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1. In $\triangle ABC$, we have $AB = 16$, $BC = 14$, and $\angle A = 60^\circ$. Find the sum of all possible values of $AC$.
2. In $\triangle ABC$, we have $AB = 14$, $BC = 16$, and $\angle A = 60^\circ$. Find the sum of all possible values of $AC$.
3. Let
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hi Hate Number Theory
Let's have a go at number 1 first.
There are two possible places for C, I've marked them C and C'.
The line BD bisects CC' at right angles.
AC + AC' = 2.AD, so you can use trig to get AD and hence answer the question.
Please have a go and post back your answer. If that is OK we can progress to the others.
Bob
Children are not defined by school ...........The Fonz
You cannot teach a man anything; you can only help him find it within himself..........Galileo Galilei
Sometimes I deliberately make mistakes, just to test you! …………….Bob
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Hi, I would just use the cosine rule for the first two.
For problem one, say c=14 and b=16. For problem two, say c=16 and b=14.
Bob is right that there are two solutions for problem one. For problem two there are two solutions algebraically, but because one of them is negative there is actually only one answer.
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For problem three, I think the algebra is extremely tedious to do the old-fashioned way. I have the computed answer with no working if that is all you wanted (I'm not sure it is).
Last edited by Relentless (2016-01-26 04:54:18)
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1. In , we have , , and . Find the sum of all possible values of .2. In , we have , , and . Find the sum of all possible values of .3. Let be positive real numbers. Prove that[list=*]
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[/list]Under what conditions does equality occur? That is, for what values of , , and are the two sides equal?
3. Rewrite
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Let
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and consider the complex numbers
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Then the given inequality follows from the triangle inequality for complex numbers:
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Equality occurs if and only if the two complex numbers and the origin are collinear with the origin not in between, i.e. if and only if
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Last edited by Nehushtan (2016-01-27 00:58:16)
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