Question

determining the number of possible triangles how many distinct triangles can be formed for which ( mangle x = 51^circ ), ( x = 5 ), and ( y = 2 )? options: zero, one, two

Explanation:

Step1: Recall the Law of Sines

The Law of Sines states that $\frac{\sin A}{a}=\frac{\sin B}{b}=\frac{\sin C}{c}$, where $A,B,C$ are angles and $a,b,c$ are the lengths of the sides opposite those angles respectively. For our problem, we have $\angle X = 51^\circ$, side $x = 5$ (opposite $\angle X$), and side $y=2$ (opposite $\angle Y$). So we can write $\frac{\sin X}{x}=\frac{\sin Y}{y}$.

Step2: Solve for $\sin Y$

Substitute the known values into the Law of Sines formula: $\frac{\sin 51^\circ}{5}=\frac{\sin Y}{2}$. First, calculate $\sin 51^\circ\approx0.7771$. Then we have $\sin Y=\frac{2\times\sin 51^\circ}{5}=\frac{2\times0.7771}{5}=\frac{1.5542}{5} = 0.31084$.

Step3: Analyze the value of $\sin Y$

We know that the range of the sine function is $[- 1,1]$, and $0.31084\in[-1,1]$. Now, we also know that if we have $\sin\theta = k$, then $\theta=\arcsin(k)$ or $\theta = 180^\circ-\arcsin(k)$ (for $k\in[0,1]$). But we also need to check the triangle angle - sum and the side - angle relationship. The side opposite $\angle X$ is $x = 5$ and the side opposite $\angle Y$ is $y = 2$. Since $x>y$, then $\angle X>\angle Y$ (in a triangle, the larger side is opposite the larger angle). We found that $\sin Y=0.31084$, so $\angle Y=\arcsin(0.31084)\approx18.1^\circ$ (the other solution would be $180 - 18.1=161.9^\circ$, but if $\angle Y = 161.9^\circ$, then $\angle X+\angle Y=51 + 161.9=212.9^\circ>180^\circ$, which is not possible for a triangle). Also, since $x = 5>y = 2$, and we have a valid acute angle for $\angle Y$ (less than $\angle X$), we can form exactly one triangle.

Answer:

one