The distance from the tower at point Z and X to the fire at point Y is approximately 0.0005779 meters and 1416.15 meters respectively
To find the distance from the tower at point Z to the fire at point Y, we can use the law of sines.
The law of sines states that the ratio of the length of a side of a triangle to the sine of the opposite angle is the same for all three sides.
In this case, we can set up the following equation:
sin(YZX) / XY = sin(YXZ) / XZ
Now we can plug in the known values:
sin(62°) / XY = sin(16°) / 442
Solving for XY, we get:
XY = 0.88295 * 442 / 0.2756
XY≈1416.15 meters.
Solving for ZY :
sin(102°) / ZY = sin(16°) / 442
0.927 / ZY = 0.2756 / 442
ZY=0.927* 0.2756 / 442
ZY≈0.0005779 meters
Using a calculator, we can calculate XY to be approximately 1416.15 meters and ZY to be approximately 0.0005779 meters.
Therefore, the distance from the tower at point Z and X to the fire at point Y is approximately 0.0005779 meters and 1416.15 meters. respectively
The probable question may be:
In triangle YZX ,Two lookout towers have been set up as X and Z to keep a watch on forest fires. The two towers are represented by the points X and Z in the figure. A fire is spotted at point Y and each tower measures the angle from the opposite tower to the fire as shown in the figure. If the distance between the two towers XZ is 442 meters, what is the distance, in meters, from the tower at point Z to the fire as YZ and YX as the angle of YZX= 62 degree and angle of YXZ=102 degree?