which triangle is similar to △abc if sin(a) = 1/4, cos(a) = √15/4, and tan(a) = 1/√15?

Explanation:

Step1: Recall sine - cosine - tangent definitions

In a right - triangle, $\sin\theta=\frac{\text{opposite}}{\text{hypotenuse}}$, $\cos\theta = \frac{\text{adjacent}}{\text{hypotenuse}}$, and $\tan\theta=\frac{\text{opposite}}{\text{adjacent}}$. Given $\sin(A)=\frac{1}{4}$, $\cos(A)=\frac{\sqrt{15}}{4}$, $\tan(A)=\frac{1}{\sqrt{15}}$, the ratio of the opposite side to the hypotenuse is $1:4$, and the ratio of the adjacent side to the hypotenuse is $\sqrt{15}:4$, and the ratio of the opposite side to the adjacent side is $1:\sqrt{15}$.

Step2: Check each triangle

For $\triangle RST$:

The sides are $5,12,13$. $\sin\theta=\frac{5}{13}$, $\cos\theta=\frac{12}{13}$, $\tan\theta=\frac{5}{12}$, so it is not similar.

For $\triangle IJK$:

The sides are $3,3\sqrt{15},12$. $\sin\theta=\frac{3}{12}=\frac{1}{4}$, $\cos\theta=\frac{3\sqrt{15}}{12}=\frac{\sqrt{15}}{4}$, $\tan\theta=\frac{3}{3\sqrt{15}}=\frac{1}{\sqrt{15}}$.

For $\triangle LMN$:

The sides are $\sqrt{6},3,\sqrt{15}$. $\sin\theta=\frac{3}{\sqrt{15}}=\frac{\sqrt{15}}{5}$, $\cos\theta=\frac{\sqrt{6}}{\sqrt{15}}=\frac{\sqrt{10}}{5}$, $\tan\theta=\frac{3}{\sqrt{6}}=\frac{\sqrt{6}}{2}$, so it is not similar.

For $\triangle XYZ$:

The sides are $6,6\sqrt{15},24$. $\sin\theta=\frac{6}{24}=\frac{1}{4}$, $\cos\theta=\frac{6\sqrt{15}}{24}=\frac{\sqrt{15}}{4}$, $\tan\theta=\frac{6}{6\sqrt{15}}=\frac{1}{\sqrt{15}}$.

Answer:

$\triangle IJK$ and $\triangle XYZ$