the length of the shadow of a pole on level ground increases by 90m when the angle of elevation of the sun changes from 58 degree to 36 degree calculate correct to three significant figure the height of the pole

We have a triangle formed by the change in the length of the shadow and the rays from the two angle of elevation to the top of the pole giving an angle 22° opposite to the increase in the length of the shadow

We have by sin rule;

90/(sin (22°) = (Initial ray from the top of the pole to the end of the shadow's length)/(sin(122°)

Let the initial ray from the top of the pole to the end of the shadow's length = l₁

90/(sin (22°) = l₁/(sin(122°)

l₁ = 90/(sin (22°) ×(sin(122°) = 283.3 m

Therefore;

The height of the pole = 283.3 m × sin(36°) = 166.52 m

The height of the pole= 167 m to three significant figures.

proz proz · Answer 2 · 2020-08-21T14:44:06+02:00

Answer:

height of pole = 121 m (to 3 significant figures

Step-by-step explanation:

From the diagram attached to this solution:

Let the length of the shadow = b

Let the height of the triangle = h

[tex]Triangle\ 1\\\\Tan\ (58)\ =\ \frac{h}{b}\\1.60\ =\ \frac{h}{b} - - - - - -(1)\\\\[/tex]

[tex]Tan\ (36) = \frac{h}{b+90} \\0.73\ =\ \frac{h}{b+90} - - - - (2)[/tex]

[tex]From\ equation\ (1)\\1.60 = \frac{h}{b}\\ cross-multiplying\\1.60b = h - - - - - (3)\\\\\\from\ equation\ (2)\\0.73\ =\ \frac{h}{b+90}\\ corss-multiplying\\0.73b\ +\ 65.7 = h\\h\ =\ 0.73b\ +\ 65.7- - - - - -(4)\\[/tex]

Notice that equation (3) = equation (4) = h

[tex]1.60b = 0.73b\ + 65.7\\1.60b\ -\ 0.73b\ = 65.7\\0.87b\ =\ 65.7\\b\ =\ \frac{65.7}{0.87} = 75.52\ m\\[/tex]

Now let us find the height of pole from equation (3)

1.60b = h - - - - - - (3)

where b = 75.52

h = 1.60 × 75.52

h = 120.83 m

= 121 m (to three significant figure