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