8.1. If x is the distance from P to the flagpole and h is the height of the flagpole, then
tan(53°) = h/x
tan(28°) = h/(x + 20)
Solve for the first equation for h in terms of x :
h = x tan(53°)
Substitute this into the second equation and solve for x :
tan(28°) = x tan(53°) / (x + 20)
(x + 20) tan(28°) = x tan(53°)
x tan(28°) + 20 tan(28°) = x tan(53°)
x (tan(28°) - tan(53°)) = -20 tan(28°)
x = 20 tan(28°) / (tan(53°) - tan(28°))
x ≈ 13.4 m
8.2. By the law of cosines,
BD² = 7² + 11² - 2 • 7 • 11 cos(110°)
⇒ BD = √(170 - 154 cos(110°))
⇒ BD ≈ 14.9 cm
and
BC² = BD² + 16² - 2 • BD • 16 cos(40°)
⇒ BC = √(426 - 154 cos(110°) - 32 cos(40°) √(170 - 154 cos(110°)))
⇒ BC ≈ 10.6 cm
Now use Heron's formula. If S is the semiperimeter of ∆ABD and S' is the semiperimeter of ∆BCD, i.e.
S = (AB + BD + AD)/2 ≈ 16.5 cm
S' = (BC + CD + BD)/2 ≈ 20.7 cm
then
area ∆ABD = √(S (S - AB) (S - BD) (S - AD)) ≈ 36.2 cm²
and
area ∆BCD = √(S' (S' - BC) (S' - CD) (S' - BD)) ≈ 76.7 cm²
so that the total area of ABCD is approximately 112.9 cm².