Question

Which triangle is similar to ABC if sin(A) =cos(A) = vs., and tan(A) = VIS?1_JXR3кMол51312V153246V15のS3v15L V6 N12YTN6

Answer 1

Suppose that triangle ABC is a right triangle; therefore, we can use the following trigonometric identities,

$\sin \theta=\frac{O}{H},\cos \theta=\frac{A}{H},\tan \theta=\frac{O}{A}$

Then, in our case,

$\begin{gathered} \frac{1}{4}=\frac{O}{H},\frac{A}{H}=\frac{\sqrt[]{15}}{4},\frac{O}{A}=\frac{1}{\sqrt[]{15}} \\ \Rightarrow O=1.A=\sqrt[]{15},H=4 \end{gathered}$

In a diagram,

Comparing the ratio of the corresponding sides of triangles ABC and RST

$\begin{gathered} \frac{13}{4}=3.25 \\ \frac{12}{1}=12,\frac{12}{\sqrt[]{15}}=3.09833 \end{gathered}$

The ratio between corresponding sides is not constant; thus, triangle RTS is not similar to ABC.

On the other hand, as for triangle KJI

$\begin{gathered} \frac{12}{4}=3 \\ \frac{3\sqrt[]{15}}{\sqrt[]{15}}=3 \\ \frac{3}{1}=3 \end{gathered}$

The ratio is constant; therefore, the answer is triangle JKI