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goblinko [34]
2 years ago
9

Two buildings are facing each other on a road of width 12 metre . From the top of the first building, which is 10 metre high , t

he angle of elevation of the top of the second is found to be 60°. What is the height of the second building ?
So easy question
Can u people solve ?
I dont think so​
Mathematics
1 answer:
Karo-lina-s [1.5K]2 years ago
8 0

Answer:

\displaystyle 10 + 12\sqrt{3} <em>metres</em>

Step-by-step explanation:

You are finding the <em>height</em><em> </em>of the building with an angle of elevation, therefore we need to solve for <em>EC</em><em> </em>to add it to <em>BA</em><em> </em>[10 metres] and use TRIGONOMETRIC RATIOS to arrive at our conclusion. Just in case you have forgotten what they were, here they are:

\displaystyle \frac{OPPOSITE}{HYPOTENUSE} = sin\:θ \\ \frac{ADJACENT}{HYPOTENUSE} = cos\:θ \\ \frac{OPPOSITE}{ADJACENT} = tan\:θ \\ \frac{HYPOTENUSE}{ADJACENT} = sec\:θ \\ \frac{HYPOTENUSE}{OPPOSITE} = csc\:θ \\ \frac{ADJACENT}{OPPOSITE} = cot\:θ

We can now solve for <em>EC</em>:

\displaystyle cot\:60 = \frac{12}{EC} → ECcot\:60 = 12 \\ \\ \frac{ECcot\:60}{cot\:60} = \frac{12}{cot\:60} → 12\sqrt{3} = EC

<em>OR</em>

\displaystyle tan\:60 = \frac{EC}{12} → 12tan\:60 = EC → 12\sqrt{3} = EC

Now that you have solved for <em>EC</em>, you can now add it to your original 10 metres to get \displaystyle 10 + 12\sqrt{3} <em>metres</em>. As a decimal, you would get \displaystyle 30,78460969... <em>metres</em>. You can go ahead and round this off if necessary.

** The reason why the <em>cotangent</em><em> </em>[or <em>tangent</em>] ratio was used was because <em>EA</em><em> </em>is equivalent to <em>DB</em> by the definition of a rectangle. It has two pairs of parallel and congruent sides with <em>four</em><em> </em><em>right</em><em> </em><em>angles</em>. Plus, that is the <em>adjacent</em><em> </em><em>side</em><em> </em>of the triangle, while <em>EC</em><em> </em>is the <em>opposite</em><em> </em><em>side</em><em> </em>of the triangle, so we knew our ratios were correct.

I am joyous to assist you at any time.

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