The-angle-of-elevetion-of-the-top-of-a-tower-from-a-point-A-in-the-north-is-and-the-angle-of-the-top-of-the-tower-from-the-point-B-in-the-east-of-the-point-A-is-prove-that-the-height-of-the-tow

Question Number 177507 by peter frank last updated on 06/Oct/22

The angle of elevetion of the

top of a tower from a point

A in the north is α and the

angle of the top of the tower

from the point B in the east of the point

A is β . prove that the height

of the tower is

\frac{ABsin α \sin B}{\sqrt{\sin^{2} α - \sin^{2} β}}

Answered by som(math1967) last updated on 06/Oct/22

l e t O P = h e i g h t = x

O A = x c o t α

O B = x c o t β

A B = \sqrt{{O B}^{2} - {O A}^{2}}

= \sqrt{x^{2} {c o t}^{2} β - x^{2} {c o t}^{2} α}

A B = x \sqrt{\frac{{c o s}^{2} β {s i n}^{2} α - {s i n}^{2} β {c o s}^{2} α}{{s i n}^{2} α {s i n}^{2} β}}

s i n α s i n β A B = x \sqrt{{s i n}^{2} α - {s i n}^{2} α {s i n}^{2} β - {s i n}^{2} β + {s i n}^{2} α {s i n}^{2} β}

∴ x = \frac{A B s i n α s i n β}{\sqrt{{s i n}^{2} α - {s i n}^{2} β}}

Commented by som(math1967) last updated on 06/Oct/22

Commented by peter frank last updated on 06/Oct/22

thank you