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Please solve with explanation (high points)

Please solve with explanation (high points)-example-1

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Answer:

8.1. h = 17.743514416718 m

8.2. Area of ABCD = 112.912444233369 cm²

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

User Joalcego
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