Question

In triangle ABC, AC=10.4, CB=x andAB=12 (hypotenuse), how do you find BC, m<A, and m<B?

Answer 1

Using the Pythagoras theorem,
let us consider a right angled triangle ABC, right angled at "C"

AB = 12
BC = 'x'
AC = 10.4
`AC^2 + BC^2 = AB^2`
`10.4^2` + `x^2` = `12^(2)`
`x^(2)` = `12^(2)` - `10.4^(2)`
`x^(2)` = 144-108.16
x = `sqrt 35.84`
BC = x = 5.98

calculating the angles:
AB = 12 = side "c"
BC = 5.98 = side "a"
AC = 10.4 = side "b"

by law of sines

a/Sin A = b/Sin B = c/Sin C

`sin (C) = sin (90) = 1`
`c/sin C = 12/1 = 12`
`b/sin B= 10.4/sin B`

and by law: `b/sin B = c/sin C`
so `10.4/sin B = 12`
`sin B = (10.4 /12)`
B = `arcsin (10.4/12)`
`= arcsin (0.86)`
`approx 59.31^o`

Similarly

`a/sin A = c/sin C`

`5.98/sin A = 12`

`sin A = (5.98/12)`

`A= arcsin (0.498) approx 29.86^o`