A triangle has two corners with angles of # (3 pi ) / 4 # and # ( pi )/ 12 #. If one side of the triangle has a length of #4 #, what is the largest possible area of the triangle?

Answer 1

Largest possible area of the triangle is

#color(brown)(A_t = 10.93# sq units

#hat A = (3pi)/4, hat B = pi/12, hat C -= pi/6#
To get the largest area, side 4 should correspond to the least angle #(pi/12)#

As per the Law of Sines,

#a = (sin A * b) / sin B = (sin ((3pi)/4) * 4) / sin (pi/12)#
#a = 10.93#

Largest possible area of the triangle is

#A_t = (1/2) a b sin C = (1/2) * 4 * 10.93 * sin (pi/6) = 10.93#
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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 given two sides and the angle between them:

[ \text{Area} = \frac{1}{2} \times \text{side}_1 \times \text{side}_2 \times \sin(\theta) ]

Given one side with a length of 4 and two angles, you can find the lengths of the other two sides using the law of sines. Then, you can calculate the area using the formula above. Finally, you can maximize the area by adjusting the lengths of the other sides while keeping the given side fixed.

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Answer 3

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 the lengths of two sides and the sine of the included angle. Given one side length and two angles of the triangle, you can find the lengths of the other two sides using trigonometric ratios. Then, you can calculate the area using the formula mentioned earlier.

Let's denote the three angles of the triangle as A, B, and C, and their corresponding side lengths as a, b, and c, respectively.

Given: Angle A = ( \frac{3\pi}{4} ) Angle B = ( \frac{\pi}{12} ) Side length a = 4

First, find the third angle C using the fact that the sum of angles in a triangle is ( \pi ) (radians).

( \text{Angle C} = \pi - \text{Angle A} - \text{Angle B} ) ( \text{Angle C} = \pi - \frac{3\pi}{4} - \frac{\pi}{12} ) ( \text{Angle C} = \pi - \frac{9\pi}{12} - \frac{\pi}{12} ) ( \text{Angle C} = \pi - \frac{10\pi}{12} ) ( \text{Angle C} = \frac{2\pi}{12} ) ( \text{Angle C} = \frac{\pi}{6} )

Now, using the Law of Sines, find the lengths of the other two sides: ( \frac{a}{\sin A} = \frac{b}{\sin B} = \frac{c}{\sin C} )

( \frac{4}{\sin(\frac{3\pi}{4})} = \frac{b}{\sin(\frac{\pi}{12})} = \frac{c}{\sin(\frac{\pi}{6})} )

From ( \sin(\frac{3\pi}{4}) = \frac{\sqrt{2}}{2} ), ( \sin(\frac{\pi}{12}) ), and ( \sin(\frac{\pi}{6}) = \frac{1}{2} ), you can solve for b and c.

( b = 4 \times \frac{\sin(\frac{\pi}{12})}{\sin(\frac{3\pi}{4})} ) ( b = 4 \times \frac{\sin(\frac{\pi}{12})}{\frac{\sqrt{2}}{2}} ) ( b = 4 \times \frac{2\sin(\frac{\pi}{12})}{\sqrt{2}} ) ( b = 2\sqrt{2}\sin(\frac{\pi}{12}) )

Similarly, solve for c:

( c = 4 \times \frac{\sin(\frac{\pi}{6})}{\sin(\frac{3\pi}{4})} ) ( c = 4 \times \frac{\frac{1}{2}}{\frac{\sqrt{2}}{2}} ) ( c = 4 \times \frac{1}{\sqrt{2}} ) ( c = 2\sqrt{2} )

Now, you have all three side lengths. Use Heron's formula to find the area of the triangle:

( \text{Area} = \sqrt{s(s-a)(s-b)(s-c)} )

where ( s = \frac{a+b+c}{2} ).

Calculate ( s ), then the area using the side lengths found earlier.

Finally, compare the areas obtained by changing ( \frac{\pi}{12} ) to ( \frac{5\pi}{12} ) and ( \frac{7\pi}{12} ) to find the largest possible area.

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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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