According to the solution of 1/ radical sign10/10 = c/sin150, c= radical sign 10/2 is obtained.
2. According to the half-angle formula, the square of cosb/2 = (1+COSB)/2 ∴ COSB = 3/5 b = 53.
∫ s =1/2absinc =1/2× 2× (2× sin53/sin82) radical number 2/2=
Inappropriate ... the data is strange ...
3.(2a-c)cosB=bcosC sine theorem:
(2 Sina-sinC)cosB = sinb cosc 2 Sina cosB = sinb cosc+sinC cosB
2 sinA cosb = sin(B+C)2 sinA cosb = sinA
Cosb = 1/2 b = 60 b = (incompetent again)
4. According to sine theorem, tanb/tanc = (2a-c)/c = (2sina-sibc)/sinc, which is tangent to a chord, that is, sinB*cosC=2sinA*cosB-sinC*cosB. Therefore, sin (b+c) = 2sina * is obtained by using sine sum angle formula. So B=60. SinA/sinC= radical number 3- 1, so sin( 120-C)/sinC= radical number 3- 1, so cotC=2- radical number 3. So C=75 degrees and A=45 degrees. B=60 degrees.