sqrt(cot^2(36°)+cot^2(43°50') book tells me answer is 30°5' but I cant figure out how they got that

$${\sqrt{{{cot}{\left({\mathtt{36}}\right)}}^{\,{\mathtt{2}}}{\mathtt{\,\small\textbf+\,}}{{cot}{\left({\mathtt{43}}{\mathtt{\,\small\textbf+\,}}{\frac{{\mathtt{50}}}{{\mathtt{60}}}}\right)}}^{\,{\mathtt{2}}}}} = {\mathtt{1.726\: \!067\: \!549\: \!808\: \!880\: \!2}}$$

I think you might mean the angle, A, such that cot(A) = 1.726...

$${\mathtt{A}} = {acot}{\left({\sqrt{{{cot}{\left({\mathtt{36}}\right)}}^{\,{\mathtt{2}}}{\mathtt{\,\small\textbf+\,}}{{cot}{\left({\mathtt{43}}{\mathtt{\,\small\textbf+\,}}{\frac{{\mathtt{50}}}{{\mathtt{60}}}}\right)}}^{\,{\mathtt{2}}}}}\right)} \Rightarrow {\mathtt{A}} = {\mathtt{30.085\: \!926\: \!410\: \!525^{\circ}}}$$

 This is A = 30° + 0.085926*60' 

$${\mathtt{0.085\: \!926}}{\mathtt{\,\times\,}}{\mathtt{60}} = {\mathtt{5.155\: \!56}}$$

So 

A = cot^-1( √[cot²(36°) + cot²(43°50')] ) = 30°5'

(Note that acot is the same thing as cot^-1)

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