SAT · Math · Geometry and Trigonometry
Teorema de Pitágoras y razones trigonométricas (seno, coseno, tangente) en triángulos rectángulos.
Triángulo $3-4-5$, con A opuesto al lado 3: $\operatorname{sen} A=\frac{3}{5}$, $\cos A=\frac{4}{5}$, $\tan A=\frac{3}{4}$.
Intenta responder antes de abrir la solución. Las preguntas están en el idioma del examen.
In right triangle $ABC$ shown, what is the length of $\overline{AB}$?
Respuesta correcta: B. $10$
By the Pythagorean theorem, $AB2=62+82=36+64=100$, so $AB=10$.
$14$ adds the legs. $7$ is their average. $48$ multiplies them.
In right triangle $RST$, angle $T$ is the right angle, $RS=26$, and $ST=10$. What is the value of $\cos R$?
Respuesta numérica (escribes el número).
Respuesta: 12/13 · 24/26 · .9230 · .9231 · 0.923
$RS$ is the hypotenuse because it is opposite the right angle. By the Pythagorean theorem, $RT=\sqrt{262-102}=\sqrt{676-100}=\sqrt{576}=24$. The cosine of $R$ is the adjacent side over the hypotenuse: $\cos R=\frac{RT}{RS}=\frac{24}{26}=\frac{12}{13}$.
In triangle $JKL$, angle $K$ is a right angle, $JK=20$, and $\cos J=\frac{4}{5}$. What is the area of triangle $JKL$?
Respuesta correcta: A. $150$
$\cos J=\frac{JK}{JL}$, the adjacent side over the hypotenuse, so $\frac{20}{JL}=\frac{4}{5}$ and $JL=25$. Then $KL=\sqrt{252-202}=\sqrt{225}=15$. The legs are $JK=20$ and $KL=15$, so the area is $\frac{1}{2}(20)(15)=150$.
$300$ forgets the $\frac{1}{2}$. $250$ uses the hypotenuse $25$ as a leg: $\frac{1}{2}(20)(25)$. $200$ uses $20$ for both legs.