Question

find formula of s in terms of a, b, cos(x)

Answer 1

Answer:

$\displaystyle s = \frac{2ab\cos x}{a+b}$

Step-by-step explanation:

We want to find a formula for s in terms of a, b, and cos(x).

Let the point where s intersects AB be D.

Notice that s bisects ∠C. Then by the Angle Bisector Theorem:

$\displaystyle \frac{a}{BD} = \frac{b}{AD}$

We can find BD using the Law of Cosines:

$\displaystyle BD^2 = a^2 + s^2 - 2as \cos x$

Likewise:

$\displaystyle AD^2 = b^2+ s^2 - 2bs \cos x$

From the first equation, cross-multiply:

$bBD = a AD$

And square both sides:

$b^2 BD^2 =a^2 AD^2$

Substitute:

$\displaystyle b^2 \left(a^2 + s^2 - 2as \cos x\right) = a^2 \left(b^2 + s^2 - 2bs \cos x\right)$

Distribute:

$a^2b^2 + b^2s^2 - 2ab^2 s\cos x = a^2b^2 + a^2s^2 - 2a^2 bs\cos x$

Simplify:

$b^2 s^2 - 2ab^2 s \cos x = a^2 s^2 - 2a^2 b s \cos x$

Divide both sides by s (s ≠ 0):

$b^2 s -2ab^2 \cos x = a^2 s - 2a^2 b \cos x$

Isolate s:

$b^2 s - a^2s = -2a^2 b \cos x + 2ab^2 \cos x$

Factor:

$\displaystyle s (b^2 - a^2) = 2ab^2 \cos x - 2a^2 b \cos x$

Therefore:

$\displaystyle s = \frac{2ab^2 \cos x - 2a^2 b \cos x}{b^2- a^2}$

Factor:

$\displaystyle s = \frac{2ab\cos x(b - a)}{(b-a)(b+a)}$

Simplify. Therefore:

$\displaystyle s = \frac{2ab\cos x}{a+b}$