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

1 Answer
Don't Memorise
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
#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#

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