In triangle ABC, D is the mid point of BC. If AD is the perpendicular to AC, then cosAcosC= ?[1] 2(c2+a2) / 3ac[2] (c2-a2) / 3ac[3] 3(c2+a2) /3ac[4] 2(c2-a2) /3ac
In triangle ABC, D is the mid point of BC. If AD is the perpendicular to AC, then cosAcosC= ?
[1] 2(c2+a2) / 3ac
[2] (c2-a2) / 3ac
[3] 3(c2+a2) /3ac
[4] 2(c2-a2) /3ac
Shubhashish singh , 10 Years ago
Grade 11
