Question

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

Answer 1

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,