Question

question 7(multiple choice worth 1 points) (05.02 mc) if sin(y°) = cos(x°), which of the following statements is true? y = x and △abc ≅ △fed w = x and △abc ≅ △fed y = x and △abc ~ △fed w = x and △abc ~ △fed

Explanation:

Step1: Recall trigonometric identity

In a right - triangle, $\sin\theta=\cos(90^{\circ}-\theta)$. Given $\sin(y^{\circ})=\cos(x^{\circ})$, then $x + y=90$. Also, in right - triangle $\triangle ABC$ and $\triangle FED$, $\angle B=\angle E = 90^{\circ}$.
For right - triangles, if two angles of one right - triangle are equal to two angles of another right - triangle, the triangles are similar.
We know that in $\triangle ABC$ and $\triangle FED$, $\angle B=\angle E = 90^{\circ}$. And since $\sin(y^{\circ})=\cos(x^{\circ})$, we have that $\angle A=x^{\circ}$ and in $\triangle FED$, the non - right angle corresponding to the cosine value related to $\sin(y^{\circ})$ gives us that the non - right angles are equal.
In $\triangle FED$, if we consider the angle relationships, we know that $\angle F+\angle D = 90^{\circ}$. Since $\sin(y^{\circ})=\cos(x^{\circ})$, we can show that $\angle F=x^{\circ}$ (because of the co - function identity $\sin\alpha=\cos(90^{\circ}-\alpha)$). So $w = x$. And $\triangle ABC\sim\triangle FED$ (by the AA (angle - angle) similarity criterion for right - triangles).

Answer:

$w = x$ and $\triangle ABC\sim\triangle FED$