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Firstly, Solve \(\triangle ABC\) where \(\hat { A } =55^{ \circ },\quad b_{ 1 }=b_{ 2 }=14.3㎝\quad and\quad b=16.3cm\).

*Find out the sum of all the angles and add it to the sum of all the sides.** (Correct to 1 decimal places.)*

i.e.,

\[\hat { A } +\hat { B_{ 1 } } +\hat { B_{ 2 } } +\hat { C_{ 1 } } +\hat { C_{ 2 } } +AB_{ 1 }+AB_{ 2 }+B_{ 1 }C+B_{ 2 }C+AC\\ or\quad \\ \hat { A } +\hat { B_{ 1 } } +\hat { B_{ 2 } } +\hat { C_{ 1 } } +\hat { C_{ 2 } } +c_{ 1 }+c_{ 2 }+a_{ 1 }+a_{ 2 }+b\\ \]

*(They both have the same value.)*

For clarification, \(\hat { A } =55^{ \circ }\\ B_{ 1 }C=a_{ 1 }=14.3\\ B_{ 2 }C=a_{ 2 }=14.3\\ AC=b=16.3\\ AB_{ 1 }=c_{ 1 }\\ AB_{ 2 }=AB_{ 1 }+B_{ 1 }B_{ 2 }=c_{ 2 }\\ \)

*Details and Assumptions:*

If your answer is \(\theta ^{ \circ }+\alpha ^{ \circ }+\beta ^{ \circ }+x\) cm \(+y\) cm, write your answer as \(\theta +\alpha +\beta +x +y \).

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