Finding angle in triangle and semicircle

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The discussion revolves around solving a geometry problem involving a semicircle and an isosceles triangle with a shared base AB and equal areas. The goal is to find the angle x, which is given as 57.5°. Participants suggest calculating the area of the semicircle using the formula 0.5 x pi x (0.5AB)^2, while also considering the triangle's area. To simplify the problem, one participant recommends splitting the triangle into two right-angled triangles. The conversation highlights the need for understanding the relationship between the areas of the shapes to solve for the angle.
jacky50
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Dear all,
right now, I am doing my IGCSE and this question is in my coursebook. I can't think of any way to calculate it...
It's part of the section circumference and area of circles. So must be some how solved using this.

The semi-circle and the isosceles triangle have the same base AB and the same area. Find the angle x.
The answer is: 57.5°

Thanks a lot in advance
Jacky
 

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Welcome to PF!

Hi Jacky! Welcome to PF! :wink:

Call the angle x, and the length AB 1 …

what are the areas of the triangle and the circle? :smile:
 


Hey tiny-tim, thanks :)

the area is not given, just that its the same
 
I know they're not given …

I'm asking you what they are
 


Area of semi-circle = 0.5 x pi x (0.5AB)^2
But I don't have a clue about the triangle
 
jacky50 said:
But I don't have a clue about the triangle

Split it into two right-angled triangles.
 

Thread 'Family of lines that are at a distance of 5 from the origin'

I picked up this problem from the Schaum's series book titled "College Mathematics" by Ayres/Schmidt. It is a solved problem in the book. But what surprised me was that the solution to this problem was given in one line without any explanation. I could, therefore, not understand how the given one-line solution was reached. The one-line solution in the book says: The equation is ##x \cos{\omega} +y \sin{\omega} - 5 = 0##, ##\omega## being the parameter. From my side, the only thing I could...
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