Saved Bookmarks
| 1. |
In the isosceles triangle `ABC, |vec(AB)| = |vec(BC)| = 8`,a point E divide AB internally in the ratio `1:3`, then the cosine of the angle between `vec(CE)` and `vec(CA)` is (where `|vec(CA)| = 12`) |
|
Answer» `abs(AB)=abs(BC)=8` `abs(vecb)=8` `abs(vecc)=12` `abs(vecb-vecc)=8` `vec(CE)=vecE-vecC` `=vecb/4-vecC` Therefore, `abs(vecCE)=sqrt(148-(vecb*vecc)/2)` `a/sina=b/sinb=c/sinc` `cosA=3/4` Therefore `vecb*vecc=72` Put it in equation 1, we get, `sqrt(148-36)=sqrt112` In `triangle AEC`, `(CE)/sinA=(AE)/sinalpha` `sin A= sqrt7/4` Therefore, `(sqrt112*4)/sqrt3=2/sinalpha` Hence, `sinalpha=1/8` and `cosalpha=3sqrt7/8` |
|