In right triangle JKL, the right angle is at vertex K. Line segment KM is perpendicular to… — ACT

In right triangle $JKL$ , the right angle is at vertex $K$ . Line segment $KM$ is perpendicular to hypotenuse $JL$ , with point $M$ lying on $JL$ . If $\sin(\angle KJL) = \frac{4}{5}$ and the length of segment $JM$ is $9$ units, what is the length, in units, of side $KL$ ?

A
$9$
B
$12$
C
$15$
$20$ Answer
E
$25$

Answer

By using SOHCAHTOA,

\sin(\angle KJL) = \frac{4}{5}

implies that

\cos(\angle KJL) = \frac{3}{5}

and

\tan(\angle KJL) = \frac{4}{3}

. In the smaller right triangle

\triangle JMK

\cos(\angle KJL) = \frac{\text{adjacent}}{\text{hypotenuse}} = \frac{JM}{JK}

, so

\frac{3}{5} = \frac{9}{JK}

, which yields

JK = 15

. Next, in the large right triangle

\triangle JKL

\tan(\angle KJL) = \frac{\text{opposite}}{\text{adjacent}} = \frac{KL}{JK}

, so

\frac{4}{3} = \frac{KL}{15}

, giving

KL = 20

Step-by-Step Solution

Determine

\cos(\angle KJL)

and

\tan(\angle KJL)

using SOHCAHTOA.

\cos(\angle KJL) = \frac{3}{5}

and

\tan(\angle KJL) = \frac{4}{3}

Since

\sin(\angle KJL) = \frac{\text{opposite}}{\text{hypotenuse}} = \frac{4}{5}

in a right triangle, the adjacent side ratio is

\sqrt{5^2 - 4^2} = 3

, giving

\cos(\angle KJL) = \frac{3}{5}

and

\tan(\angle KJL) = \frac{4}{3}

Apply the cosine ratio in right triangle

\triangle JMK

(where

\angle JMK = 90^\circ

JK = 15

\triangle JMK

\cos(\angle KJL) = \frac{JM}{JK}

. Substituting the given values gives

\frac{3}{5} = \frac{9}{JK} \implies 3 \cdot JK = 45 \implies JK = 15

Apply the tangent ratio in main right triangle

\triangle JKL

to calculate

KL

KL = 20

\triangle JKL

\tan(\angle KJL) = \frac{KL}{JK}

. Substituting

JK = 15

gives

\frac{4}{3} = \frac{KL}{15} \implies 3 \cdot KL = 60 \implies KL = 20

Key Concept

Right Triangle Trigonometry (SOHCAHTOA) across Similar Right Triangles

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