angle ACB is in fact 90°, since AB is a diameter.
So the perimeter is
p = AB + CB + CA
= 6 + 6sin35° + 6sin55°
Given: ∆ABC, m∠A = 35°, Circle k(O, r=3), O∈ AB, AB is a diameter in the circle passing through point O. AC and CB are chords which intersect. Not stated that <ACB is 90 degrees.
Find: Perimeter of ∆ABC
2 answers
Thank you