Question 162648
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Okay, we'll mark point "D" along line "AB" where it extends to point "C" on the opposite bank. So line {{{highlight(CD)}}} is what we're looking for right?
Given:
AB=652ft
Points "A", "B", & "C" forms a triangle, where:
{{{A=(53^o)18min=53.3^o}}}, *note {{{18/60=0.30^o}}}
{{{B=(48^o)36min=48.6^o}}}, *note {{{36/60=0.60^o}}}
{{{C=180-A-B=180-(53.3^o)-(48.6^o)}}}
{{{C=78.10^o}}}
We need to get line "CB", by SINE LAW:
{{{sinA/CB=sinC/AB}}}
{{{CB=(sinA*AB)/sinC=(sin53.3^o)(652ft)/(sin78.1^o)}}}
{{{CB=534.239ft}}}
Now, points "B", "C", & "D" forms a right triangle, and we can solve for line {{{highlight(CD)}}} by trigo functions. But 1st we solve for line DB:
{{{cosB=adj/hyp=DB/CB}}}
{{{DB=(cosB)(CB)=(cos48.6^o)(534.239)}}}
{{{DB=353.30ft}}} ----> REMEMBER
Now, for CD,
{{{sinB=opp/hyp=CD/CB}}}
{{{CD=(sinB)(CB)=(sin48.6^o)(534.239)}}}
{{{highlight(CD=400.71ft)}}}, ANSWER
To check, we use trigo function TANGENT;
{{{tanB=opp/adj=CD/DB}}}
{{{CD=(tanB)(DB)=(tan48.6^o)(353.30ft)}}}
{{{CD=400.74}}}, close enough! Bec. we're rounding off
Thank you,
Jojo</pre>