Let n be a positive integer. In triangle ABC, AB = 3n, AC = 2n + 15, BC = n + 30, and angle \(A > \angle B > \angle C\). How many possible values of n are there?
The possible values of n are 8, 9, 10, and 11, giving us 4 possible values of n.
There are more than 4 values.
1) First draw up a triangle that looks possible.
All the angles are different so if you use a triangle that looks a bit like a 30,60, 90 degree triangle that will be good to work with.
2) Label the 3 angles correctly according to their relative angle size
3) Label the 3 sides
4) Now the smallest side is opposite the smallest angle etc so write 'smallest' and 'biggest' near the correct sides.
5) Now you know that smallest side < middle side Solve the inequality
6) You also know that the middle side < longest side. Solve the inequality.
7) So what do you know so far?
8) Now to be a triangle at all the sum of the 2 smallest sides must be longer than the long side.
Form the inequality and solve it.
9) So what have you found.
If you have questions then ask. You can take a photo of your pic on your phone and post it if need be.
I am very happy to help more if necessary but i want to be convinced that you are trying to help yourself first.
Please no one else answer
Thank you for you help!
Basically what I did was I made all the inequalities.
3n+2n+15>n+30 --------> n>15/2
3n+n+30> 2n+15 -------> n>15/4
2n+15+n+30>3n (no use)
I had to make a diagram for this next part.
We know BC>AC because largest side is opposite largest angle, smallest side is opposite smallest angle. We also know AC>AB from similar reasoning. This means
n+30>2n+15 -------> n<15
2n+15>3n ------> n<15
There are 11 values of n that satisfy all of the inequalities