Answer:
As per the given statement:
The distance from jack eyes to the top of the tower = 50.64 m
Also, jack and john eyes are 1.5 m from the ground.
Labelled diagram as shown in the attachment
Let BC = x m be the distance of the jack from the base of the tower and
AG = y m be the distance above 1.5 m from the the ground.
Using tangent ratio:
[tex]\tan \theta = \frac{\text{Opposite side}}{\text{Adjacent side}}[/tex]
In triangle AGE;
[tex]\tan 40^{\circ} = \frac{AG}{EG}[/tex]
Here, AG = y m and EG = 20+x m
then;
[tex]\tan 40^{\circ} = \frac{y}{20+x}[/tex]
or we can write this as;
[tex]y = \tan 40^{\circ} (20+x)[/tex] ........[1]
Similarly, in AGF;
[tex]\tan 60^{\circ} = \frac{AG}{FG}[/tex]
then;
[tex]\tan 60^{\circ} = \frac{y}{x}[/tex]
[tex]\sqrt{3} = \frac{y}{x}[/tex]
or
[tex]y = \sqrt{3}x[/tex] ......[2]
Substitute equation [2] into the equation [1], to solve for x;
[tex]\sqrt{3}x= \tan 40^{\circ}(20+x)[/tex]
or
[tex]\sqrt{3}x= 0.84(20+x)[/tex]
Using distributive property: [tex]a\cdot (b+c) = a\cdot b+ a\cdot c[/tex]
we have;
[tex]1.732x = 16.8 + 0.84x[/tex]
Subtract 0.84 x from both sides we get;
[tex]0.892x = 16.8[/tex]
Divide both sides by 0.892 we get;
[tex]x \approx 18.8 m[/tex]
Therefore, 18.8 meter johns from the base of the tower.
