Question

In the ambiguous case, if b and A are given and $c_1,c_2$ are the two values of
third side, prove that
(i) $c_1+c_2=2b\cos A$ and $c_1{c}_2=b^2-a^2$.
(ii) $c_1^2-2c_1{c}_2\cos 2A+c_2^2=4a^2{\cos }^{2}A.$
(iii) $(c_1-c_2{)}^2+(c_1+c_2{)}^2{\tan }^{2}A=4a^2$.

Answer 1

the quadratic given the two values $c_1.c_2$ of $c^2-2bccosA+(b^2-a^2)=0$
Then we have
(i) $c_1+c_2=2bcosa$ ........(1)
and $c_1{c}_2=b^2-a^2$ ........(2)
(ii) From (1) and (2),
$(c_1+c_2{)}^2=4b^{2}co{s}^{2}A=4(c_1{c}_2+a^2)co{s}^{2}A$
or $c_1^2+2c_1{c}_2+c_z^2-4c_1{c}_{2}co{s}^{2}A=4a^{2}co{s}^{2}A$
or $c_1^2-2c_1{c}_2(2co{s}^{2}A-1)+c_2^2=4a^{2}co{s}^{2}A$
or $c_1^2-2c_1{c}_{2}cos2A+c_2^2=4a^{2}co{s}^{2}A$