# A triangle has two corners with angles of # (2 pi ) / 3 # and # ( pi )/ 6 #. If one side of the triangle has a length of #1 #, what is the largest possible area of the triangle?

Maximum area:

All solutions rely on recognizing that the third angle must be

There are two possible configurations to consider.

For brevity, you might want to skip to version 2 which has the larger area.

In Version 1, the side with length

In Version 2, the side with length

Version 1

Solution 1: Determine the area using side with length 1 as the base and a calculated height.

By definition of the

The angle

So

and

This gives an approximate area of

Solution 2: Determine the length of the missing sides and use Heron's Formula

perimeter is

and

semiperimeter,

By Heron's Formula

~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~

Version 2

Solution 1: Determine the area using one of the sides with length 1 as the base and a calculated height

By definition of the

Note that

So Version 2 gives the larger area and thus is the version being asked for.

Solution 2: Determine the length of the missing side (using Law of Sines) and apply Heron's Formula.

(highlights only)

missing side:

semiperimeter

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I got an area of (approximately)

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To find the largest possible area of the triangle, we can use the formula for the area of a triangle, which is given by ( \frac{1}{2} \times \text{base} \times \text{height} ). Since we know one side length is 1, let's call it the base.

The height of the triangle can be found using trigonometry. Considering the angle ( \frac{\pi}{6} ), the height opposite to this angle can be calculated as ( \sin\left(\frac{\pi}{6}\right) \times \text{base} ).

Similarly, for the angle ( \frac{2\pi}{3} ), the height opposite to this angle can be calculated as ( \sin\left(\frac{\pi}{3}\right) \times \text{base} ).

Now, to find the largest possible area, we maximize the area formula ( \frac{1}{2} \times \text{base} \times \text{height} ). We know the base length is 1, so we just need to maximize the product of the base and the heights.

The product of the two heights is ( \sin\left(\frac{\pi}{6}\right) \times \sin\left(\frac{\pi}{3}\right) ).

Substituting the values of ( \sin\left(\frac{\pi}{6}\right) ) and ( \sin\left(\frac{\pi}{3}\right) ), we get ( \frac{1}{2} \times \frac{\sqrt{3}}{2} = \frac{\sqrt{3}}{4} ).

So, the largest possible area of the triangle is ( \frac{1}{2} \times 1 \times \frac{\sqrt{3}}{4} = \frac{\sqrt{3}}{4} ).

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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.

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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