Suppose that triangle ABC is a right triangle; therefore, we can use the following trigonometric identities,
![\sin \theta=(O)/(H),\cos \theta=(A)/(H),\tan \theta=(O)/(A)](https://img.qammunity.org/2023/formulas/mathematics/college/bnz2yd6pnyb74jy96dwp8rb3oast8o83ce.png)
Then, in our case,
![\begin{gathered} (1)/(4)=(O)/(H),(A)/(H)=\frac{\sqrt[]{15}}{4},(O)/(A)=\frac{1}{\sqrt[]{15}} \\ \Rightarrow O=1.A=\sqrt[]{15},H=4 \end{gathered}](https://img.qammunity.org/2023/formulas/mathematics/college/ko2bzia17b8qirrtl54l1cnyabsx2e06m2.png)
In a diagram,
Comparing the ratio of the corresponding sides of triangles ABC and RST
![\begin{gathered} (13)/(4)=3.25 \\ (12)/(1)=12,\frac{12}{\sqrt[]{15}}=3.09833 \end{gathered}](https://img.qammunity.org/2023/formulas/mathematics/college/v77jw1avetoseycoka81pdm7hqa51oa3bx.png)
The ratio between corresponding sides is not constant; thus, triangle RTS is not similar to ABC.
On the other hand, as for triangle KJI
![\begin{gathered} (12)/(4)=3 \\ \frac{3\sqrt[]{15}}{\sqrt[]{15}}=3 \\ (3)/(1)=3 \end{gathered}](https://img.qammunity.org/2023/formulas/mathematics/college/kgqioinau7g4oiet5ska9jo94mqa4ye8wk.png)
The ratio is constant; therefore, the answer is triangle JKI