Originally Posted by
Shyam here you see, a<b, so calculate, $\displaystyle b \sin \alpha$
$\displaystyle \alpha= A$,
$\displaystyle b \sin \alpha= 8\times \sin 35$
$\displaystyle =4.5<6$
$\displaystyle a > b \sin \alpha $
so, two solutions, two triangles. Draw the diagram. CB = CB' = 6 cm = a
we will use two triangles ABC and AB'C
use sine law in triangle AB'Cto find angle B'
$\displaystyle \frac{8}{\sin B'}= \frac{6}{\sin \alpha}$
$\displaystyle \frac{8}{\sin B'}= \frac{6}{\sin 35}$
$\displaystyle 6 \sin B' = 8 \sin 35$
$\displaystyle sin B' = 0.764768581$
B'= 49.88 = 50 degrees.
In triangle CBB', angle B = angle B' = 50 degrees.
So, supplementary angle B = 180 - 50 = 130 degrees.
In triangle ACB, angle C = 180 - (35 + 130) = 15 degrees.
In triangle ACB', angle C = 180 - (35 + 50) = 95 degrees.
Use sin law again in triangle ACB,
$\displaystyle \frac {AB}{\sin 15}=\frac{6}{\sin 35} $
AB = 2.7 cm
Use sin law again in triangle ACB',
$\displaystyle \frac {AB'}{\sin 95}=\frac{6}{\sin 35} $
AB = 10.4 cm
Solution is :
For triangle ABC,
angle A= $\displaystyle \alpha$ = 35 degrees, B = 130 degrees, C = 15 degrees
a= 6 cm, b = 8 cm, c = AB = 2.7 cm
For triangle AB'C,
angle A= $\displaystyle \alpha$ = 35 degrees, B' = 50 degrees, C = 95 degrees
a= 6 cm, b = 8 cm, c = AB' = 10.4 cm