IB Maths-teacher took a long time to try to solve it
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Dr.Peterson
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Presumably, C is where theta is marked, right?
You can find alpha by simple trig. From that, you can find theta. The arc is not needed for these steps, just triangles.
Then find the area.
Please show your work, as far as you get; that's part of what we ask in our submission guidelines.
The question as presented is rather obscure. The AREA of the wall?
Presumably, the angles at the corners are 90 degrees. And presumably C is where [MATH]\theta[/MATH] is indicated, but it would be nice to be told that.
If all those guesses are correct, I am missing what is difficult about the problem. The radius of the circle is
[MATH]\sqrt{15^2 + 5^2} = \sqrt{225 + 25} = \sqrt{25 * 10} = 5\sqrt{10}.[/MATH]
[MATH]\theta = \pi - 2 \alpha.[/MATH]
[MATH]\alpha = arctan \left ( \dfrac{5}{15} \right ) [/MATH],
and so on.
Presumably, the angles at the corners are 90 degrees. And presumably C is where [MATH]\theta[/MATH] is indicated, but it would be nice to be told that.
If all those guesses are correct, I am missing what is difficult about the problem. The radius of the circle is
[MATH]\sqrt{15^2 + 5^2} = \sqrt{225 + 25} = \sqrt{25 * 10} = 5\sqrt{10}.[/MATH]
[MATH]\theta = \pi - 2 \alpha.[/MATH]
[MATH]\alpha = arctan \left ( \dfrac{5}{15} \right ) [/MATH],
and so on.