Question

Find the area of the triangle with the given measurements. Round the solution to the nearest hundredth if necessary.

A = 50°, b = 30 ft, c = 14 ft

Answer 1

And what do A, b, and c represent? Sides?

Answer 2

Usually, lowercase letters represent sides and their corresponding uppercase letters represent the angle opposite them...

Answer 3

By the way, isn't this ambiguous?

Answer 4

yeah what terenzrignz said

Answer 5

Never mind, one of them is angle. I see that.

Ok, so just sketch, it may help you.

Answer 6

I don't think it is ambiguous, since you can use the law of cosines.

Answer 7

The first time i tried it i got 134.99 ft2

Answer 8

I hope you know what it is.

Answer 9

Oh right, sorry... my bad... My trig's getting rusty..

Answer 10

Find the unknown side then use Heron's formula. You know this formula, right?

Answer 11

ahh yeah i think i got it in my notes

Answer 12

Yeah, so you know what to do now?

Answer 13

i tried it a second time and i got 160.87 ft

Answer 14

Well, state the formula, then.

Answer 15

(Or, if you're looking to have a little more challenge, derive the formula)

Answer 16

I got 160.87 ft² too.

Don't forgot ' ² '

Answer 17

the formula is s=1/2(a+b+c)

Answer 18

And \(A = \sqrt{s(s-a)(s-b)(s-c)}\)

All you need to do is find A.

Answer 19

thnx

Answer 20

But you shouldn't have to resort to that... there's a way to directly get the area of a triangle given two sides and the angle between them

Answer 21

what would s be in that equation

Answer 22

You can do this way, personally I think this way is easier to do. There are several approaches in this problem.

Answer 23

Just use law of cosines to find unknown side.

Answer 24

No, you don't even need it, all you have to is compute
\[\huge bc \sin A\]
You'll see you get the area.

Answer 25

Sorry, I meant
\[\huge \frac{bc \sin A}{2}\]

Answer 26

Hmm, I don't know this formula, and it worked, so you can do it this way.

Do you know this formula, @TNNG ?

Answer 27

Oh well, there are many ways to skin a cat...

Answer 28

no i do not but it did work

Answer 29

Okay, let's play a game...

Answer 30

Yeah, it did.

What's this formula called, @terenzreignz ?

Answer 31

Btw, I don't know what it's called, I just learned it from my teacher, last year :D

Answer 32

OK, I see. Thanks.

Answer 33

clearly
h = b sin A
k = b cos A
So... the area of the triangle on the right side is
(1/2)(b^2 sin A cos A)
But what about the triangle on the left side?

Answer 34

Its area would be
the base (c - k)
times the height (h)
divided by 2
But h = bsin A
k = bcos A

So...
its area is given by
(1/2)(c - bcos A)(bsin A)

Now add this to the area of the right side triangle, you get
(1/2)(bc sin A + b^2sin A cos A - b^2sin A cos A)
=(1/2)(bc sin A)

Proved! :)