Two corners of a triangle have angles of # (2 pi )/ 3 # and # ( pi ) / 6 #. If one side of the triangle has a length of # 4 #, what is the longest possible perimeter of the triangle?

Answer 1

Longest possible perimeter = 14.928

Sum of the angles of a triangle #=pi#
Two angles are #(2pi)/3, pi/6# Hence #3^(rd) #angle is #pi - ((2pi)/3 + pi/6) = pi/6#
We know# a/sin a = b/sin b = c/sin c#
To get the longest perimeter, length 2 must be opposite to angle #pi/24#
#:. 4/ sin(pi/6) = b/ sin((pi)/6) = c / sin ((2pi)/3)#
#b = (4 sin((pi)/6))/sin (pi/6) = 4#
#c =( 4* sin((2pi)/3))/ sin (pi/6) = 6.9282#
Hence perimeter #= a + b + c = 4 + 4 + 6.9282 = 14.9282#
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Answer 2

To find the longest possible perimeter of the triangle, we need to consider the case where the remaining angle is the largest possible. Since the sum of angles in a triangle is always ( \pi ) radians (or 180 degrees), we can find the measure of the third angle by subtracting the given angles from ( \pi ):

[ \text{Third angle} = \pi - \frac{2\pi}{3} - \frac{\pi}{6} ]

Then, we can calculate the perimeter of the triangle using the given side lengths and angles, considering the triangle to be equilateral:

[ \text{Perimeter} = 3 \times \text{side length} ]

Substituting the given side length of 4:

[ \text{Perimeter} = 3 \times 4 ]

[ \text{Perimeter} = 12 ]

Therefore, the longest possible perimeter of the triangle is 12 units.

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Answer from HIX Tutor

When evaluating a one-sided limit, you need to be careful when a quantity is approaching zero since its sign is different depending on which way it is approaching zero from. Let us look at some examples.

When evaluating a one-sided limit, you need to be careful when a quantity is approaching zero since its sign is different depending on which way it is approaching zero from. Let us look at some examples.

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When evaluating a one-sided limit, you need to be careful when a quantity is approaching zero since its sign is different depending on which way it is approaching zero from. Let us look at some examples.

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