Question

which triangle is similar to $\triangle abc$ if $\sin(a) = \frac{1}{4}$, $\cos(a) = \frac{\sqrt{15}}{4}$, and $\tan(a) = \frac{1}{\sqrt{15}}$?

Explanation:

Step1: Recall trigonometric ratios

For 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 opposite, adjacent, hypotenuse to angle \(A\) are in ratio \(1:\sqrt{15}:4\) (since \(\sin(A)=\frac{1}{4}\) implies opposite = 1, hypotenuse = 4; \(\cos(A)=\frac{\sqrt{15}}{4}\) implies adjacent = \(\sqrt{15}\)).

Step2: Analyze each triangle

- **Triangle RST**: Right-angled at S. Sides: 5, 12, 13. Ratios: \(5:12:13\) (not \(1:\sqrt{15}:4\)).
- **Triangle IJK**: Right-angled at J. Sides: 3, \(3\sqrt{15}\), 12. Simplify ratios: Divide by 3: \(1:\sqrt{15}:4\) (matches the ratio of \(\triangle ABC\)).
- **Triangle LMN**: Right-angled at N. Sides: \(\sqrt{6}\), 3, \(\sqrt{15}\). Check ratios: \(\sqrt{6}:3:\sqrt{15}\) (not \(1:\sqrt{15}:4\)).
- **Triangle XYZ**: Right-angled at Z. Sides: 6, \(6\sqrt{15}\), 24. Simplify ratios: Divide by 6: \(1:\sqrt{15}:4\)? Wait, 6: \(6\sqrt{15}\):24 = 1:\(\sqrt{15}\):4? Wait, 24/6 = 4, \(6\sqrt{15}/6=\sqrt{15}\), 6/6=1. Wait, but wait, in triangle IJK, sides are 3 (opposite), \(3\sqrt{15}\) (adjacent), 12 (hypotenuse). In XYZ: 6 (opposite), \(6\sqrt{15}\) (adjacent), 24 (hypotenuse). Wait, but let's re - check triangle IJK: J is right angle. So angle at I: opposite to JK (3), adjacent to JK is IJ (12)? Wait no, right angle at J, so angle at I: opposite side is JK = 3, adjacent side is IJ = 12? Wait no, \(\sin(\text{angle at I})=\frac{JK}{IK}=\frac{3}{12}=\frac{1}{4}\), \(\cos(\text{angle at I})=\frac{IJ}{IK}=\frac{12}{12}\)? No, wait IK is 12? Wait no, the sides: IJ = 12, JK = 3, IK = \(3\sqrt{15}\)? Wait no, by Pythagoras: \(IJ^{2}+JK^{2}=12^{2}+3^{2}=144 + 9=153\), and \((3\sqrt{15})^{2}=9\times15 = 135\). Wait, that's a mistake! Wait, \(12^{2}+3^{2}=153 eq(3\sqrt{15})^{2}=135\). Oh no, miscalculation. Let's recalculate triangle IJK: Right - angled at J, so \(IJ^{2}+JK^{2}=IK^{2}\). \(IJ = 12\), \(JK = 3\), so \(IK^{2}=12^{2}+3^{2}=144 + 9 = 153\), \(IK=\sqrt{153}=3\sqrt{17}\), not \(3\sqrt{15}\). So my earlier mistake. Let's check triangle XYZ: Right - angled at Z. \(YZ = 6\), \(XZ = 6\sqrt{15}\), \(XY = 24\). Check Pythagoras: \(YZ^{2}+XZ^{2}=6^{2}+(6\sqrt{15})^{2}=36 + 540 = 576\), \(XY^{2}=24^{2}=576\). So it's right - angled. Now, \(\sin(\text{angle at X})=\frac{YZ}{XY}=\frac{6}{24}=\frac{1}{4}\), \(\cos(\text{angle at X})=\frac{XZ}{XY}=\frac{6\sqrt{15}}{24}=\frac{\sqrt{15}}{4}\), \(\tan(\text{angle at X})=\frac{6}{6\sqrt{15}}=\frac{1}{\sqrt{15}}\). Wait, but earlier triangle IJK was miscalculated. Wait, the problem is in the diagram: triangle IJK has \(JK = 3\), \(IJ = 12\), \(IK = 3\sqrt{15}\)? But \(3^{2}+12^{2}=9 + 144 = 153\), \((3\sqrt{15})^{2}=135\), so that's not a right triangle. So triangle XYZ: \(YZ = 6\), \(XZ = 6\sqrt{15}\), \(XY = 24\). \(6^{2}+(6\sqrt{15})^{2}=36+540 = 576 = 24^{2}\), so it's right - angled. And \(\sin(\text{angle at X})=\frac{6}{24}=\frac{1}{4}\), \(\cos(\text{angle at X})=\frac{6\sqrt{15}}{24}=\frac{\sqrt{15}}{4}\), \(\tan(\text{angle at X})=\frac{6}{6\sqrt{15}}=\frac{1}{\sqrt{15}}\). Wait, but also check triangle IJK again. Wait, maybe the right angle is at J, so sides: \(JK = 3\), \(IJ = 12\), \(IK\) should be \(\sqrt{3^{2}+12^{2}}=\sqrt{153}\), but the diagram says \(IK = 3\sqrt{15}=\sqrt{135}\), which is wrong. So the correct one is triangle XYZ? Wait no, wait the first triangle IJK: maybe I misread the sides. Wait the problem's diagram: triangle IJK has \(JK = 3\), \(IJ = 12\), \(IK = 3\sqrt{15}\)? No, that's incorrect. Wait triangle XYZ: \(YZ = 6\), \(XZ = 6\sqrt{15}\), \(XY = 24\). Let's check the ratios: opposite (YZ = 6), adjacent (XZ = 6\sqrt{15}), hypotenuse (XY = 24). So ratio 6:6\sqrt{15}:24 = 1:\sqrt{15}:4 (divide by 6). Now check triangle IJK: if \(JK = 3\), \(IJ = 12\), \(IK\) should be \(\sqrt{3^{2}+12^{2}}=\sqrt{153}\), but the diagram says \(3\sqrt{15}\), which is wrong. So maybe the correct triangle is XYZ? Wait no, wait the initial ratio: opposite = 1, adjacent = \(\sqrt{15}\), hypotenuse = 4. So a triangle with sides \(k, k\sqrt{15}, 4k\) for some \(k\). Let's check triangle XYZ: \(k = 6\), so sides 6, 6\sqrt{15}, 24 (4*6 = 24). Triangle IJK: if \(k = 3\), sides 3, 3\sqrt{15}, 12 (4*3 = 12). But \(3^{2}+(3\sqrt{15})^{2}=9 + 135 = 144 = 12^{2}\)! Oh! I made a mistake earlier. \(3^{2}+(3\sqrt{15})^{2}=9 + 9\times15=9+135 = 144 = 12^{2}\). So triangle IJK is right - angled at J (since \(JK^{2}+IJ^{2}=IK^{2}\)? Wait no, \(JK = 3\), \(IJ = 12\), \(IK = 3\sqrt{15}\)? No, \(3^{2}+(3\sqrt{15})^{2}=9 + 135 = 144 = 12^{2}\), so actually the right angle is at K? Wait, no, the right angle is marked at J. So maybe the labels are different. Anyway, the ratio of sides in triangle IJK: 3 (one leg), \(3\sqrt{15}\) (other leg), 12 (hypotenuse). Divide by 3: 1, \(\sqrt{15}\), 4. Which matches the ratio of \(\triangle ABC\) (opposite = 1, adjacent = \(\sqrt{15}\), hypotenuse = 4). Triangle XYZ: 6, 6\sqrt{15}, 24. Divide by 6: 1, \(\sqrt{15}\), 4. Also matches. Wait, but the problem is to find which is similar. Similar triangles have proportional sides. Let's check the angles. For \(\triangle ABC\), angle \(A\) has \(\sin(A)=\frac{1}{4}\), \(\cos(A)=\frac{\sqrt{15}}{4}\), \(\tan(A)=\frac{1}{\sqrt{15}}\). For triangle IJK: let's take angle at I. \(\sin(\text{angle at I})=\frac{JK}{IK}=\frac{3}{12}=\frac{1}{4}\), \(\cos(\text{angle at I})=\frac{IJ}{IK}=\frac{12}{12}\)? No, \(IJ = 12\), \(IK = 12\)? No, \(IK = 3\sqrt{15}\approx11.62\), \(IJ = 12\). Wait, I think I messed up the right angle. The right angle is at J, so angle at I: opposite side is JK = 3, adjacent side is IJ = 12, hypotenuse is IK = \(\sqrt{3^{2}+12^{2}}=\sqrt{153}\)? No, earlier calculation was wrong. \(3^{2}+12^{2}=9 + 144 = 153\), \((3\sqrt{15})^{2}=135\), so \(3^{2}+12^{2} eq(3\sqrt{15})^{2}\). So my mistake. The correct right triangle with sides in ratio 1:\(\sqrt{15}\):4 is triangle XYZ: 6, 6\sqrt{15}, 24 (since \(6^{2}+(6\sqrt{15})^{2}=36 + 540 = 576 = 24^{2}\)) and triangle IJK is incorrect. Wait, no, let's recalculate triangle IJK: if \(JK = 3\), \(IK = 3\sqrt{15}\), \(IJ = 12\), then \(JK^{2}+IK^{2}=9 + 135 = 144 = IJ^{2}\), so right angle at K. So angle at I: opposite side is JK = 3, adjacent side is IK = 3\sqrt{15}, hypotenuse is IJ = 12. So \(\sin(\text{angle at I})=\frac{3}{12}=\frac{1}{4}\), \(\cos(\text{angle at I})=\frac{3\sqrt{15}}{12}=\frac{\sqrt{15}}{4}\), \(\tan(\text{angle at I})=\frac{3}{3\sqrt{15}}=\frac{1}{\sqrt{15}}\), which matches angle \(A\). So triangle IJK (right - angled at K) has angle at I with same trigonometric ratios as angle \(A\), so they are similar (AA similarity, since both are right triangles and one acute angle has same trig ratios).

Wait, but also triangle XYZ: right - angled at Z, angle at X: \(\sin(X)=\frac{6}{24}=\frac{1}{4}\), \(\cos(X)=\frac{6\sqrt{15}}{24}=\frac{\sqrt{15}}{4}\), \(\tan(X)=\frac{6}{6\sqrt{15}}=\frac{1}{\sqrt{15}}\), also matches. But wait, the options: let's check the diagram again. The triangle with sides 3, \(3\sqrt{15}\), 12 (triangle IJK) and 6, \(6\sqrt{15}\), 24 (triangle XYZ). But maybe the intended answer is triangle IJK (or XYZ). Wait, but in the problem, the first triangle IJK has sides 3, \(3\sqrt{15}\), 12, which is \(k = 3\) times \(1,\sqrt{15},4\). Triangle XYZ is \(k = 6\) times. But let's check the answer options. Wait, the triangle with vertices I, J, K: right - angled (at J or K), sides 3, \(3\sqrt{15}\), 12. So the correct triangle is the one with sides in ratio \(1:\sqrt{15}:4\), which is triangle IJK (or XYZ). But let's see the options: the second triangle (IJK) has sides 3, \(3\sqrt{15}\), 12, which is \(3(1,\sqrt{15},4)\), so it's similar.

Answer:

The triangle similar to \(\triangle ABC\) is the triangle with vertices \(I\), \(J\), \(K\) (right - angled, sides \(3\), \(3\sqrt{15}\), \(12\)) (or the triangle \(XYZ\) with sides \(6\), \(6\sqrt{15}\), \(24\), but based on the given diagram and side labels, the triangle \(IJK\) (or the one with sides \(3\), \(3\sqrt{15}\), \(12\)) is the correct one. However, re - checking the Pythagorean theorem for triangle \(IJK\): \(3^{2}+(3\sqrt{15})^{2}=9 + 135 = 144 = 12^{2}\), so it's a right triangle, and the ratios match. So the answer is the triangle with sides \(3\), \(3\sqrt{15}\), \(12\) (triangle \(IJK\)).