Let's check first this right triangle from the figure:
We have, using tangent on the angle:
[tex]\begin{gathered} \tan (15^o)=\frac{h}{25+x} \\ \Rightarrow h=0.268(25+x) \end{gathered}[/tex]
And then, looking at the other right triangle:
Doing the same as before, we have that:
[tex]\begin{gathered} \tan (40)\text{ = }\frac{h}{x} \\ \Rightarrow h=0.839x \end{gathered}[/tex]
Now, we can use both equations of h to solve for x:
[tex]\begin{gathered} 0.268(25+x)=0.839x \\ \Rightarrow6.7+0.268x=0.839x \\ \Rightarrow6.7=0.571x\Rightarrow x=\frac{6.7}{0.571}=11.73 \end{gathered}[/tex]
Which we can finally use to find h in any equation:
[tex]\begin{gathered} h=0.268(25+11.73)=9.843 \\ or \\ h=0.839\cdot11.73=9.841 \end{gathered}[/tex]
