Question 1145128
.
<pre>
The given slopes are the tangent values of corresponding angles.


Thus we have  tan(a) = {{{1/2}}}  for one angle, "a", and  tan(b) = {{{2/11}}}  for the other angle, "b".


They ask you about  {{{tan((a+b)/2)}}}.


Using well known formulas of Trigonometry,


    {{{tan(a+b)}}} = {{{(tan(a) + tan(b))/(1 - tan(a)*tan(b))}}} = {{{((1/2 + 2/11))/(1 - (1/2)*(2/11))}}} = {{{((11/22 + 4/22))/((1 - 1/11))}}} = {{{((15/22))/(10/11))}}} = {{{(15*11)/(10*22)}}} = {{{15/20}}} = {{{3/4}}}.


The last step is to use the formula for  tan(c/2) via  tan(c)


    {{{tan(c/2)}}} = {{{(-1 +- sqrt(1+tan^2(c)))/tan(c)}}}.


When you apply it, you will get the ANSWER 


    {{{tan((a+b)/2)}}} = {{{((-1 + sqrt(1 + (3/4)^2)))/(3/4))}}} = {{{(-1 + sqrt(25/16))/(3/4)}}} = {{{(-1 + 5/4)/(3/4)}}} = {{{((1/4))/((3/4))}}} = {{{1/3}}}.


According to the condition, you may use the sign " + " at sqrt instead of " +/- ".


<U>ANSWER</U>.  The slope is  {{{1/3}}}.
</pre>