Saved Bookmarks
| 1. |
In an acute angled triangle ABC, if tan(A+B-c) = I and sec(B + C-A)=2, find the values ofA, B and C. |
|
Answer» Tan ( A + b - c) = 1 tan ( a+ b - c) = tan 45°a+b+-c = 45°. {1}By angle sum property a +b+ c = 180°. { 2}also, sec( b +c - a) = 2b + c - a = 60°. { 3}from eqn 1 we get a + b = 45 + Cput this value in eqn 245+c + c = 1802c = 180 - 45 c = 135/2 = 67.5 °put this value in eqn 1 a +b = 45+67.5a+ b = 112.5 (4)also in eqn 3 B + 67.5 - A = 60-A + B = - 7.5A - B = 7.5 (5)eliminating (4) from (5)A = 105/2 ° = 52.5° B = 45° |
|