Question

Suppose a triangle has sides a, b, and c, and that a^2 + b^2 < c^2. Let be the measure of the angle opposite the side of length c. Which of the following must be true?

***choose ALL that apply!!***

A. cos theta < 0

B. cos theta > 0

C. The triangle is not a right triangle.

D. a^2 + b^2 - c^2 = 2abcos theta

***my answer: C and A... What do u think @terenzreignz? :)

Answer 1

since \[c^2\ge b^2+a^2\]

\[c^2=a^2+b^2+\text{something}\]

Answer 2

Well obviously a and b cannot both be true, right? Only one of them...

Answer 3

yes

Answer 4

yeah haha :P

well i think that this is an acute triangle.. am i right when i think that? :)

Answer 5

Start with this...a\[\large a^2 + b^2 < c^2 \quad \rightarrow\quad a^2+b^2-c^2 < 0 \]

Answer 6

So \[\large a^2+b^2-c^2 \]
is negative okay?

Answer 7

so this something is has to be negetive in order for the quanity to be plus
so
\[\cos\theta \le0\]

Answer 8

ok :) how can i verify if it'll work? :D

Answer 9

Let's be formal :D
Since \(\large a^2+b^2-c^2\) is negative, and 2ab is positive, then
\[\large \cos \theta \]

which the law of cosines dictates to be equal to
\[\large \frac{a^2+b^2-c^2}{2ab}\] should be... (positive/negative) ?

Answer 10

negative?

Answer 11

That's right :D
So A is one of them...

Answer 12

\[\cos \theta=\frac{ a^2+b^2-c^2 }{ 2ab } \le 0\]

Answer 13

positive @iheartfood ?
It has a negative numerator and a positive denominator -.-

Answer 14

okay yay! :)

so it would be A and C then? since this isn't a right triangle? :D

if so, what about D?! :D

Answer 15

Can't be a right triangle since cos theta is not 0...
now...
as for D...

Answer 16

Doesn't that like follow directly from the law of cosines?
\[\Large c^2 = a^2 + b^2 -2ab \ \cos(\theta)\]

Answer 17

yeah haha sorry... totally over thought it and over thought it wrongly ahha.. put yeah i see that its negative @terenzreignz cuz of that!

Answer 18

You can just bring \(c^2\) to the other side and do the same to 2ab cos(theta)

Answer 19

\[\Large c^2\color{red}{-c^2}\color{green}{+2ab \ \cos (\theta)} = a^2 + b^2 -2ab \ \cos(\theta)\color{red}{-c^2}\color{green}{+2ab \ \cos(\theta)}\]

Answer 20

okay! so it's A and C then? :)

B and D just aren't the answers here?!

Answer 21

D is just the law of Cosines rearranged...

Answer 22

so D is also an answer then? :O

Answer 23

yes

Answer 24

okay! so A,C, and D it is! :) thx!! :D

Answer 25

okay! so A,C, and D it is! :) thx!! :D