Trigonometry! #10

Let, $CE = EA= y$ . Using cosine rule, $4^{2} = 4^{2} + y^{2} - 2(4)(y).\cos{C}$ and, $4^{2} = 4^{2} + y^{2} - 8y.\cos{C}.$ We observe that, $y^{2} - 8y.\cos{C} = 0. \cdots (1)$ Again, $5^{2} = (2y)^{2} + 2^{2} - 2(2)(2y).\cos{C}.$ $5^{2} = 4y^{2} + 2^{2} - 8y.\cos{C}.$ This can be written as, $5^{2} = 3y^{2} + 2^{2} + y^{2} - 8y.\cos{C}. \cdots\ (2)$ Substituting equation (1) in (2) we get, $25 = 3y^{2} + 4$ $3y^{2} = 21$ $y^{2} = 7$ $y = \sqrt{7}$ We had assumed that $CE = EA = y$ .

The question is to find the value of $CA$ , which is $CE + EA$ .

$CA = y + y$

$\boxed{CA = 2\sqrt{7}}$

Let $\angle EBC = \theta$

In $\Delta BGD$ , using cosine rule, we get $\cos \theta = \frac{25}{32}$

In $\Delta EBC$ , using cosine rule, we get $EC=\sqrt{7}$

$\therefore AC=\boxed{2\sqrt{7}}$

Using cosine formula 2 times.

AE = EC = x Angle BCA= a

In triangle BCE: 16= 16+ x^2 -8x cos a X= 8. Cos a

In triangle ADC 25= 4x^2 + 4 -8x cos a 25=4x^2 + 4 - x^2 21=3x^2 X^2 = 7 hence, x= sqrt 7,

AC= 2x= 2.sqrt7

Same as Omkar Kulkarni but putting little detail.
Let G be the centroid, intersection of AD and BE.
$In \ \triangle \ DBG, \ \ CosDBG=\dfrac{(DB)^2+(2/3*BE)^2-(AD/3)^2}{2*2*8/3}=\dfrac{2^2+(8/3)^2-(5/3)^2}{2*2*8/3}=\dfrac{25}{32}.\\ In\ \triangle\ CBE\ \ CosCBE=CosDBG=\dfrac{4^2+4^2-CE^2}{2*4*4}=\dfrac{32-CE^2}{32}.\\ \therefore\ CE=\sqrt7.\\ So\ AC=2\sqrt7.$

We can solve it using geometry also. Applying Apollonius thm to triangle ABC twice.Once considering AD as a median and nxt BE as a median.

Extreme shortcut: given that AD = 5 and BC = 4, CA must be close to AD. The only value close to 5 amongst the available choices is 2 \sqrt {7}.

Use formula for length of median drawn to a side in terms of lengths of sides.it can be referred in chapter solution of triangles. Apply that to BE & AD to get two equations in AB & AC . Solve them to get the answer as 2sq.root7..