How to Find the Area of a Triangle Using Trigonometry

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Math teacher John Schnatterly and I made a series of videos with the goal of "flipping" the classroom. They students watch the video and attempt the challenge question at home, and then do what would typically be homework in the classroom, where the teacher can help.

Check out our other videos in this series:

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Thank you so much, you explained this so much better than my teacher, I finally understand it now. :)

amandabrown

The area for the last problem was 4.386
The angles for the bonus were 49.8 and the other two were 65.1

evollove

I've solved your challenge question, it wasn't easy like I thought but here's how I did it: considering that the area and two of the sides of the triangle were given, I used heron's formula to determine the length of the last side, rearranging it into a 4th degree equation, which was very hairy but gave me the result of the last side, last side ≈ 10.10643 yds.
Then I determined the measure of the angle opposite to the last side with the cosine rule, obtaining ≈ 49.81°, hence the measure of the corresponding angles is equivalent to: (180° - 49.81°) / 2 ; which is ≈ 65.1°
To wrap it up, the angles are 65.1° ; 65.1° and 49.81° ; the last side is approximately 10.10643 yds.
The method that I've use is pretty rough, but it works... I wish you told us how to solve it in an easier way!

Lou-ryyq

you have saved me from failing my class thank 😭

kailibmendo

Shout out to all my British people who use the metric system

TheHolyBaker

Emily is close enough ..I think I got one angle 49.8 deg and two at 65.1 deg which adds up to 180 deg. ....as it is an isosceles triangle ..2 equal sides..2 equal angles

fcar

its approximately 66 degrees for the equal side and 48 for the other one,

highfall

thank you, this video was very nice to learn... :)
but can we have the answer after the questions you gave us like the challenge question and the other question before?

minzu