# Thread: [SOLVED] Finding Area of a Scalene Triangle

1. ## [SOLVED] Finding Area of a Scalene Triangle

A scalaene triangle has 3 sides: RS = 12.1, ST = 19.8, RT = 17.4. NOw i have to find the area of the triangle.
It didn't say which side was the base, so I made side ST the base. Then, I made a dotted line for the height, H. So now there's 2 triangles: RSY, and RYT. Triangle RSY has RS which is 12.1, and angle Y, which is 90 degrees. How can I figure out other information for triangle RSY with only that info? I thought i'd use trigonometric ratios. So, the sin of Y = o/h = i'm not sure. the hypotenous is 12.1, but what is the opposite side for angle Y, which is a right angle?
Sorry, I know this post must seem rambling.... but thats what progress I've made on the qu., lol. I would appreciate any help you may have. I know, I'm not asking for the answer. I just need some step-by-step instructions to guide me though. thank you.

2. Originally Posted by ria3
A scalaene triangle has 3 sides: RS = 12.1, ST = 19.8, RT = 17.4. .
Heron's formula!

RonL

3. ## Law of Cosines

Originally Posted by ria3
A scalaene triangle has 3 sides: RS = 12.1, ST = 19.8, RT = 17.4. NOw i have to find the area of the triangle.
It didn't say which side was the base, so I made side ST the base. Then, I made a dotted line for the height, H. So now there's 2 triangles: RSY, and RYT. Triangle RSY has RS which is 12.1, and angle Y, which is 90 degrees. How can I figure out other information for triangle RSY with only that info? I thought i'd use trigonometric ratios. So, the sin of Y = o/h = i'm not sure. the hypotenous is 12.1, but what is the opposite side for angle Y, which is a right angle?
Sorry, I know this post must seem rambling.... but thats what progress I've made on the qu., lol. I would appreciate any help you may have. I know, I'm not asking for the answer. I just need some step-by-step instructions to guide me though. thank you.

$\displaystyle \cos^{-1}\alpha=1.056$

So you end up with

$\displaystyle \frac{1}{2}19.8(12.1)\sin(1.056) \approx 104.3$

4. Hello, ria3!

A scalaene triangle has sides: $\displaystyle RS = 12.1,\;ST = 19.8,\; RT = 17.4$
Find the area of the triangle.

Someone will mention Heron's Formula . . . which is the fastest method.
But, assuming you know Trigonometry, I'll show you another way . . .

We have: .$\displaystyle \Delta RST$ with angles $\displaystyle R,S,T$
. . and their respective opposite sides: $\displaystyle r=19.8,\;s=17,4,\;t=12.1$

The area of the triangle is: .$\displaystyle A \;=\;\frac{1}{2}rs\sin T$
. .
(One-half the product of two sides times the sine of the included angle.)

To find $\displaystyle \angle T$, we use a variation of the Law of Cosines: .$\displaystyle \cos T \;=\;\frac{r^2 + s^2 - t^2}{2rs}$

We have: .$\displaystyle \cos T \;=\;\frac{19.8^2 + 17.4^2 - 12.1^2}{2(19.8)(17,4)} \;=\;0.795875421$

. . Then: .$\displaystyle T \:\approx\:37.3^o$

Hence: .$\displaystyle A \;=\;\frac{1}{2}rs\sin T \;=\;\frac{1}{2}(19.8)(17.4)\sin37.3^o \;=\;104.2918726$

. . Therefore: .$\displaystyle A \;\approx\;104.3\text{ units}^2$

5. Thanks EmptySet! I have a qu. about how you did it though:
you did:
h=12.1 sin (2)
h = 12.1 sin(0.514)
How did you get sin2? In the triangle you drew, you wrote something for one of the angles, but i can't make out what number it is.

6. Thanks Soroban! When finding cosT, can I use the cosine law (instead of the variation, cuz i think that's what is expected of me)? i did, and I got
cosT = 25.4167.
cos 25.4167 = 0.9032
so angle T = 25 degrees... which is very different from your answer. i'm not sure why i'm getting a different answer?

7. never mind those 2 qu., lol.
i've got it!!
thank you guys soo much!!!