In right triangle DEF, the right angle is located at vertex E. The length of leg DE is 32 … — ACT

In right triangle $DEF$ , the right angle is located at vertex $E$ . The length of leg $DE$ is $32$ meters. If $\cos(D) = \frac{8}{17}$ , what is the length of leg $EF$ , in meters?

A
15
B
36
60Cevap
D
68
E
136

Cevap

The length of leg

EF

is 60 meters.

Using the definition of cosine,

\cos(D) = \frac{\text{adjacent}}{\text{hypotenuse}} = \frac{DE}{DF}

. Substituting

DE = 32

gives

\frac{32}{DF} = \frac{8}{17}

, so

DF = 68

. Using the Pythagorean theorem,

EF = \sqrt{68^2 - 32^2} = 60

meters.

Adım Adım Çözüm

Set up the cosine ratio for angle D

\cos(D) = \frac{\text{adjacent}}{\text{hypotenuse}} = \frac{DE}{DF} = \frac{8}{17}

By SOHCAHTOA, cosine of an acute angle in a right triangle is the ratio of the adjacent leg to the hypotenuse.

Calculate the length of hypotenuse DF

\frac{32}{DF} = \frac{8}{17} \implies DF = \frac{32 \times 17}{8} = 68\text{ meters}

Substitute the given value

DE = 32

into the ratio and solve for

DF

Calculate the length of leg EF using the Pythagorean theorem

EF = \sqrt{DF^2 - DE^2} = \sqrt{68^2 - 32^2} = \sqrt{4624 - 1024} = \sqrt{3600} = 60\text{ meters}

In right triangle

DEF

DE^2 + EF^2 = DF^2

Anahtar Kavram

Right Triangle Trigonometry (SOHCAHTOA) & Pythagorean Theorem

Soru

Cevap

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