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$\alpha = 90^\circ - 59^\circ
$\alpha = 90^\circ - 59^\circ 23' = 30^\circ 37'$
$\beta = 90^\circ + 29^\circ 42' = 119^\circ 42'$
$\theta = 180^\circ - \alpha - \beta = 29^\circ 41'$
$\dfrac{x}{\sin \beta} = \dfrac{31.2}{\sin \theta}$
$x = 54.73 ~ \text{m}$
$y = x \sin 59^\circ 23'$
$y = 47.098 ~ \text{m}$ answer
Thanks a lot sir its a great
In reply to $\alpha = 90^\circ - 59^\circ by Jhun Vert
Thanks a lot sir its a great help.
Thanks a lot sir its a great
In reply to $\alpha = 90^\circ - 59^\circ by Jhun Vert
Thanks a lot sir its a great help. I had another solution using tangent got the same answer..
(No subject)
AB=A'B'=31.2m
AA'=BB'=X
A'C=y=31.2 + y'
y=31.2 + y'
tan 29042'=y'/x
x = y'/ tan 29042'
tan 59023'=y/x
tan 59023'=(31.2+y')/(y'/(tan 29042')
(tan 59023')(y')/(tan29042')=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
In reply to (No subject) by esmilitar
Yes. also an straightforward solution.