Question

right triangles hjk and xyz, shown in the diagram, are similar. diagram: triangle hjk (right angle at j, hj = 15, jk = 10); triangle xyz (right angle at y, xy = 9, yz = 6) which set of steps correctly shows the relationship between sinh and sinx?
a. $\frac{10}{6} = \frac{hk}{xz}$; $\frac{10}{hk} = \frac{6}{xz}$; $sin h = sin x$
b. $\frac{15}{9} = \frac{hk}{xz}$; $\frac{15}{hk} = \frac{9}{xz}$; $sin h = sin x$
c. $\frac{10}{6}
eq \frac{hk}{xz}$; $\frac{10}{hk}
eq \frac{6}{xz}$; $sin h
eq sin x$
d. $\frac{15}{9}
eq \frac{hk}{xz}$; $\frac{15}{hk}
eq \frac{9}{xz}$; $sin h
eq sin x$

Explanation:

Step1: Recall Similar Triangles Property

Similar triangles have corresponding angles equal. So, \(\angle H=\angle X\) (since \(\triangle HJK \sim \triangle XYZ\) and they are right - angled, the non - right angles correspond).

Step2: Recall Sine of an Angle in a Right Triangle

In a right triangle, \(\sin\theta=\frac{\text{opposite}}{\text{hypotenuse}}\). For \(\angle H\) in \(\triangle HJK\), \(\sin H=\frac{JK}{HK}\). For \(\angle X\) in \(\triangle XYZ\), \(\sin X = \frac{YZ}{XZ}\).
Since \(\triangle HJK\sim\triangle XYZ\), the ratios of corresponding sides are equal. That is, \(\frac{JK}{YZ}=\frac{HK}{XZ}\), so \(\frac{10}{6}=\frac{HK}{XZ}\), and cross - multiplying gives \(\frac{10}{HK}=\frac{6}{XZ}\).
Also, since \(\angle H = \angle X\), \(\sin H=\sin X\) (because the sine of equal angles is equal). Looking at the options, option A follows these steps.

Answer:

A. \(\frac{10}{6}=\frac{HK}{XZ}\), \(\frac{10}{HK}=\frac{6}{XZ}\), \(\sin H = \sin X\)