Question 1062997
.
5 cos @ + 12 sin @ = 13
Find  tan @ = ?
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<U>Answer</U>. {{{tan(alpha)}}} = {{{12/5}}}.


<U>Solution</U>


<pre>
Let x = {{{cos(alpha)}}}.  Then {{{sin(alpha)}}} = {{{sqrt(1-cos^2(alpha))}}} = {{{sqrt(1-x^2)}}}.


Then the original equation is equivalent to

{{{5x + 12*sqrt(1-x^2)}}} = 13,   or

5x - 13 = {{{-12*sqrt(1-x^2)}}}.   (1)

To solve (1), square both sides. You will get

{{{(5x - 13)^2}}} = 144*(1-x^2),

25x^2 - 130x + 169 = 144 - 144x^2,

25x^2 + 144x^2 - 130x + 169 - 144 = 0,

169x^2 - 130x + 25 = 0,

{{{(13x - 5)^2}}} = 0,

---->  13x - 5 = 0  ---->  13x = 5  ---->  x = {{{5/12}}}.


Thus you get {{{cos(alpha)}}} = {{{5/13}}}.


Then {{{sin(alpha)}}} = {{{sqrt(1-(5/13)^2)}}} = {{{sqrt(1-25/169)}}} = {{{sqrt((169-25)/169)}}} = {{{sqrt(144/169)}}} = {{{12/13}}}.


Now  {{{tan(alpha)}}} = {{{sin(alpha)/cos(alpha)}}} = {{{((12/13))/((5/13))}}} = {{{12/5}}}.
</pre>