Answer:
[tex]\displaystyle 10 + 12\sqrt{3}[/tex] metres
Step-by-step explanation:
You are finding the height of the building with an angle of elevation, therefore we need to solve for EC to add it to BA [10 metres] and use TRIGONOMETRIC RATIOS to arrive at our conclusion. Just in case you have forgotten what they were, here they are:
[tex]\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[/tex]
We can now solve for EC:
[tex]\displaystyle cot\:60 = \frac{12}{EC} → ECcot\:60 = 12 \\ \\ \frac{ECcot\:60}{cot\:60} = \frac{12}{cot\:60} → 12\sqrt{3} = EC[/tex]
OR
[tex]\displaystyle tan\:60 = \frac{EC}{12} → 12tan\:60 = EC → 12\sqrt{3} = EC[/tex]
Now that you have solved for EC, you can now add it to your original 10 metres to get [tex]\displaystyle 10 + 12\sqrt{3}[/tex] metres. As a decimal, you would get [tex]\displaystyle 30,78460969...[/tex] metres. You can go ahead and round this off if necessary.
** The reason why the cotangent [or tangent] ratio was used was because EA is equivalent to DB by the definition of a rectangle. It has two pairs of parallel and congruent sides with four right angles. Plus, that is the adjacent side of the triangle, while EC is the opposite side of the triangle, so we knew our ratios were correct.
I am joyous to assist you at any time.
