Math Help - trigonometry problems help please

A flagpole 6 meters tall stands on top of a building. From a point on the same horizontal as the base of the building, the angle of elevation of the top and the bottom of the flagpole are 60°30' and 46°20', respectively. How high is the building

3. Or: Let "d" be the distance from the point at which the angles are measured to the base of the building (presumably directly below the flag pole) and "h" the height of the building to the bottom of the flag. Since the angle to the bottom of the flag pole is 46°20', you have $\frac{h}{d}= tan(46°20')$. Since the angle to the top of the flagpole is 60°30', you also have [tex]\frac{h+ 6}{d}= tan(60°30').
Those can be written as tan(46°20')d= h and tan(60°30') d= h+ 6. Since you are only asked for h, I would recommend solving the first equation for $d= \frac{h}{tan(46°20')}$ and putting that into the second equation: $\frac{tan(60°30')}{tan(46°20')}h= h+ 6$.