Question

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

Explanation:

Step1: Recall the sine - cosine - tangent definitions in a right - triangle

In a right - triangle, if \(\theta\) is one of the acute angles, \(\sin\theta=\frac{\text{opposite}}{\text{hypotenuse}}\), \(\cos\theta=\frac{\text{adjacent}}{\text{hypotenuse}}\), and \(\tan\theta=\frac{\text{opposite}}{\text{adjacent}}\). Given \(\sin(A)=\frac{1}{4}\), \(\cos(A)=\frac{\sqrt{15}}{4}\), we can assume the opposite side \(=k\), the adjacent side \(=\sqrt{15}k\), and the hypotenuse \( = 4k\) (\(k>0\)).

Step2: Check the ratios of the sides of each triangle

• For triangle \(IJK\):

The sides are \(IJ = 12\), \(JK=3\), \(IK = 3\sqrt{15}\). Let's find the ratios. If we consider the right - angle at \(J\), \(\frac{JK}{IK}=\frac{3}{3\sqrt{15}}=\frac{1}{\sqrt{15}}\), \(\frac{JK}{IJ}=\frac{3}{12}=\frac{1}{4}\), \(\frac{IJ}{IK}=\frac{12}{3\sqrt{15}}=\frac{4}{\sqrt{15}}\).

• For triangle \(RST\):

The sides are \(RS = 5\), \(ST=12\), \(RT = 13\). \(\sin\) (angle at \(S\))\(=\frac{5}{13}\), \(\cos\) (angle at \(S\))\(=\frac{12}{13}\), \(\tan\) (angle at \(S\))\(=\frac{5}{12}\)

• For triangle \(LMN\):

The sides are \(MN = 3\), \(LN=\sqrt{6}\), \(LM=\sqrt{15}\). \(\sin\) (angle at \(N\))\(=\frac{\sqrt{15}}{\sqrt{15 + 6}}
eq\frac{1}{4}\)

• For triangle \(XYZ\):

The sides are \(YZ = 6\), \(XZ=6\sqrt{15}\), \(XY = 24\).
\(\frac{YZ}{XY}=\frac{6}{24}=\frac{1}{4}\), \(\frac{XZ}{XY}=\frac{6\sqrt{15}}{24}=\frac{\sqrt{15}}{4}\), \(\frac{YZ}{XZ}=\frac{6}{6\sqrt{15}}=\frac{1}{\sqrt{15}}\)

Answer:

The triangle \(XYZ\) is similar to \(\triangle ABC\)