Question

With usual notations, in $\triangle ABC$ if a = 2, b = 3, c = 5 and $\frac{cos A}{a}+\frac{cos B}{b}+\frac{cos C}{c}=\frac{k+7}{30}$ then k =

Answer 1

12

Answer 2

The given problem involves a triangle ABC with side lengths a=2, b=3, and c=5. We need to find the value of k from the given equation:

$\frac{\cos A}{a}+\frac{\cos B}{b}+\frac{\cos C}{c}=\frac{k+7}{30}$

First, let's analyze the given side lengths. For a triangle to be formed, the triangle inequality must hold: the sum of the lengths of any two sides must be greater than the length of the third side. Here, a+b = 2+3 = 5. Since c=5, we have a+b=c. This indicates a degenerate triangle, where the three vertices are collinear. Specifically, if A and B are the endpoints of the side c (length 5), then C must lie on the line segment AB such that AC=b=3 and CB=a=2. In this configuration, the angle C (opposite side c) is $180^\circ$, and angles A and B are $0^\circ$.

We can use the cosine rule for each angle: $\cos A = \frac{b^2+c^2-a^2}{2bc}$ $\cos B = \frac{a^2+c^2-b^2}{2ac}$ $\cos C = \frac{a^2+b^2-c^2}{2ab}$

Substitute these into the expression $\frac{\cos A}{a}+\frac{\cos B}{b}+\frac{\cos C}{c}$: $\frac{\cos A}{a}+\frac{\cos B}{b}+\frac{\cos C}{c} = \frac{1}{a}\left(\frac{b^2+c^2-a^2}{2bc}\right) + \frac{1}{b}\left(\frac{a^2+c^2-b^2}{2ac}\right) + \frac{1}{c}\left(\frac{a^2+b^2-c^2}{2ab}\right)$ To combine these terms, we find a common denominator, which is $2abc$: $= \frac{b^2+c^2-a^2}{2abc} + \frac{a^2+c^2-b^2}{2abc} + \frac{a^2+b^2-c^2}{2abc}$ $= \frac{(b^2+c^2-a^2) + (a^2+c^2-b^2) + (a^2+b^2-c^2)}{2abc}$ $= \frac{a^2+b^2+c^2}{2abc}$ This is a general identity for any triangle.

Now, substitute the given values a=2, b=3, c=5 into this identity: $a^2 = 2^2 = 4$ $b^2 = 3^2 = 9$ $c^2 = 5^2 = 25$ $a^2+b^2+c^2 = 4+9+25 = 38$

$2abc = 2 \times 2 \times 3 \times 5 = 60$

So, the expression becomes: $\frac{\cos A}{a}+\frac{\cos B}{b}+\frac{\cos C}{c} = \frac{38}{60} = \frac{19}{30}$

We are given that $\frac{\cos A}{a}+\frac{\cos B}{b}+\frac{\cos C}{c}=\frac{k+7}{30}$. Equating the two expressions: $\frac{19}{30} = \frac{k+7}{30}$ Multiply both sides by 30: $19 = k+7$ Solve for k: $k = 19 - 7$ $k = 12$