# Puzzle Solutions 16-30

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

solve-my-maths-solutions-16-30

## 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. Fixed. Many thanks for the spot ðŸ™‚

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 90Âº 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 360Âº. Then:
90+90+2delta+theta=360. If theta = delta – 10, you do not get delta = 65Âº and theta = 55Âº.
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 90Âº 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 360Âº. Then:
90+90+2delta+theta=360.
If theta =delta-10 you don’t get 55Âº and 65Âº.
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 (55Âº and 65Âº).