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fall 2025 geometry b wwva right triangle relationships and trigonometry…

Question

fall 2025 geometry b wwva right triangle relationships and trigonometry which equation is correct and can be used to solve for the value of z? $\frac{sin(51^{circ})}{2.6}=\frac{sin(53^{circ})}{z}$ $\frac{sin(76^{circ})}{2.6}=\frac{sin(53^{circ})}{z}$ $\frac{sin(51^{circ})}{2.6}=\frac{sin(76^{circ})}{z}$

Explanation:

Step1: Recall the Law of Sines

The Law of Sines states that in a triangle $\frac{\sin A}{a}=\frac{\sin B}{b}=\frac{\sin C}{c}$, where $A$, $B$, $C$ are angles of the triangle and $a$, $b$, $c$ are the lengths of the sides opposite to those angles respectively.

Step2: Identify angles and sides

In the given triangle, we have side - length $2.6$ opposite the angle of $76^{\circ}$ and side - length $z$ opposite the angle of $51^{\circ}$.

Step3: Apply the Law of Sines

According to the Law of Sines, $\frac{\sin(51^{\circ})}{z}=\frac{\sin(76^{\circ})}{2.6}$, which can be rewritten as $\frac{\sin(76^{\circ})}{2.6}=\frac{\sin(51^{\circ})}{z}$.

Answer:

$\frac{\sin(76^{\circ})}{2.6}=\frac{\sin(51^{\circ})}{z}$