The angle of elevation of the top of a vertical tower from a point A, due east of it is 45 degrees. The angle of elevation of the top of the same tower from a point B, due south of A is 30 degrees. If the distance between A and B is $54\sqrt{2} m$, then the height of the tower(in metres), is?
Answer given : $54 m$.
They have proceeded by doing $\tan 30 =\frac{h}{OB}$. And upon finding $OB$, they use Pythagoras theorem to find the value of $h$.
How are they doing $\tan 30 =\frac{h}{OB}$ to begin with? If that occurs, should $\sin 30 = \frac{h}{BC}$? Shouldn't angle COB be 90 degrees in order to do that(But clearly it's greater than 90 degrees)?
Best Answer
I think you are viewing the diagram incorrectly. The diagram is 3-dimensional. So standing at $A$, you look up at the tower, so $AC$ is a line 'rising into the air'. Then you walk 'down' to $B$ and 'look up' at the tower again. So you have two triangles, one right triangle $COA$, and another right triangle $COB$. Applying trig to the triangle $COB$, you find $\tan 30^\circ= \dfrac{h}{OB}$. Think of the diagram as the picture of a 'cheese wedge'-like object and you start to see what was drawn.