Answer:
8.1. h = 17.743514416718 m
8.2. Area of ABCD = 112.912444233369 cm²
Step-by-step explanation:
Q8.
Let h be the height of the flagpole
x be the distance between point P and the flagpole.
8.1.
tan(53) = h/x
⇒ h = xtan(53) (1)
On the other hand ,
tan(28) = h/(x+20)
⇒ h = (x+20)tan(28) (2)
Let’s equate equation so (1) and (2) :
⇒ xtan(53) = (x+20)tan(28)
⇒ x = 20tan(28) ÷ (tan(53) - tan(28)) = 13.370697151775
from (1) , we have :
h = xtan(53)
= 13.370697151775×tan(53)
= 17.743514416718
8.2.
Area ABCD = area ABD + area BCD
area ABD = (1/2) AB AD sin(BAD)
= (1÷2)×11×7×sin(110)
=36.178165900258
area BCD = (1/2) DB DC sin(BDC)
firsly , we need to calculate DB (using the law of cosines I need triangle ABD) :
DB² = AB² + AD² - 2AB AD cos(110)
= 11^2+7^2-2×11×7×cos(110)
= 222.671102072153
Then
DB = √(222,671102072153)
= 14.92216814247
Going back ,
area BCD = (1/2) DB DC sin(BDC)
= (1÷2)× (14,92216814247)×16×sin(40)
=76.734278333111
Conclusion :
Area ABCD = area ABD + area BCD
= 36.178165900258+ 76.734278333111
= 112.912444233369
