If the sides of the Triangle are $a$, $b$ and $\sqrt{(a^2+ab+b^2)}$, then find the value of the greatest Angle.
$$\cos C=\frac{a^2+b^2-c^2}{2ab}\tag{1}$$ as you nearly say, but $c$ is $\sqrt{a^2+ab+b^2}$. Substituting that into $(1)$ should give you the $C$ that you want.
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Updated on August 01, 2022Comments
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Admin over 1 year
If the sides of the Triangle are $a$, $b$ and $\sqrt{(a^2+ab+b^2)}$, then find the value of the greatest Angle.
Answer Given in my notebook - $\frac{2\pi}{3}$ radians.
The Question is from the Chapter Solutions of Triangles and I'm aware of Sine and Cosine Rule of the triangles.
My try.
In any case, $a$ or $b$, no matter which side is greater, the 3rd side (Let's denote it by c) $ \sqrt{(a^2+ab+b^2)} $ will have the greatest angle (i.e. greatest angle will be opposite to this side).
Now, let's take Opposite Angles of sides a, b and c as $A$, $B$, and $C$ respectively. Then the only way to find the angle formed at angle C of the triangle can be easily calculated using Cosine Rule (if all the angles are given, which in this case is true), so putting the formula I get -
$$\cos C = \frac{a^2+b^2-c^2}{2ab} = \sqrt{(a^2+ab+b^2)}$$
Now, in the following case, I've to get the Value of Cosine function as $120^o or \frac{2\pi}{3}$ which I'm not sure how to do.
Can anyone help me with that?
Thanks :)
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saulspatz over 5 yearsYou should have a $2$ in the denominator is your formula for the cosine.
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Admin over 5 years@saulspatz Thanks, but still it's difficult for me to solve
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Doug M over 5 yearsrather that the formual abouve, I think the law of cosines directly is more intiufitve. $c^2 = a^2 + b^2 - 2ab\cos C \implies \cos C = -\frac 12$
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Admin over 5 yearsThat's what I'm unable to do, How can you solve such equations?
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Angina Seng over 5 yearsNo need to solve: substitute!
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Admin over 5 yearsSubstitute what in where?
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saulspatz over 5 yearsSubstitute $\sqrt{a^2+ab+b^2}$ for $c$
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Admin over 5 yearsoh okay, I followed it, by mistake, I substituted that size side c, in angle C of cos..... My bad XD :)
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Admin over 5 yearsWould you have any problem if I'd edit your answer and add full Solution to it?
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Angina Seng over 5 years@AbhasKumarSinha I see you've already done that $\ddot\smile$
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Admin over 5 yearsYeah! :) XD :D I waited for your response, it was getting late so I added my answer too