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$\alpha = 90^\circ - 59^\circ
α=90∘−59∘23′=30∘37′
β=90∘+29∘42′=119∘42′
θ=180∘−α−β=29∘41′
xsinβ=31.2sinθ
x=54.73 m
y=xsin59∘23′
y=47.098 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.