# Trigonometry, area of triangle

• Feodalherren
In summary: In summary:The student is trying to figure out what side of a triangle is the hypotenuse. They are not able to do so using right triangle trigonometry and are instead told to use trigonometry of non-right triangles. Unfortunately, they are unable to solve the equation for the area of the triangle using the cosine law or the sine law. They are able to solve the equation using acute angles.

## Homework Statement

Use the given information to determine the area of each triangle.

Two of the sides are 5cm and 7cm, and the angle between those sides is 120 degrees.

Trig functions

## The Attempt at a Solution

I tried drawing up a picture but it's of no help to me without me knowing which side is the hypotenuse, how do I determine which side is the hypotenuse?

It is not a right triangle, there is no hypotenuse. The sides, 5 cm and 7 cm long, enclose 120°angle. Show your drawing.

ehild

"Which side is the hypotenuse"? You understand that only right triangles have "hypotenuses", don't you? And no angle in a right triangle can be larger than 90 degrees. You are told that this triangle has angle with measure 120 degrees so this is NOT a right triangle and does not have a "hypotenuse"!

Unfortunately, that leaves me with no idea about what you do know about trigonometry of non-right triangles. Seeing that you are given two sides and the angle between them, I would be inclined to use the "cosine law" to find the length of the third side. Do you know that law? Once you have that, you can use the "sine law" to find the other two angles. Do you know that law? And once you know those, you can choose any side you like as a base and use right triangle trigonometry to find the length of a perpendicular to that side. Area= (1/2)(base)(height).

Cosine law: if two sides of a triangle have lengths a and b and the angle between them has measure C, then the third side has length c satisfying $c^2= a^2+ b^2- 2ab cos(C)$.

Sine law: if we denote the lengths of the sides of a triangle by a, b, and c and the angles opposite each by A, B, and C, then $\frac{sin(A)}{a}= \frac{sin(B)}{b}= \frac{sin(C)}{c}$.

There is a formula for the area of a triangle that is related to the Law of Sines. The formula assumes that the known parts of the triangle are SAS.
$A = \frac{1}{2}ab \sin C = \frac{1}{2}bc \sin A = \frac{1}{2}ca \sin B$.
So all the OP has to do is to plug in the numbers into one of the formulas above and get the area.

Thanks guys, I solved it by using acute angles.

1/2ab (Sin θ)
Sin 120 = Sin 60

(35sqrt3)/4 cm^2

Correct?

Feodalherren said:
Thanks guys, I solved it by using acute angles.

1/2ab (Sin θ)
Sin 120 = Sin 60

(35sqrt3)/4 cm^2

Correct?

The answer is right, but why are you saying that the sine of 120 radians equals the sine of 60 radians? That's not true.

Degrees, not radians. By using reference angles we can find that Sin 120 = Sin 60.

My point was that if you mean degrees, use the degree symbol. Without it, I assume you mean radians. So you should have written
$\sin 120^{\circ} = \sin 60^{\circ}$.

Last edited:
eumyang said:
My point was that if you mean degrees, use the degree symbol. Without it, I assume you mean radians. So you should have written
$\sin 120^{\circ} = \sin 60^{\circ}$.

Let's not be picky about semantics now. Clearly he meant degrees.

It's one of my pet peeves. I can't tell you how many times I've seen trig expressions without the degree symbol on forums like this one or at school.

That's why I wrote degrees, I was too lazy to find the degree symbol. Writing on paper is a different thing.