drawbridge at the entrance to an ancient castle is raised and lowered by a pair of chains. The figure represents the drawbridge when flat. Find the height of the suspension point of the chain, to the nearest tenth of a meter, and the measures of the acute angles the chain makes with the wall and the drawbridge, to the nearest degree.
Answers
Answer:
4.0 meters, ∠C = 39°, ∠A = 51°
Step-by-step explanation:
Firstly, our diagram shows us that the given triangle is actually a right triangle. So we can use the Pythagorean Theorem to solve for the height of the chain:
[tex]a^{2} +b^{2} =c^{2}[/tex]
[tex](3.3)^{2} +b^{2} =(5.2)^{2}[/tex]
[tex]b^{2} =(5.2)^{2}-(3.3)^{2}[/tex]
[tex]b =\sqrt{(5.2)^{2}-(3.3)^{2}}[/tex]
[tex]b=4.0187...[/tex]
[tex]b=4.0 m[/tex]
Now, we can use the Law of Cosines to figure out one of the acute angles:
[tex]c^{2} =a^{2} +b^{2} -2ab(cos(C))[/tex]
[tex](3.3)^{2} =(4.0)^{2} +(5.2)^{2} -2(4.0)(5.2)(cos(C))[/tex]
[tex]cos(C)=\frac{(3.3)^{2}-(4.0)^{2} -(5.2)^{2}}{-2(4.0)(5.2)}[/tex]
[tex]C=cos^{-1}( \frac{(3.3)^{2}-(4.0)^{2} -(5.2)^{2}}{-2(4.0)(5.2)})[/tex]
[tex]C=39.3915...[/tex]
∠C = 39°
And since we know that all angles in a triangle add up to 180°:
∠A + ∠B + ∠C = 180
∠A + 90 + 39 = 180
∠A = 180 - 90 - 39
∠A = 51°
However, you should always review any answers on the Internet and make sure they are correct! Check my work to see if I made any mistakes!
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Step-by-step explanation:
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