help with a trigonometry question please???

Question

In triangle ABC, tan (angle CAB) = 22/7 and the altitude from A divides BC into segments of length 3 and 17. what is the area of triangle ABC?

Answer 1

For discussion purposes, let us denote the foot of the altitude from A to BC to be point D. So we know that DC = 17 and DB = 3. Also, let x = angle CAD and y = angle DAB, and h to be the length of the altitude AD. Hence x + y = angle CAB.
22/7 = tan (angle CAB) = tan(x + y)
= (tan x + tan y)/(1 - tan x tan y)
= (17/h + 3/h)/[1 - (17/h)(3/h)]
= 20/(h - 51/h)
or 11h -11*51/h = 70
This is a quadratic equation and can be solved readily to get solutions h = 11 and h = -51/11.
Since the height can not be negative, we take h = 11 and hence the area of triangle ABC = (3 + 17)h/2 = 10h = 110

Answer 2

Draw the height (let its length be "h") from A to BC, it divides angle A into 2 angles, say α' and a", you have then tan(α') = 3/h, tan(α") = 17/h, apply the additional formula for the tan-function:
tan(α) = tan(α' + α") = (tan(α') + tan(α")) /(1 - tan(α')*tan(α")), so
22/7 = (3/h + 17/h)/(1 - (3/h)*(17/h)), or
22/7 = 20h/(h^2 - 51), solve this 2nd degree equation, its positive root is h = 11 and then the area is 20*11/2 = 110