A triangle has two corners with angles of # (2 pi ) / 3 # and # ( pi )/ 6 #. If one side of the triangle has a length of #3 #, what is the largest possible area of the triangle?

Answer 1

Largest possible area of the triangle is 3.8971

Given are the two angles #(2pi)/3# and #pi/6# and the length 3

The remaining angle:

#= pi - (((2pi)/3) + pi/6) = pi/6#

I am assuming that length AB (3) is opposite the smallest angle.

Using the ASA

Area#=(c^2*sin(A)*sin(B))/(2*sin(C)#
Area#=( 3^2*sin(pi/6)*sin((2pi)/3))/(2*sin(pi/6))#
Area#=3.8971#
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Answer 2

To find the largest possible area of the triangle, you can use the formula for the area of a triangle, which is 1/2 times the product of two sides times the sine of the included angle. Given that one side has a length of 3, and the two angles are (2π)/3 and π/6, you can use trigonometric ratios to find the lengths of the other two sides. Then, plug these values into the area formula to calculate the largest possible area, which is approximately 3.897.

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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.

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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