Question

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8. Find the angle of elevation of the sun when a 12.5 meter tall pole casts an 18 meter shadow.

Answer 1

`\displaystyle 35\degree`

Explanation:

When drawn out, you will notise you must use the inverce cotangent trigonometric ratio:

`\displaystyle ? = cot^(-1)\:(18)/(12,5); 34,777831366...\degree \\ \\ \boxed{35\degree \approx\:?} \\ \\ \\ OR \\ \\ \\ ? = tan^(-1)\:(12,5)/(18); 34,777831366...\degree \\ \\ \boxed{35\degree \approx\:?}`

I am joyous to assist you at any time.

Extended information on trigonometric ratios

`\displaystyle (OPPOCITE)/(HYPOTENUSE) = sin\:\theta \\ (ADJACENT)/(HYPOTENUSE) = cos\:\theta \\ (OPPOCITE)/(ADJACENT) = tan\:\theta \\ (HYPOTENUSE)/(ADJACENT) = sec\:\theta \\ (HYPOTENUSE)/(OPPOCITE) = csc\:\theta \\ (ADJACENT)/(OPPOCITE) = cot\:\theta`

Answer 2

Answer: 34.8 degrees (approximate)

The diagram is shown below.

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Work Shown:

tan(angle) = opposite/adjacent

tan(theta) = 12.5/18

theta = arctan(12.5/18)

theta = 34.7778313663639

theta = 34.8 degrees

Round this value however your teacher instructs. I decided to round to one decimal place because the 12.5 is to one decimal place. Make sure your calculator is in degree mode.

The notation arctan refers to arctangent. It is the same as inverse tangent, and it is often shown as
`\tan^(-1)` on many calculators.