Question

which triangle is similar to $\triangle abc$ if $\sin(a) = \frac{1}{4}$, $\cos(a) = \frac{\sqrt{15}}{4}$, and $\tan(a) = \frac{1}{\sqrt{15}}$?\
(images of four right triangles: first with legs 5, 12, hypotenuse 13; second with legs 3, $3\sqrt{15}$, hypotenuse 12; third with legs $\sqrt{6}$, 3, hypotenuse $\sqrt{15}$; fourth with legs 6, $6\sqrt{15}$, hypotenuse 24)

Explanation:

Step1: Recall trigonometric ratios

In a right triangle, \(\sin(A)=\frac{\text{opposite}}{\text{hypotenuse}}\), \(\cos(A)=\frac{\text{adjacent}}{\text{hypotenuse}}\), \(\tan(A)=\frac{\text{opposite}}{\text{adjacent}}\). Given \(\sin(A)=\frac{1}{4}\), \(\cos(A)=\frac{\sqrt{15}}{4}\), \(\tan(A)=\frac{1}{\sqrt{15}}\), so the sides of \(\triangle ABC\) (right - angled at the angle corresponding to \(A\)'s right - triangle) should have opposite side \(=1k\), adjacent side \(=\sqrt{15}k\), hypotenuse \( = 4k\) for some positive real number \(k\).

Step2: Analyze each triangle

• Triangle \(RST\): Right - angled at \(S\). \(RS = 5\), \(ST=12\), \(RT = 13\). \(\sin(\text{angle at }R)=\frac{12}{13}

eq\frac{1}{4}\), \(\cos(\text{angle at }R)=\frac{5}{13}
eq\frac{\sqrt{15}}{4}\), so not similar.

• Triangle \(IJK\): Right - angled at \(J\). \(JK = 3\), \(IJ = 12\), \(IK=3\sqrt{15}\). \(\sin(\text{angle at }I)=\frac{JK}{IK}=\frac{3}{3\sqrt{15}}=\frac{1}{\sqrt{15}}\), \(\cos(\text{angle at }I)=\frac{IJ}{IK}=\frac{12}{3\sqrt{15}}=\frac{4}{\sqrt{15}}

eq\frac{\sqrt{15}}{4}\), so not similar.

• Triangle \(LMN\): Right - angled at \(N\). \(MN = 3\), \(LN=\sqrt{6}\), \(LM=\sqrt{15}\). \(\sin(\text{angle at }L)=\frac{MN}{LM}=\frac{3}{\sqrt{15}}=\frac{\sqrt{15}}{5}

eq\frac{1}{4}\), \(\cos(\text{angle at }L)=\frac{LN}{LM}=\frac{\sqrt{6}}{\sqrt{15}}=\frac{\sqrt{10}}{5}
eq\frac{\sqrt{15}}{4}\), so not similar.

• Triangle \(XYZ\): Right - angled at \(Z\). \(YZ = 6\), \(XZ=6\sqrt{15}\), \(XY = 24\). \(\sin(\text{angle at }X)=\frac{YZ}{XY}=\frac{6}{24}=\frac{1}{4}\), \(\cos(\text{angle at }X)=\frac{XZ}{XY}=\frac{6\sqrt{15}}{24}=\frac{\sqrt{15}}{4}\), \(\tan(\text{angle at }X)=\frac{YZ}{XZ}=\frac{6}{6\sqrt{15}}=\frac{1}{\sqrt{15}}\). All the trigonometric ratios match with the ratios of \(\triangle ABC\) (scaled by \(k = 6\), since \(1\times6 = 6\) (opposite), \(\sqrt{15}\times6=6\sqrt{15}\) (adjacent), \(4\times6 = 24\) (hypotenuse)).

Answer:

The triangle \(XYZ\) (the fourth triangle with \(X\), \(Y\), \(Z\) where \(Z\) is the right - angle, \(YZ = 6\), \(XZ=6\sqrt{15}\), \(XY = 24\))