Using the Sine Rule for Ambiguous Cases

STEP 1: SOLVING $\triangle ABC$ :

A set of given information that would give two possible sets of solutions is said to be ambiguous .

Thus, for ambiguous cases , two solution sets will be obtained.

From the diagram, the given angle is acute and the side facing the given angle is less than the other given side. The two sets of answers are obtained using the Sine Rule which implies that:

$\frac { a }{ \sin { A } } =\frac { b }{ \sin { B } } =\frac { c }{ \sin { C } }$

Now,

$\quad \quad \frac { 14.3 }{ \sin\quad 55° } =\frac { 16.3 }{ \sin\quad B } \\ \therefore \quad \sin\quad B\quad =\quad \frac { 16.3\quad \sin\quad 55° }{ 14.3 } \\ \quad \quad \quad \quad \quad \quad =0.9337\\ \hat { B } \quad =\quad \sin^{ -1 }\quad 0.9337\\ \quad \quad =\quad 69.0°\quad or\quad 111.0°$

When $\hat { B } =\quad 69.0°,\quad \hat { C } =180°-(69°+55°)=56°$

Now,

$\quad \quad \frac { 16.3 }{ \sin { \quad 69° } } =\frac { c }{ \sin { \quad 56° } } \\ \therefore \quad \quad \quad c\quad =\frac { 16.3\quad \sin { \quad 56° } }{ \sin { \quad 69° } } \approx 14.5cm\quad \quad$

When $\hat { B } =\quad 111.0°,\quad \hat { C } =180°-(111°+55°)=14°$

Now,

$\quad \quad \frac { 16.3 }{ \sin { \quad 69° } } =\frac { c }{ \sin { \quad 14° } } \\ \therefore \quad \quad \quad c\quad =\frac { 16.3\quad \sin { \quad 14° } }{ \sin { \quad 69° } } \approx 4.2cm\quad \quad$

Thus the two sets of solutions are:

$\hat { B_{ 1 } } =111°,\quad \hat { C_{ 1 } } =14°,\quad { c_{ 1 } }=4.22\\ and\\ \hat { B_{ 2 } } =69°,\quad \hat { C_{ 2 } } =56°,\quad { c_{ { 2 } } }=14.47$

STEP 2: ADDING THE SUM OF THE ANGLES TO THE SUM OF THE SIDES OF $\triangle ABC$ : Therefore,

$\quad A+B_{ 1 }+B_{ 2 }+C_{ 1 }+C_{ 2 }+a_{ 1 }+a_{ 2 }+b+c_{ 1 }+c_{ 2 }\\ =55+111+69+14+56+14.3+14.3+16.3+4.2+14.4\\ =\boxed { 368.6 }$