagorean theorem.
STRT+RTST=ST2+RT2RTST=100RTST=52\frac{ST}{RT} + \frac{RT}{ST} = \frac{ST^2 + RT^2}{RT \cdot ST} = \frac{100}{RT \cdot ST} = \frac{5}{2}.
Finding a common denominator yields the sum of the squares of the legs in the numerator, which equals the square of the hypotenuse (RS2=102=100RS^2 = 10^2 = 100).
3
Solve for the product of the legs RTSTRT \cdot ST and calculate the area of the triangle.
RTST=40RT \cdot ST = 40, so Area=12(RTST)=12(40)=20\text{Area} = \frac{1}{2}(RT \cdot ST) = \frac{1}{2}(40) = 20.
The area of a right triangle is half the product of its perpendicular legs.

Key Concept

Using trigonometric ratios, complementary angles, and the Pythagorean theorem to calculate the area of a right triangle.
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