Question

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

Explanation:

Step1: Recall trigonometric - ratio 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 $\sin(A)=\frac{1}{4}$, $\cos(A)=\frac{\sqrt{15}}{4}$, and $\tan(A)=\frac{1}{\sqrt{15}}$. The hypotenuse of the right - triangle corresponding to angle $A$ is $4$, the opposite side is $1$, and the adjacent side is $\sqrt{15}$.

Step2: Check the ratios for each triangle

For the first triangle $\triangle SRT$ with sides $5$, $12$, and $13$ (a Pythagorean triple), it is not relevant as the ratios of its sides do not match the given trigonometric ratios.
For $\triangle IJK$: $\sin(I)=\frac{3}{12}=\frac{1}{4}$, $\cos(I)=\frac{3\sqrt{15}}{12}=\frac{\sqrt{15}}{4}$, $\tan(I)=\frac{3}{3\sqrt{15}}=\frac{1}{\sqrt{15}}$.
For $\triangle LMN$ with sides $\sqrt{6}$, $3$, and $\sqrt{15}$, $\sin(L)=\frac{\sqrt{6}}{\sqrt{15}}
eq\frac{1}{4}$, so it is not the correct triangle.
For $\triangle XYZ$ with sides $6$, $6\sqrt{15}$, and $24$, $\sin(Z)=\frac{6}{24}=\frac{1}{4}$, $\cos(Z)=\frac{6\sqrt{15}}{24}=\frac{\sqrt{15}}{4}$, $\tan(Z)=\frac{6}{6\sqrt{15}}=\frac{1}{\sqrt{15}}$.

Answer:

The triangle similar to $\triangle ABC$ is $\triangle XYZ$.