If CD is square root 5 isn't BD half of square root 5?
I know the numbers are wrong because they have to be integers but I'm just asking about the rates.
In triangle ABC, angle C is a right angle and CB>CA. Point D is located on BC so that angle CAD is twice angle DAB.
If AC/AD = 2/3, then CD/BD = m/n, where m and n are relatively prime positive integers. Find m+n
All I have is the sketch.
I'm working on it now but I think I'll really need some hints.
Veronica, this is not an "easy" problem; and we don't know "where you're at"...
OK, let's see:
start with the smaller right triangle ACD: make AC = 2 , AD = 3 ; so CD = sqrt(5) (as you show!).
now, can you calculate angle CAD?
extend CD to CB, such that angle CAB = 1.5 * angle CAD (makes angle CAD twice angle BAD);
now, can you calculate length of BC?
I hope you'll be patient with me because my math is very fragmented.
I learn theorems through the problems so certain areas I'm strong but others I hardly know the basics. But I'm a very quick learner.
First, I changed the rate to 2square root 5 and 3 square root 5 so I can at least work with one integer 5 for CD.
I calculated the value for arccos (0.666666667) =48.1896
This makes the angle CAB 72.2796.
tan 72.2796 =3.12956
BD + 5 / 2square root5 = 3.129
BD=8.992888 which I changed BD=9
and got CD/BD =5/9.
Am I allowed to change the value of 8.992888 to 9?
[QUOTE=Veronica1999;587931]Am I allowed to change the value of 8.992888 to 9[QUOTE]
No need to: it is already 9!
Need to use more precise angles: 48.1896851042214... * (3/2) = 72.2845276563321...
10/18, 15/27, 20/36 ...... also solutions, but not coprimes.