Question

From the foot of a building I have to look upwards at an angle of 22 degrees to sight the top of a tree. From the top of the building, 150 meters above ground level, I have to look down at an angle of depression of 50 degrees to look at the top of the tree. How tall is tree? How far from the building is the tree?

Answer 1

@shinalcantara

Answer 2

yo! :D

Answer 3

hold on

Answer 4

OK

Answer 5

since it can be equated to both x then:
\[x = \frac{ y }{ \tan 20^{o} } ; x = \frac{ 150 - y }{ \tan 50^{o} }\]
equate:
\[\frac{ y }{ \tan 20 ^{o} } = \frac{ 150 - y }{ \tan 50^{o} }\]
cross multiply:
\[y \tan 50^{o} = \tan 20 ^{o} (150 - y)\]
distribute tan 20
\[ytan 50 ^{o} =150 \tan 20^{o} - ytan 20^{o}\]
combine the terms with 'y'
\[y \tan 50 ^{o} + y \tan 20 ^{o} = 150 \tan 20^{o}\]
simplify:
\[1.5557 y = 54.5955\]
\[y = 35.0938 meters\]

substitute y value
\[x = \frac{ y }{ \tan 20^{o} }\]
\[x = \frac{ 35.0938 }{ \tan 20^{o} }\]
\[x = 96.4196 meters\]