Question

which triangle is similar to $\triangle abc$ if $\sin(a) = \frac{1}{4}$, $\cos(a) = \frac{\sqrt{15}}{4}$, and $\tan(a) = \frac{1}{\sqrt{15}}$? (four right triangles are shown: $\triangle rst$ with right angle at $s$, $rs = 5$, $st = 12$, $rt = 13$; $\triangle ijk$ with right angle at $j$, $jk = 3$, $ij = 12$, $ik = 3\sqrt{15}$; $\triangle lmn$ with right angle at $n$, $ln = \sqrt{6}$, $mn = 3$, $lm = \sqrt{15}$; $\triangle xyz$ with right angle at $z$, $yz = 6$, $xz = 6\sqrt{15}$, $xy = 24$)

Explanation:

Step1: Identify sides for ∠A

For $\triangle ABC$, let the right triangle have:
Opposite to $\angle A$: $1$, Adjacent: $\sqrt{15}$, Hypotenuse: $4$
(From $\sin(A)=\frac{1}{4}$, $\cos(A)=\frac{\sqrt{15}}{4}$, $\tan(A)=\frac{1}{\sqrt{15}}$)

Step2: Check triangle RST

Sides: $5,12,13$. Ratios: $\frac{5}{13},\frac{12}{13},\frac{5}{12}$ → Not matching.

Step3: Check triangle IJK

Sides: $3,12,3\sqrt{15}$. Simplify ratios:
$\sin(\angle K)=\frac{12}{3\sqrt{15}}=\frac{4}{\sqrt{15}}$ (no), $\sin(\angle JIK)=\frac{3}{3\sqrt{15}}=\frac{1}{\sqrt{15}}$, $\cos(\angle JIK)=\frac{12}{3\sqrt{15}}=\frac{4}{\sqrt{15}}$ → Not matching.

Step4: Check triangle LMN

Sides: $3,\sqrt{6},\sqrt{15}$. Ratios: $\frac{3}{\sqrt{15}}=\frac{\sqrt{15}}{5}$, $\frac{\sqrt{6}}{\sqrt{15}}=\frac{\sqrt{10}}{5}$ → Not matching.

Step5: Check triangle XYZ

Sides: $6,6\sqrt{15},24$. Simplify ratios (divide by 6):
Opposite: $1$, Adjacent: $\sqrt{15}$, Hypotenuse: $4$.
$\sin(\angle X)=\frac{6}{24}=\frac{1}{4}$, $\cos(\angle X)=\frac{6\sqrt{15}}{24}=\frac{\sqrt{15}}{4}$, $\tan(\angle X)=\frac{6}{6\sqrt{15}}=\frac{1}{\sqrt{15}}$ → Matches.

Answer:

The triangle similar to $\triangle ABC$ is $\triangle XYZ$ (the rightmost triangle with sides $6$, $6\sqrt{15}$, and $24$).