Question 1147423
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<pre>

Given triangle ABC is isosceles; therefore, the altitude AH is the median, at same time.


It means that  AH = BH = 6/2 = 3 units of length.


Then  tan(A) = {{{abs(AH)/3}}},   or


      tan(71°) = {{{abs(AH)/3}}},


which implies  | AH | = 3*tan(71°).      <U>ANSWER</U>
</pre>

Solved.


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<U>Be aware</U> : &nbsp;&nbsp;the solution by&nbsp; @josgarithmetic giving the answer &nbsp;{{{AH/sin(71)=6/sin(90)}}} &nbsp;is &nbsp;&nbsp;<U>I N C O R R E C T</U> (!)


I came to bring the correct solution.