Question

Find the area of the triangle with A = 41°, b = 22 in, and c = 36 in.

Answer 1

Ok, I may need a moment to ponder this one.

Answer 2

ok :) ill be here

Answer 3

alrighty and how do u do tht lol

Answer 4

We can find "a" with law of cosines.
\[a^2 = b^2 + c^2 - 2ac\cos(A)\]

Answer 5

2bc cos(a)*

Answer 6

Hell, Lets get an expert in here. I am drawing a blank for some reason.

Answer 7

area = 1/2bc sin A
= 1/2 (55.88cm)(91.44cm) sin 41
= 1676.12 cm2
= 659.89 in2

Answer 8

i wasnt rlly asking for the answert though... but ok..

Answer 9

its wrong anyway though were useing cos

Answer 10

why not just using sin's rule where Area =1/2bcsinA

Answer 11

to find the area we need height and base ....

use height = 22 sin(41) and base = 36

Answer 12

Or according to @Ash_Hlovate 's attachment, we could find the missing side and calculate area that way?

Answer 13

i think tht could work

Answer 14

i think so too.
btw if im wrong im so sorry..

Answer 15

lol its okay

Answer 16

\[a^2 = b^2 + c^2 - 2ac\cos(A)\]
\[K = area\]
\[K = \sqrt{s(s-a)(s-b)(s-c)}\]
\[s = \frac{a+b+c}{2}\]
b=22
c=36
A=41deg

Answer 17

Or @amistre64 's method.
\[A = \frac{1}{2}\times b\times h\]
\[h = 22\sin(41)\]
\[b = 36\]

Answer 18

Herons method is soooo brutish

Answer 19

its only use is when you dont know anything else about the triangle other than the side lenghts

Answer 20

Yeah true. That is why I posted yours further down.
However, if you wouldn't mind is if you could explain how you figured height as that? I am unfamiliar with how you would have calculated that.