Question

in triangle abc if tan(a+b-c)=1 and sec (b+c-a)=2 find the valu of a b and c​

Answer 1

Tan (A+B-C) = 1 Tan 45° = 1 Hence, A+B-C = 45° ........(i) Sec (B+C-A) = 2 Sec 60° = 2 Hence, B+C-A = 60° .......(ii) Adding equation (i) and (ii), we get, A+B-C + B+C-A = 45° + 60° 2B = 105° B = 52.5° Hence, A+B-C = 45° A+52.5° -C = 45° A-C = 45° - 52.5° A-C = -7.5° ............(iii) Now, A+B+C = 180° Hence, A + 52.5° + C = 180° (Angles of the triangle) A + C = 180° - 52.5° A + C = 127.5° .........(iv) Now, Adding equation (iii) and (iv), we get, A-C + A + C = -7.5° + 127.5° 2A = 120° A = 120°/2 A = 60° C = 127.5° - 60° C = 67.5° Hence, A = 60°, B = 52.5° and C = 67.5°.Read more on Sarthaks.com - https://www.sarthaks.com/282905/in-an-acute-angled-triangle-abc-if-tan-a-b-c-1-and-sec-b-c-a-2-find-the-values-of-a-b-c