# Thread: The Ambiguous Case...Two Solutions

1. ## The Ambiguous Case...Two Solutions

Solve the triangle (Two Solutions)

a = 6, b = 8, alpha = 35 degrees

2. Originally Posted by magentarita
Solve the triangle (Two Solutions)

a = 6, b = 8, alpha = 35 degrees
Is alpha the angle opposite side c or opposite one of the other sides (side a or side b)? And what are you trying to find ..... side c, or the other two angles, or side c and the other two angles?

3. Originally Posted by magentarita
Solve the triangle (Two Solutions)

a = 6, b = 8, alpha = 35 degrees
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

4. ## look...

Originally Posted by mr fantastic
Is alpha the angle opposite side c or opposite one of the other sides (side a or side b)? And what are you trying to find ..... side c, or the other two angles, or side c and the other two angles?

Alpha = side a

A for a, get it?

5. ## Perfect

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
Perfectly done!

6. Originally Posted by magentarita
Alpha = side a

A for a, get it?

2. When you're getting free professional help from someone it's wise not to get too cute with them. You cut them a bit of slack when they ask you a question.

Get it?

7. Originally Posted by magentarita
Alpha = side a

A for a, get it?
Alpha is the angle opposite side a.

Show some respect for others trying to help by making your statements clear, and avoid potentialy offensive comments like: "A for a, get it?". If someone more experienced with this stuff has trouble understanding what you write perhaps you should consider that the fault may lie with you rather than with them.

RonL