Suppose a triangle has the sides a, b, c. Let theta be the side opposite of b. If costheta0 What must be true a^2+b^2c^2 a^2+c^2b^2 a^2+c^2

Question

Suppose a triangle has the sides a, b, c. Let theta be the side opposite of b. If costheta>0 What must be true
a^2+b^2>c^2
a^2+c^2>b^2
a^2+c^2<b^2
a^2+b^2=c^2

owen3 · Answer

a^2 + b^2 - 2ab cos theta = c^2 

solve for a^2 + b^2
a^2 + b^2 = c^2 + 2ab cos theta

now if cos theta >0 , then
what can we say , assuming a,b >0

owen3 · Answer

you there?

cutiecomittee123 · Answer

Yes, I am lost. @owen3

cutiecomittee123 · Answer

it wouldn't make sense for any side lengths or angles to be less than zero, so of course they have to be greater than 0.

michele_laino · Answer

hint:
if we apply the theorem of Carnot, to such triangle, we get:
$$\Large  {b^2} = {a^2} + {c^2} - 2ac\cos \theta $$
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cutiecomittee123 · Answer

Okay I with you, but still lost with the equation part. I am a visual learner so its hard for me to learn when I just see the equations and laws that we need to use to figure this out

michele_laino · Answer

since $\cos \theta >0$, then we have:
$$\Large 2ac\cos \theta>0 $$

michele_laino · Answer

so:
$$\Large  {a^2} + {c^2} - 2ac\cos \theta  < {a^2} + {c^2}$$

cutiecomittee123 · Answer

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