Question

- there are angles A,B and C ,angles of triangle ABC
- given this statement :

8sin(A)cos(B)cos(C) = sqrt3

- how is this triangle ABC : obtuse,right or acute ?

Answer 1

any idea how can be this decided ?

Answer 2

@agent0smith opinion please ?

Answer 3

Idk. Maybe you can use some identities?

Answer 4

sin(A)cos(B)cos(C)

We know it's positive... so angle B or C cannot be obtuse, since then cosine would be negative. A could be obtuse.

It still could be right or acute though.

\[\large \frac{ \sqrt 3 }{ 2}*\frac{ 1 }{ 2 }*\frac{ 1 }{2 }= \frac{ \sqrt 3 }{ 8 }\]
A=B=C=60 degrees

Acute only?

Answer 5

Except that A could be obtuse, 90, or acute without changing the sign.
In fact, the following sets of {A,B,C} will give a product of sqrt(3), within round-off errors.
{90,12.82945,77.18055}
{80,23.91573,76.08427}
{100,2.71451,77.28549}
All set should give supplement sums. If not, adjust the third number.

Answer 6

@jhonyy9
Would it be possible to check if there are other restrictions or conditions to this problem? As demonstrated above, all three answers are possible. The asymmetry of the expression opens up to different possibilities.

Answer 7

@mathmate I mentioned A could be obtuse. How will those angles listed give exactly sqrt 3, though?

Answer 8

@agent0smith

The angles are all there and it's not hard to show that the given product gives sqrt(3) for each of the three sets. => There is no unique solution for the question. So the answer is all three, acute, obtuse or 90 degrees.

Perhaps there are other conditions imposed that we have not seen, or perhaps there is a typo. I am waiting for the response from the OP.

Answer 9

@agentsmith
I agree with {60,60,60} being a perfect solution, and is probably what OP is looking for!

Answer 10

@jhonny9
@agentsmith gave an exact acute solution where A=B=C=\(\pi\)/3.
To find other "exact" solutions, I took the easier case of A=90\(^\circ\).
Then
\(8sin(\pi/2)cos(B)cos(C)=\sqrt 3\) which reduces to
\(2cos(B)cos(C)=\sqrt 3 /4\)
But since B and C are complementary, we write cos(B) as sin(C)
\(2sin(C)cos(C)=\sqrt 3 /4\) or
\(sin(2C)=\sqrt 3 /4 \Rightarrow \)
\(C=(1/2)tan^{-1}(\sqrt 3/\sqrt{13})\)

B & C being complementary, so 90-C is also a solution for right triangles that satisfy the given condition
8sin(A)cos(B)cos(C) = sqrt3

Answer 11

Very nice.

Answer 12

@mathmate how did you get these ones, though?
{90,12.82945,77.18055}
{80,23.91573,76.08427}
{100,2.71451,77.28549}

Answer 13

Brute force, bisection method.
I don't mind resorting to brute force when I only need a counter-example! lol

Answer 14

But they give exactly sqrt3 as is, or are they a rounded value of some angle you found? I guess i could plug them into Wolfram to check...

Answer 15

They are just numerical values. However, exact solution exists because I interpolate values between sqrt(3)+ and sqrt(3)-.
I was only looking for existence of the solution to demonstrate that all three cases are answers.

Answer 16

@ganeshie8 please your opinion about this ? ty.