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In triangle ABC, Ab is equal to 3, and bc is equal to 5. Find the number of possible INTEGER values of CA.
*Note: No degenerate triangles, so no triangle area equal to zero, and triangle ABC can be acute, right, or obtuse, Bc can be larger than Ca too.)
If you can, find the number of possible integer values of Ca in an acute ABC triangle.*
No pictures are given for this problem.

Mar 12, 2021

#1
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Let x = CA. By the triangle inequality, $$5-3 . There are \(\boxed{5}$$ integer solutions.

Mar 12, 2021
#2
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Thanks so much. If I don't mind asking, do you know how to do the acute one too?

Guest Mar 12, 2021
#3
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idk what happened to my answer, but I meant $$3\leq x \leq 7$$

anyway, for the second part of the problem, the angle opposite of x will be acute only if $$3^2+5^2>x^2\rightarrow\sqrt{34}>x$$. For integer values, that means that it must be smaller or equal to 5.

So that means that if x is 3, 4, or 5, it will make the angle opposite of x be acute, but that doesn't mean that the triangle itself is acute. If x = 4, then it will for a 3-4-5 triangle, which is right. That means that only 3 and 5 works, so the answer is $$\boxed{2}$$

By the way, if you want to manually check if a triangle is obtuse or not, and you're given sides x, y, and z, then all of these these inequalities must hold true:

$$x^2+y^2>z^2\\ y^2+z^2>x^2\\ x^2+z^2>y^2$$

textot  Mar 12, 2021