# Puzzle Solutions 16-30

The PowerPoint below highlights solutions and some possible approaches to problems 16-30. solve-my-maths-solutions-16-30

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## 5 thoughts on “Puzzle Solutions 16-30”

1. Jeremy Warnes says:

Slide 23?

[image: Inline images 1] Fantastic set of problems though. Thank you for sharing.

Jeremy

2. solvemymaths says:

Fixed. Many thanks for the spot 🙂

3. Mar says:

Let’s draw two radius from the circle center to the points where the circle and the triangle are tangent. The angle between those lines (radius) is 2 times delta.
The angle between the radius and the triangle side is 90º since they are tangent.
Let’s consider now the quadrilateral formed by the circle center, the tangent points and the vertex of the triangle (angle theta).
The addition of the four inner angles of a quadrilateral is supposed to be 360º. Then:
90+90+2delta+theta=360. If theta = delta – 10, you do not get delta = 65º and theta = 55º.
Could you please tell me what is wrong in this reasoning? I can’t find the mistake.

4. Mar says:

Let’s draw two radius from the circle center to the points where the circle and the triangle are tangent. The angle between these radius is 2 times delta.
The angle between the radius and the triangle side is 90º since they are tangent.
Let’s consider the quadrilateral formed by the circle center, the tangent points and the triangle vertex (angle theta).
The addition of the inner angles in any quadrilateral is supposed to be 360º. Then:
90+90+2delta+theta=360.
If theta =delta-10 you don’t get 55º and 65º.
Could you please tell me what’s wrong in this reasoning? I can’t see the mistake.
Thanks

5. Mar says:

Forget about that. I found the mistake and I got the right solution (55º and 65º).