Question

how many distinct triangles can be formed for which ( mangle e = 64^circ ), ( g = 9 ), and ( e = 10 )?
1 ✔️ triangle(s)
how many distinct triangles can be formed for which ( mangle j = 120^circ ), ( k = 8 ), and ( j = 3 )?
1 ❌ triangle(s)

Explanation:

First Problem (m∠E = 64°, g = 9, e = 10)

Step1: Recall Law of Sines

The Law of Sines states $\frac{\sin G}{g}=\frac{\sin E}{e}$. We know $m\angle E = 64^\circ$, $g = 9$, $e = 10$.

Step2: Solve for $\sin G$

Substitute values into Law of Sines: $\sin G=\frac{g\sin E}{e}=\frac{9\sin64^\circ}{10}$. Calculate $\sin64^\circ\approx0.8988$, so $\sin G=\frac{9\times0.8988}{10}\approx0.8089$.

Step3: Analyze possible angles for G

Since $\sin G\approx0.8089$, $G\approx54^\circ$ or $G\approx180^\circ - 54^\circ = 126^\circ$. But check if $G + E<180^\circ$. For $G = 126^\circ$, $126^\circ+64^\circ = 190^\circ>180^\circ$, so only one valid angle for G. Thus, 1 triangle.

Second Problem (m∠J = 120°, k = 8, j = 3)

Step1: Recall Law of Sines

Law of Sines: $\frac{\sin K}{k}=\frac{\sin J}{j}$. Given $m\angle J = 120^\circ$, $k = 8$, $j = 3$.

Step2: Solve for $\sin K$

Substitute: $\sin K=\frac{k\sin J}{j}=\frac{8\sin120^\circ}{3}$. $\sin120^\circ=\frac{\sqrt{3}}{2}\approx0.8660$, so $\sin K=\frac{8\times0.8660}{3}\approx2.309$.

Step3: Analyze validity

Since the range of sine is $[-1,1]$, $\sin K\approx2.309$ is impossible. So 0 triangles.

Answer:

First problem: 1 triangle.
Second problem: 0 triangles.