From a point O on the ground of a square courtyard of area 160,000 sq.ft., the angles of elevation of three flagstaffs of equal heights at three consecutive corners of the yard at A, B and C are 45°, 60°, and 60° respectively. Find the height of the each flagstaff in meters.
The illustration below is my attempt to visualize the problem together with my initial try.
Let H be the height of flagstaff.
tan (45) = H/OA
OA = Hcot(45)
tan (60) = H/OB = H/OC
OB = Hcot(60) = OC
Using cosine law:
400^2 = 2(OB^2)-2(OB^2)cos(X)
400^2 = 2(OC^2)-2(OC^2)cos(X)
See height of flagstaffs is same therefore let it to be $h$ units.
Consider tower at $B$
$OB=h\cot60$ and similarly $OB=h\cot60$ and hence triangle $BOC$ is isosceles
also $OA=h\cot45$ Now drop a perpendicular $OX$ on $AB$.
The length of $OX$ will be $200$ because since triangle $OBC$ is isosceles therefore perpendicular from $O$ to side $BC$ bisects it.
Now by Pythagoras theorem,
$\sqrt{OA^2-OX^2}+\sqrt{OB^2-OX^2}=400$
Now you know $OA, OX, OB $ .Plug in to find value of $h$
Or if you want to save yourself from this exhausting equation, you may use similarity. Triangles $OXA$ and $BXO$ are similar therefore ratio of their corresponding sides are same. Hence
$\frac{OA}{BO}=\frac{OX}{BX}$