The problem given is one that involves measurements made in all the three dimensions and this should be kept in mind while solving the problem.

The tower is a vertical line with the base being X and Y being the topmost point. The line AB is a horizontal line. It...

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The problem given is one that involves measurements made in all the three dimensions and this should be kept in mind while solving the problem.

The tower is a vertical line with the base being X and Y being the topmost point. The line AB is a horizontal line. It is given that the angle of elevation of the line from the point A to the point Y is 42.6 degrees. The height of the tower can be determined if the distance of the point A from the point X is known. This is determined using the length of the line AB and the measure of angle XAB = 60 degrees and ABX = 81.65 degrees. The measure of angle AXB = 180 - 60 - 81.65 = 38.35. As AB = 50 m, use the law of sines to determine the length AX.

`(AX)/(sin 81.65) = (50)/(sin 38.35)`

=> `AX ~~ 79.73`m

The height of the tower is equal to h where `h/79.73 = tan 42.6`

=> `h ~~ 73.31` m

**The height of the tower is 73.31 m**

XY is the height of the tower (use X as the base)

AB is a 50 m baseline, some distance away from the tower.

Angle XAY = 42.6 (which is the vertical angle from ground level to the top of the tower)

Angle XAB = 60 (and is the horizontal angle between the line segment AX and AB

Angle ABX = 81.65 (the horizontal angle between the line segment AB and BX.

Find Angle AXB = 180 degrees - 60 degrees - 81.65 degrees = 38.35 degrees

Use law of sines to find distance AX:

sin 38.35/50 = sin 81.65 / AX and AX = 79.7 m

Use AX and the vertical angle to find the height of the tower:

42.6 degrees is the vertical angle between line AX and XY

Therefore,

Tan 42.6 = XY/79.7

**XY, the height of the tower is 73.3 m Answer**