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From her eye which stands 1.69 m above the ground and sitting measures the angle of elevation to the top of a prominent size creeper to be 36° if she's standing at a horizontal distance of 275 m from the peso's ice cream what is the height of the size of a skyscrape scraper 110 years 100th of a meter if necessary

Answer :

The height of the skyscraper is approximately 198.36 meters.

To find the height of the skyscraper, we can use trigonometry. We have a right triangle formed by the observer (her eye), the top of the skyscraper, and the base of the skyscraper (the ground).

Given:

- The angle of elevation from her eye to the top of the skyscraper is 36°.

- The horizontal distance from her eye to the base of the skyscraper is 275 m.

- The height of her eye above the ground is 1.69 m.

We can use the tangent function, which relates the angle of elevation [tex](\( \theta \))[/tex] to the opposite side (height of the skyscraper, ( h ) and the adjacent side (horizontal distance, ( d ):

[tex]\[ \tan(\theta) = \frac{\text{opposite}}{\text{adjacent}} \][/tex]

Substituting the given values:

[tex]\[ \tan(36^\circ) = \frac{h + 1.69}{275} \][/tex]

Now, we solve for ( h ):

[tex]\[ h + 1.69 = 275 \times \tan(36^\circ) \][/tex]

[tex]\[ h = 275 \times \tan(36^\circ) - 1.69 \][/tex]

Let's calculate ( h ):

[tex]\[ h = 275 \times \tan(36^\circ) - 1.69 \][/tex]

[tex]\[ h \approx 275 \times 0.726542528 - 1.69 \][/tex]

[tex]\[ h \approx 200.0496948 - 1.69 \][/tex]

[tex]\[ h \approx 198.3596948 \][/tex]

Therefore, the height of the skyscraper is approximately 198.36 meters.

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Rewritten by : Barada