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Solve for area of triangle using differentials
- Thread starter Jtgs5249
- Start date
D
Deleted member 4993
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An isosceles triangle has equal sides of length 12m. If the angle theta between these sides is increased from 30 to 33 degrees, use differentials to approximate the change in the area of the triangle.
A = 1/2 * (12 * cos(Θ/2)] * [2 * 12 * sin(Θ/2)]
Hello, Jtgs5249!
The area of the triangle is given by: .\(\displaystyle A \:=\:\frac{1}{2}(12^2)\sin\theta \:=\:72\sin\theta\)
Then the change in area \(\displaystyle \Delta A\) due to a change in angle \(\displaystyle \Delta\theta\)
. . is approximated by: .\(\displaystyle dA \;=\;72\cos\theta\,d\theta\)
We are given: .\(\displaystyle \theta = \frac{\pi}{6}\)
. . and: -\(\displaystyle d\theta \:=\:33^o - 30^o \:=\:\frac{11\pi}{60} - \frac{\pi}{6} \:=\:\frac{\pi}{60}\)
Therefore: .\(\displaystyle dA \:=\:72\left(\cos\frac{\pi}{6}\right)\frac{\pi}{6} \:=\:72\left(\frac{\sqrt{3}}{2}\right)\frac{\pi}{60} \)
. . . . . . . . \(\displaystyle dA \:=\:\dfrac{34\sqrt{3}\pi}{5} \;\approx\;3.265\text{ m}^2\)
\(\displaystyle \text{An isosceles triangle has equal sides of length 12 m.}\)
\(\displaystyle \text{If the angle }\theta\text{ between these sides is increased from }30^o\text{ to }33^o,\)
\(\displaystyle \text{use differentials to approximate the change in the area of the triangle.}\)
The area of the triangle is given by: .\(\displaystyle A \:=\:\frac{1}{2}(12^2)\sin\theta \:=\:72\sin\theta\)
Then the change in area \(\displaystyle \Delta A\) due to a change in angle \(\displaystyle \Delta\theta\)
. . is approximated by: .\(\displaystyle dA \;=\;72\cos\theta\,d\theta\)
We are given: .\(\displaystyle \theta = \frac{\pi}{6}\)
. . and: -\(\displaystyle d\theta \:=\:33^o - 30^o \:=\:\frac{11\pi}{60} - \frac{\pi}{6} \:=\:\frac{\pi}{60}\)
Therefore: .\(\displaystyle dA \:=\:72\left(\cos\frac{\pi}{6}\right)\frac{\pi}{6} \:=\:72\left(\frac{\sqrt{3}}{2}\right)\frac{\pi}{60} \)
. . . . . . . . \(\displaystyle dA \:=\:\dfrac{34\sqrt{3}\pi}{5} \;\approx\;3.265\text{ m}^2\)
Split the isosceles triangle I into 2 congruent triangles II and III.So far, I think figured out the base is 12sin(X/2) and height is 12cos(X/2). So for the equation I would put A=1/2(12sin(x/2))(12cos(x/2)) but I don't know how to solve that and for an answer because I'm not suppose to plug in numbers that aren't constants. I placed X in for theta.
The height of triangle II = the height of triangle I \(\displaystyle = 12 cos\left(\dfrac{\theta}{2}\right).\)
The base of triangle II \(\displaystyle = 12 sin\left(\dfrac{\theta}{2}\right).\)
The base of triangle I = the base of triangle II + the base of triangle III = 2 * the base of triangle II \(\displaystyle = 2 * 12 sin\left(\dfrac{\theta}{2}\right).\)
The area of triangle I \(\displaystyle = \dfrac{1}{2} * height * base = \dfrac{1}{2} * \left\{12cos\left(\dfrac{\theta}{2}\right)\right\} * \left\{2 * 12 sin\left(\dfrac{\theta}{2}\right)\right\}.\)
Now what?
D
Deleted member 4993
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Why is it [2 * 12 * sin(Θ/2)]? Where does the multiplying by 2 come from?
See the answer above..
So I do the derivative of that and multiply it by TT/60 and that gives me 3.26. Is that correct?
Someone has done the entire problem for you...