Question

applying similarity of right triangles
which triangle is similar to △abc if sin(a) = 1/4, cos(a) = √15/4, and tan(a) = 1/√15?

Explanation:

Step1: Recall sine - cosine - tangent definitions

In 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 $\sin(A)=\frac{1}{4}$, $\cos(A)=\frac{\sqrt{15}}{4}$, $\tan(A)=\frac{1}{\sqrt{15}}$, we know that the ratio of the opposite side to the hypotenuse is $1:4$ and the ratio of the adjacent side to the hypotenuse is $\sqrt{15}:4$, and the ratio of the opposite side to the adjacent side is $1:\sqrt{15}$.

Step2: Check the first triangle $\triangle IJK$

For $\triangle IJK$ with right - angle at $J$, if we consider the sides, $\sin(\angle I)=\frac{3}{12}=\frac{1}{4}$, $\cos(\angle I)=\frac{3\sqrt{15}}{12}=\frac{\sqrt{15}}{4}$, $\tan(\angle I)=\frac{3}{3\sqrt{15}}=\frac{1}{\sqrt{15}}$.

Step3: Check the second triangle $\triangle LMN$

For $\triangle LMN$ with right - angle at $N$, $\sin(\angle L)=\frac{3}{\sqrt{15 + 9}}=\frac{3}{\sqrt{24}}
eq\frac{1}{4}$, so it is not similar.

Step4: Check the third triangle $\triangle XYZ$

For $\triangle XYZ$ with right - angle at $Z$, $\sin(\angle X)=\frac{6}{24}=\frac{1}{4}$, $\cos(\angle X)=\frac{6\sqrt{15}}{24}=\frac{\sqrt{15}}{4}$, $\tan(\angle X)=\frac{6}{6\sqrt{15}}=\frac{1}{\sqrt{15}}$.

Answer:

The triangles $\triangle IJK$ and $\triangle XYZ$ are similar to $\triangle ABC$.