Step-by-step explanation:
so, we have a large triangle made of the 2 cables as legs and the ground distance AB as baseline.
the tower is the height to the baseline of that large triangle.
let's call the top of the tower T.
and remember, the sum of all angles in a triangle is always 180°.
we know the angle A = 62°, and angle B = 72°.
assuming that AB is a truly horizontal line that means that the 2 legs (cables) have different lengths, the triangle is not isoceles, and the tower is not in the middle of the baseline.
so, the height (tower) splits the baseline into 2 parts. let's call them p and q.
p + q = 12 m
p = 12 - q
let's simply define that p is the part of the baseline on the A side, and q is the part of the baseline on the B side.
we have now 2 small right-angled triangles the large height (tower) splits the large triangle into.
one has the sides
AT, height (tower), p
angle A = 62°
angle T = 180 - 90 - 62 = 28°
the other has the sides
BT, height (tower), q
angle B = 72°
angle T = 180 - 90 - 72 = 18°
now remember the law of sine :
a/sin(A) = b/sin(B) = c/sin(C)
with the sides and the associated angles being opposite.
p/sin(28) = height/sin(62)
q/sin(18) = height/sin(72)
we know from above that
p = 12 - q
so,
(12 - q)/sin(28) = height/sin(62)
height = (12 - q)×sin(62)/sin(28)
q/sin(18) = height/sin(72)
height = q×sin(72)/sin(18)
and therefore, as height = height we get
(12 - q)×sin(62)/sin(28) = q×sin(72)/sin(18)
(12 - q)×sin(62)×sin(18) = q×sin(72)×sin(28)
12×sin(62)×sin(18) - q×sin(62)×sin(18) =
= q×sin(72)×sin(28)
12×sin(62)×sin(18) = q×sin(72)×sin(28) + q×sin(62)×sin(18) =
= q×(sin(72)×sin(28) + sin(62)×sin(18))
q = 12×sin(62)×sin(18) / (sin(72)×sin(28) + sin(62)×sin(18))
q = 4.551603755... m
p = 12 - q = 7.448396245... m
height = q×sin(72)/sin(18) = 14.00839594... m ≈ 14 m
the cell tower is about 14 m tall.
![c^{2]](https://tex.z-dn.net/?f=c%5E%7B2%5D)
- 
, or A. 
, with a being -4c and b being 3d. -4c squares. Simplifying those values, we get 16