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: first with right angle at s, legs 5 and 12, hypotenuse 13; second with right angle at j, legs 3 and $3\sqrt{15}$, hypotenuse 12; third with right angle at n, legs $\sqrt{6}$ and 3, hypotenuse $\sqrt{15}$; fourth with right angle at z, legs 6 and 24, hypotenuse $6\sqrt{15}$

Explanation:

Step1: Recall trigonometric ratios

For a right triangle, \(\sin(\theta)=\frac{\text{opposite}}{\text{hypotenuse}}\), \(\cos(\theta)=\frac{\text{adjacent}}{\text{hypotenuse}}\), \(\tan(\theta)=\frac{\text{opposite}}{\text{adjacent}}\). Given \(\sin(A)=\frac{1}{4}\), \(\cos(A)=\frac{\sqrt{15}}{4}\), \(\tan(A)=\frac{1}{\sqrt{15}}\), so the sides relative to angle \(A\) (assuming \(A\) is an acute angle in a right triangle) are: opposite \(= 1\), adjacent \(=\sqrt{15}\), hypotenuse \(= 4\) (or scaled versions).

Step2: Analyze each triangle

- **Triangle \(RST\)**: Right-angled at \(S\). Sides: \(5, 12, 13\) (since \(5^2 + 12^2 = 13^2\)). Ratios: \(\sin(\text{angle})=\frac{5}{13}\), \(\cos(\text{angle})=\frac{12}{13}\), \(\tan(\text{angle})=\frac{5}{12}\). Not matching.
- **Triangle \(IJK\)**: Right-angled at \(J\). Sides: \(3, 3\sqrt{15}, 12\) (check \(3^2+(3\sqrt{15})^2 = 9 + 135 = 144 = 12^2\)). Ratios: \(\sin(\text{angle at } I)=\frac{3}{12}=\frac{1}{4}\), \(\cos(\text{angle at } I)=\frac{3\sqrt{15}}{12}=\frac{\sqrt{15}}{4}\), \(\tan(\text{angle at } I)=\frac{3}{3\sqrt{15}}=\frac{1}{\sqrt{15}}\). Matches the ratios of \(\angle A\).
- **Triangle \(LMN\)**: Right-angled at \(N\). Sides: \(\sqrt{6}, 3, \sqrt{15}\) (check \((\sqrt{6})^2+3^2 = 6 + 9 = 15 = (\sqrt{15})^2\)). Ratios: \(\sin(\text{angle at } L)=\frac{3}{\sqrt{15}}=\frac{\sqrt{15}}{5}\), \(\cos(\text{angle at } L)=\frac{\sqrt{6}}{\sqrt{15}}=\frac{\sqrt{10}}{5}\), \(\tan(\text{angle at } L)=\frac{3}{\sqrt{6}}=\frac{\sqrt{6}}{2}\). Not matching.
- **Triangle \(XYZ\)**: Right-angled at \(Z\). Sides: \(6, 24, 6\sqrt{15}\) (check \(6^2+(6\sqrt{15})^2 = 36 + 540 = 576 = 24^2\)). Ratios: \(\sin(\text{angle at } X)=\frac{6}{24}=\frac{1}{4}\), but \(\cos(\text{angle at } X)=\frac{6\sqrt{15}}{24}=\frac{\sqrt{15}}{4}\), wait? Wait, no: opposite to angle \(X\) is \(6\), adjacent is \(6\sqrt{15}\), hypotenuse \(24\). Wait, but in \(\triangle ABC\), angle \(A\) has opposite \(1\), adjacent \(\sqrt{15}\), hypotenuse \(4\). But in \(\triangle IJK\), the sides are scaled by \(3\) (since \(1\times3 = 3\), \(\sqrt{15}\times3 = 3\sqrt{15}\), \(4\times3 = 12\)). In \(\triangle XYZ\), scaling factor would be \(6\) ( \(1\times6 = 6\), \(\sqrt{15}\times6 = 6\sqrt{15}\), \(4\times6 = 24\) ). Wait, but earlier analysis of \(\triangle IJK\) had sides \(3, 3\sqrt{15}, 12\) (scaled by 3), and \(\triangle XYZ\) scaled by 6. Wait, but when we check the ratios for \(\triangle IJK\): \(\sin=\frac{3}{12}=\frac{1}{4}\), \(\cos=\frac{3\sqrt{15}}{12}=\frac{\sqrt{15}}{4}\), \(\tan=\frac{3}{3\sqrt{15}}=\frac{1}{\sqrt{15}}\), which matches. For \(\triangle XYZ\), \(\sin=\frac{6}{24}=\frac{1}{4}\), \(\cos=\frac{6\sqrt{15}}{24}=\frac{\sqrt{15}}{4}\), \(\tan=\frac{6}{6\sqrt{15}}=\frac{1}{\sqrt{15}}\). Wait, but maybe I made a mistake in \(\triangle IJK\) sides. Wait, \(IJ = 12\), \(JK = 3\), \(IK = 3\sqrt{15}\)? Wait, no: in right triangle, hypotenuse is the longest side. \(12\) is longer than \(3\sqrt{15}\) (since \(3\sqrt{15}\approx11.62\)), so \(IJ = 12\) (hypotenuse), \(JK = 3\) (opposite), \(IJ\) is hypotenuse. So angle at \(I\): opposite is \(JK = 3\), adjacent is \(IK = 3\sqrt{15}\), hypotenuse \(IJ = 12\). So ratios are correct. For \(\triangle XYZ\), hypotenuse \(XY = 24\), opposite \(YZ = 6\), adjacent \(XZ = 6\sqrt{15}\). Ratios also match. Wait, but maybe the triangle \(IJK\) is the one with sides scaled by 3, and \(XYZ\) scaled by 6. But let's check the first triangle again. Wait, the problem is to find which is similar. Similar triangles have proportional sides. Let's check the ratios of sides. For \(\triangle ABC\) (hypothetical) with sides \(1, \sqrt{15}, 4\) (scaled by \(k\)). For \(\triangle IJK\): sides \(3, 3\sqrt{15}, 12\) (ratio \(3:3\sqrt{15}:12 = 1:\sqrt{15}:4\) when divided by 3). For \(\triangle XYZ\): sides \(6, 6\sqrt{15}, 24\) (ratio \(6:6\sqrt{15}:24 = 1:\sqrt{15}:4\) when divided by 6). Wait, but maybe the diagram shows \(\triangle IJK\) (the second triangle) as the one with right angle at \(J\), sides \(3, 3\sqrt{15}, 12\). Wait, but let's re - check the first triangle \(RST\): sides 5,12,13 (5 - 12 - 13 triangle), ratios don't match. Triangle \(LMN\): sides \(\sqrt{6}, 3, \sqrt{15}\) ( \((\sqrt{6})^2+3^2 = 6 + 9 = 15 = (\sqrt{15})^2\) ), ratios: \(\sin=\frac{3}{\sqrt{15}}=\frac{\sqrt{15}}{5} eq\frac{1}{4}\), so no. Now, between \(\triangle IJK\) and \(\triangle XYZ\), but let's check the given triangles. Wait, the triangle \(IJK\) (second one) has sides 3, \(3\sqrt{15}\), 12. The ratios of sides (opposite:adjacent:hypotenuse) are \(3:3\sqrt{15}:12 = 1:\sqrt{15}:4\), which matches the ratios from \(\sin(A), \cos(A), \tan(A)\) (since \(\sin(A)=\frac{1}{4}\) (opposite/hypotenuse), \(\cos(A)=\frac{\sqrt{15}}{4}\) (adjacent/hypotenuse), \(\tan(A)=\frac{1}{\sqrt{15}}\) (opposite/adjacent)). So the triangle with right angle at \(J\) ( \(IJK\) ) is similar.

Answer:

The triangle with vertices \(I\), \(J\), \(K\) (the second triangle, right - angled at \(J\) with sides \(3\), \(3\sqrt{15}\), and \(12\)) is similar to \(\triangle ABC\).