TRIGONOMETRY,help!!! Draw a triangle ABC and label its sides. Let angle A=θ. Assume that a

You need to use the law of cosines, such
that:

`(BC)^2 = (AB)^2 + (AC)^2 - 2AB*AC*cos (hatA)`

`a^2 = x^2 + b^2 - 2x*b cos theta`

Moving the terms to one side
yields:

`x^2 + b^2 - 2x*b cos theta - a^2 = 0`

You need to rearrange the terms, such
that:

`x^2 - 2x*b cos theta + b^2 - a^2 = 0`

You need to use quadratic formula, such
that:

`x_(1,2) = (2bcos theta+- sqrt(4b^2cos^2 theta - 4b^2 +
4a^2))/2`

You need to notice that `x_(1,2)` exist if the radical is
valid, hence `4b^2cos^2 theta - 4b^2 + 4a^2 >= 0` such
that:

`4b^2cos^2 theta - 4b^2 + 4a^2 >= 0
`

`4b^2cos^2 theta - 4(b^2 - a^2) >=0 `

Since the problem provides that `b>a` , hence `b^2 >
a^2 => b^2 - a^2 > 0`

`cos^2 theta >= (b^2 -
a^2)/b^2 => cos theta >= +-(sqrt(b^2-a^2))/b`

`x_(1,2) = (2bcos theta+- 2sqrt(b^2cos^2 theta - b^2 +
a^2))/2`

`x_(1,2) = (bcos theta+- sqrt(b^2(cos^2 theta - 1) +
a^2))`

`x_(1,2) = (bcos theta+- sqrt(a^2 - b^2 sin^2
theta))`

Hence, evaluating x, under the given
conditions, `cos theta >= +-(sqrt(b^2-a^2))/b` , yields `x_(1,2) = (bcos
theta+- sqrt(a^2 - b^2 sin^2 theta)).`