Question

fall 2023 geometry b wwva trigonometric ratios which trigonometric ratios are correct for triangle abc? choose three correct answers. tan (c) = √3 tan (b) = 2√3 / 3 sin (c) = √3 / 2

Explanation:

Step1: Recall trigonometric ratio definitions

In a right - triangle, $\tan\theta=\frac{\text{opposite}}{\text{adjacent}}$ and $\sin\theta=\frac{\text{opposite}}{\text{hypotenuse}}$. For $\angle C$, the opposite side to $\angle C$ is $AB$, the adjacent side is $AC$, and the hypotenuse is $BC$. For $\angle B$, the opposite side is $AC$ and the adjacent side is $AB$.

Step2: Calculate $\tan(C)$

In right - triangle $ABC$, $\angle C = 60^{\circ}$, $\tan(C)=\tan(60^{\circ})=\sqrt{3}$. Since $\tan\theta=\frac{\text{opposite}}{\text{adjacent}}$, and for $\angle C$, if we assume the opposite side to $\angle C$ is the side opposite to the $60^{\circ}$ angle and adjacent side is the side next to it, $\tan(C)=\frac{AB}{AC}$. In a 30 - 60 - 90 triangle with side lengths in the ratio $1:\sqrt{3}:2$, $\tan(60^{\circ})=\sqrt{3}$.

Step3: Calculate $\tan(B)$

$\angle B = 30^{\circ}$, $\tan(B)=\tan(30^{\circ})=\frac{\sqrt{3}}{3}$, not $\frac{2\sqrt{3}}{3}$. Because $\tan(B)=\frac{AC}{AB}$, and in a 30 - 60 - 90 triangle, if the side opposite to $30^{\circ}$ is $a$, the side opposite to $60^{\circ}$ is $a\sqrt{3}$, so $\tan(30^{\circ})=\frac{1}{\sqrt{3}}=\frac{\sqrt{3}}{3}$.

Step4: Calculate $\sin(C)$

$\sin(C)=\frac{AB}{BC}$. In a right - triangle where $\angle C = 60^{\circ}$, and assuming the hypotenuse $BC = 2x$, the side opposite to $\angle C$ (i.e., $AB=\sqrt{3}x$), $\sin(C)=\frac{\sqrt{3}}{2}$.

Answer:

$\tan(C)=\sqrt{3}$, $\sin(C)=\frac{\sqrt{3}}{2}$ are correct. So the correct answers are the first and the third options.