The shadow of a post is 6m longer when the elevation of the sun is 30° than when it is 60° calculate the height of the post. Pls help me. Thank you​

The height of the pole is 5.2m

Step-by-step explanation:

Suppose that we have a triangle rectangle, where the height of the post is one cathetus, the shadow is another cathetus, and the angle of elevation of the sun is measured from the vertex at the end of the shadow.

The adjacent cathetus is the shadow.

We can use the relation:

Tan(θ) = (opposite cathetus)/(adjacet cathetus)

Then when the angle of elevation is 30°, we will have:

Tan(30°) = H/S

Where:

H = height

S = length of the shadow.

And when the elevation is 30°, the length of the shadow is 6 m longer than when the elevation is 60°.

So when the elevation is 60° the shadow will be 6 meters shorter, then we have the equation:

Tan(60°) = H/(S - 6m)

And i will rewrite this as:

Tan(60°)*(S - 6m) = H

Then we have the system of equations:

Tan(30°) = H/S

Tan(60°)*(S - 6m) = H

To solve this, we need to isolate S in one of the equations, let's isolate S in the first one.

S = H/Tan(30°)

Now we can replace this in the other equation to get:

Tan(60°)*(H/tan(30°) - 6m) = H

Now we can solve this for H.

(Tan(60°)/Tan(30°)*H  - Tan(60°)*6m = H

3*H - 10.39m = H

3*H - H = 10.39m

2*H = 10.39m

H = 10.39m/2 = 5.2m

The height of the pole is 5.2m

three-fourths of an hour

36÷48=0.75

0.75=

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step-by-step explanation:

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