A TOWER STANDS IN A FIELD WHOSE SHAPE IS THAT OF AN EQILATERAL TRIANGLE AND WHOSE SIDE IS 80m. IT SUBTENDS ANGLES AT THREE CORNERS WHOSE TANGENTS ARE RESPECTIVELY ROOT 3 + 1, ROOT 2, ROOT 2. FIND ITS HEIGHT
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Answer:
Let the tower stand at O and its height OP = h which subtends an angle of 45
o
,60
o
,60
o
at A, B, C respectively.
OA=hcot45
o
=h;
OB=hcot60
o
=
3
h
=OC
OB=OC
Also AB = AC = BC = a say.
If D be the mid-point of BC then OD and AD both are perpendicular to base BC.
∴ AD is median as well as altitude. In an isosceles or equilateral Δ, both centroid and orthocentre coincide, then
OA=
3
2
AD=
3
2
a
2
+
4
a
2
or h=OA=
3
2
.
2
1
5
a
=
3
100
5
∴a=100,
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