Trigonomatry question?

Question

in a triangle(all the angles are smaller than 90 degrees).
AB=a
BC=b
AC=c
S=the area
S/a*c=2/5
S/a*b=3/8
find the angles of the triangle

Answer 1

Using the eq'n A = 0.5abSinx
Where 'A' = Area of triangle
'a' length of one side
'b' length of base
'x' the angle between 'a' & 'b'.

Then Sinx = 2A/ab
So angles are :-
Sinx = 2 x 2/5 = 4/5
x = Sin^-1(4/5) = 53.13 degrees
y = Sin^-1(6/8) = 48.59 degrees
z = 180 - 48.59 - 53.13 = 78.28 degrees.

Answer 2

In triangle ABC ,let A be the vertex
draw AD perpendicular from A to the base BC
area of triangle S = 1/2(a) AD
AD = 2S/a ------- eqn ----(1)
from the given relation
S/(a*c) = 2/5; S = (2/5)(a)(c)
2S = (4/5)(a)(c) ------ eqn (2)
S/(a*b) - 3/8 ; S = (3/8) (a)(b)
2S = (3/4)(a)(b) ------- eqn(3)

from right angled triangle ADB
sin B = AD/c
= 2S/(a)(c) ( AD = 2S/ a from eqn (1))
= (4/5)(a)(c)/(a)(c) ( 2S = (4/5)(a)(c) from eqn(2)
= 4/5
B = 53.1 degrees
from the right angled triangle ADC
sin C = AD/b
= 2S/(a)(b)
= (3/4)(a)(b)/(a)(b)
= 3/4
C = 48.6 degrees
A = 180 -(53.1 + 48.6)
180 - 101.7
= 78.3 degrees

Answer 3

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