Please help to solve this problem:

The angle of elevation of a point C from a point B is 29^{0}42'; the angle of elevation of C from another point A 31.2 m directly below B is 59^{0}23'. How high is C from the horizontal line through A?

April 17, 2017 - 11:00pm

#1
Angle of Elevation

$\alpha = 90^\circ - 59^\circ 23' = 30^\circ 37'$

$\beta = 90^\circ + 29^\circ 42' = 119^\circ 42'$

$\theta = 180^\circ - \alpha - \beta = 29^\circ 41'$

$\dfrac{x}{\sin \beta} = \dfrac{31.2}{\sin \theta}$

$x = 54.73 ~ \text{m}$

$y = x \sin 59^\circ 23'$

$y = 47.098 ~ \text{m}$

answerThanks a lot sir its a great help.

Thanks a lot sir its a great help. I had another solution using tangent got the same answer..

^{}AB=A'B'=31.2m

AA'=BB'=X

A'C=y=31.2 + y'

y=31.2 + y'

tan 29

^{0}42'=y'/xx = y'/ tan 29

^{0}42'tan 59

^{0}23'=y/xtan 59

^{0}23'=(31.2+y')/(y'/(tan 29^{0}42')(tan 59

^{0}23')(y')/(tan29^{0}42')=31.2 +y'2.9625y'-y'=31.2

y'=31.2/1.9625

y'=15.898m

y=31.2 + 15.898

y=47.098m Ans

Yes. also an straightforward solution.