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

Apr 25, 2015

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

AB = 12
BC = 'x'
AC = 10.4
$A {C}^{2} + B {C}^{2} = A {B}^{2}$
${10.4}^{2}$ + ${x}^{2}$ = ${12}^{2}$
${x}^{2}$ = ${12}^{2}$ - ${10.4}^{2}$
${x}^{2}$ = 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

$\sin \left(C\right) = \sin \left(90\right) = 1$
$\frac{c}{\sin} C = \frac{12}{1} = 12$
$\frac{b}{\sin} B = \frac{10.4}{\sin} B$

and by law: $\frac{b}{\sin} B = \frac{c}{\sin} C$
so $\frac{10.4}{\sin} B = 12$
$\sin B = \left(\frac{10.4}{12}\right)$
B = $\arcsin \left(\frac{10.4}{12}\right)$
$= \arcsin \left(0.86\right)$
$\approx {59.31}^{o}$

Similarly

$\frac{a}{\sin} A = \frac{c}{\sin} C$

$\frac{5.98}{\sin} A = 12$

$\sin A = \left(\frac{5.98}{12}\right)$

$A = \arcsin \left(0.498\right) \approx {29.86}^{o}$