The angle of elevation of a tower from a point $A$ due south of it, is $x$, from a point $B$ due east of $A$, is $y$. If $AB=l,$ then the height $h$ of the tower is given by
- $\displaystyle \frac { l }{ \sqrt { \cot ^{ 2 }{ y } -\cot ^{ 2 }{ x } } } $
- $\displaystyle \frac { l }{ \sqrt { \tan ^{ 2 }{ y } -\tan ^{ 2 }{ x } } } $
- $\displaystyle \frac { 2l }{ \sqrt { \cot ^{ 2 }{ y } -\cot ^{ 2 }{ x } } } $
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None of these
Let the tower height be h. Distance to point A = h * cot(x). Distance to point B = h * cot(y). Since A and B are due south and east, triangle OAB is right-angled at A (where O is the base of the tower). By Pythagoras, OA^2 + AB^2 = OB^2. (h*cot(x))^2 + l^2 = (h*cot(y))^2. l^2 = h^2(cot^2(y) - cot^2(x)). h = l / sqrt(cot^2(y) - cot^2(x)).
Let the tower height be h and its base be at point T. In the right triangle formed at A, the distance AT equals h cot x. Since B is east of A and T is south of A, triangle ABT is a right triangle at A, so BT squared equals AB squared plus AT squared, giving BT squared equals l squared plus h squared cot squared x. In the right triangle at B, BT also equals h cot y. Equating the squares gives h squared cot squared y equals l squared plus h squared cot squared x. Factoring out h squared gives h squared times (cot squared y minus cot squared x) equals l squared, so h equals l divided by the square root of cot squared y minus cot squared x.