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: Define reference triangle for ∠A

We can model △ABC as a right triangle with ∠A, so:
$\sin(A)=\frac{\text{opposite}}{\text{hypotenuse}}=\frac{1}{4}$, $\cos(A)=\frac{\text{adjacent}}{\text{hypotenuse}}=\frac{\sqrt{15}}{4}$, $\tan(A)=\frac{\text{opposite}}{\text{adjacent}}=\frac{1}{\sqrt{15}}$
So the side ratios (opp:adj:hyp) are $1:\sqrt{15}:4$.

Step2: Check △RST ratios

Sides: 5, 12, 13. Ratios: $5:12:13$. Does not match $1:\sqrt{15}:4$ ($\sqrt{15}\approx3.87
eq12$).

Step3: Check △IJK ratios

Sides: 3, 12, $3\sqrt{15}$. Simplify ratios by dividing by 3: $1:4:\sqrt{15}$. Rearranged to opp:adj:hyp: $1:\sqrt{15}:4$, which matches the reference ratios.

Step4: Verify remaining triangles (optional)

△LMN: Sides 3, $\sqrt{6}$, $\sqrt{15}$. Ratios: $3:\sqrt{6}:\sqrt{15}=1:\frac{\sqrt{6}}{3}:\frac{\sqrt{15}}{3}$, does not match.
△XYZ: Sides 6, $6\sqrt{15}$, 24. Simplify by dividing by 6: $1:\sqrt{15}:4$. Wait, but check angle correspondence: $\tan(X)=\frac{6}{6\sqrt{15}}=\frac{1}{\sqrt{15}}$, but the given trig values are for ∠A. For △IJK, $\tan(K)=\frac{12}{3\sqrt{15}}=\frac{4}{\sqrt{15}}$, $\tan(I)=\frac{3}{12}=\frac{1}{4}$, $\tan(J)=\frac{3}{12}=\frac{1}{4}$? No, correct: in △IJK, right angle at J, so for angle I: $\sin(I)=\frac{3}{3\sqrt{15}}=\frac{1}{\sqrt{15}}$, no. Wait, correct check: for △XYZ, right angle at Z, $\sin(X)=\frac{6}{24}=\frac{1}{4}$, $\cos(X)=\frac{6\sqrt{15}}{24}=\frac{\sqrt{15}}{4}$, $\tan(X)=\frac{6}{6\sqrt{15}}=\frac{1}{\sqrt{15}}$. Wait, no, earlier mistake: △IJK: right angle at J, so sides: JI=12, JK=3, IK=3√15. $\sin(K)=\frac{12}{3\sqrt{15}}=\frac{4}{\sqrt{15}}$, $\cos(K)=\frac{3}{3\sqrt{15}}=\frac{1}{\sqrt{15}}$, which does not match. △XYZ: $\sin(X)=\frac{ZY}{XY}=\frac{6}{24}=\frac{1}{4}$, $\cos(X)=\frac{XZ}{XY}=\frac{6\sqrt{15}}{24}=\frac{\sqrt{15}}{4}$, $\tan(X)=\frac{ZY}{XZ}=\frac{6}{6\sqrt{15}}=\frac{1}{\sqrt{15}}$, which exactly matches the given values for ∠A. Wait, recheck △IJK: $\sin(I)=\frac{JK}{IK}=\frac{3}{3\sqrt{15}}=\frac{1}{\sqrt{15}}$, which is not $\frac{1}{4}$. So my earlier step 3 was wrong. Correct step:

Step3: Recheck △XYZ ratios

Sides: 6, $6\sqrt{15}$, 24. Divide by 6: $1:\sqrt{15}:4$, which matches the reference ratio $1:\sqrt{15}:4$. For the right angle at Z, the angle at X has $\sin(X)=\frac{6}{24}=\frac{1}{4}$, $\cos(X)=\frac{6\sqrt{15}}{24}=\frac{\sqrt{15}}{4}$, $\tan(X)=\frac{6}{6\sqrt{15}}=\frac{1}{\sqrt{15}}$, which exactly matches the given trigonometric values for ∠A.

Step4: Confirm △IJK does not match

△IJK sides: 3,12,$3\sqrt{15}$. $\sin(\text{angle})=\frac{3}{3\sqrt{15}}=\frac{1}{\sqrt{15}}
eq\frac{1}{4}$, so it does not match.

Answer:

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