The value of g across from angle G is 5feet
According to sine rule
[tex]\frac{e}{sinE}=\frac{f}{sinF}=\frac{g}{sinG}[/tex]
Given the following
∠G = 45°
∠F = 82°
f = 7feet
Required
side g
Substitute the given values into the formula
[tex]\frac{f}{sinF}=\frac{g}{sinG}\\ \frac{7}{sin82}=\frac{g}{sin45}\\\frac{7}{0.9903}=\frac{g}{0.7071}\\7.0686=\frac{g}{0.7071}\\g=7.0686*0.7071\\g=4.998\\g\approx5ft[/tex]
Hence the value of g across from angle G is 5feet
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The length of the line segment EF ([tex]g[/tex]) is approximately 5 feet.
Let be [tex]EFG[/tex] a Triangle, whose expression derived from the Law of Sines is described below:
[tex]\frac{e}{\sin E} = \frac{f}{\sin F} = \frac{g}{\sin G}[/tex] (1)
Where:
[tex]e[/tex] - Measure of the line segment FG, in feet.
[tex]f[/tex] - Measure of the line segment EG, in feet.
[tex]g[/tex] - Measure of the line segment EF, in feet.
[tex]E[/tex] - Angle at vertex E, in sexagesimal degrees.
[tex]F[/tex] - Angle at vertex F, in sexagesimal degrees.
[tex]G[/tex] - Angle at vertex G, in sexagesimal degrees.
We can determine a missing length, by knowing the length of a neighboring side and two consecutive angles. If we know that [tex]f = 7\,ft[/tex], [tex]G = 45^{\circ}[/tex] and [tex]F = 82^{\circ}[/tex], then the measure of the line segment EF is:
[tex]g = f\cdot \left(\frac{\sin G}{\sin F} \right)[/tex] (2)
[tex]g = (7\,ft)\cdot \left(\frac{\sin 45^{\circ}}{\sin 82^{\circ}} \right)[/tex]
[tex]g\approx 4.998\,ft[/tex]
[tex]g \approx 5\,ft[/tex]
The length of the line segment EF ([tex]g[/tex]) is approximately 5 feet.
