Question

(Geometry), homework please help

Answer 1

Exercise 1:

The angle in the low left corner is supplementar with the external angle. Since the external angle is 105, the inner angle is 75.

Since in every triangle the inner angles sum to 180, we have

$x = 180-75-72 = 33$

Exercise 2:

Again, you need to use the fact that in every triangle, the sum on the inner angles is always 180.

Exercise 3:

Since PQ and BC are parallel, triangles APQ e ABC are similar. This means that their side are in the same proportion, so

$AP : AB = AQ : AC$

Plug the known values of AB, AP and AC and solve for AQ:

$6: 12= AQ : 18 \implies AQ = \frac{6\cdot 18}{12} = 9$