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DaniilM [7]
3 years ago
13

Benjamin is trying to find the height of a radio antenna on the roof of a local building.

Mathematics
1 answer:
xz_007 [3.2K]3 years ago
8 0

The height of the antenna on the roof of the local building is approximately 8 meters.

The situation forms a right angle triangle.

<h3>Properties of a right angle triangle:</h3>
  • One of its angles is equals to 90 degrees
  • The sides of the triangles can be calculated using Pythagoras theorem.

Therefore, let's find the height of the building and the radio antenna from the eye point.

Using trigonometric ratios,

tan 40° = opposite / adjacent

tan 40° = x / 25

where

x = the height of the building and the radio antenna from the eye point.

x = 25 tan 40

x = 25 × 0.83909963117

x = 20.9774907794 meters

Let's find the height of the building from his eye point.

tan 28° = y / 25

where

y = height of the building from his eye point

y = 25 × tan 28°

y = 25 × 0.53170943166

y = 13.2927357915 meters

Height of the antenna = 20.9774907794 - 13.2927357915 = 7.68475498786

Height of the antenna ≈ 8 meters

learn more on elevation here: brainly.com/question/17582385?referrer=searchResults

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<h3>What is an equation?</h3>

An equation is an expression that shows the relationship between two or more numbers and variables.

Given that equation:

\frac{5-x}{x^2+3x-4} /\frac{x^2-2x - 15}{x^2+5x+4}

=\frac{5-x}{(x+4)(x-1)} /\frac{(x-5)(x+3)}{(x+4)(x+1)} \\\\=\frac{5-x}{(x+4)(x-1)} * \frac{(x+4)(x+1)}{(x-5)(x+3)}\\\\\frac{-(x-5)}{(x+4)(x-1)} * \frac{(x+4)(x+1)}{(x-5)(x+3)}\\\\=\frac{-(x+1)}{(x-1)(x+3)}

The quotient when \frac{5-x}{x^2+3x-4} /\frac{x^2-2x - 15}{x^2+5x+4} in simplified form is \frac{-(x+1)}{(x-1)(x+3)}

Find out more on equation at: brainly.com/question/2972832

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