The Ambiguous Case...Two Solutions

**magentarita** · September 13th 2008, 06:13 AM

Solve the triangle (Two Solutions)

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

**mr fantastic** · September 13th 2008, 02:30 PM

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?

Your post is well named

**Shyam** · September 13th 2008, 03:03 PM

Originally Posted by magentarita

Solve the triangle (Two Solutions)

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

here you see, a<b, so calculate, $b \sin \alpha$

$\alpha= A$ ,

$b \sin \alpha= 8\times \sin 35$

$=4.5<6$

$a > b \sin \alpha$

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'

$\frac{8}{\sin B'}= \frac{6}{\sin \alpha}$

$\frac{8}{\sin B'}= \frac{6}{\sin 35}$

$6 \sin B' = 8 \sin 35$

$sin B' = 0.764768581$

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,

$\frac {AB}{\sin 15}=\frac{6}{\sin 35}$

AB = 2.7 cm

Use sin law again in triangle ACB',

$\frac {AB'}{\sin 95}=\frac{6}{\sin 35}$

AB = 10.4 cm

Solution is :

For triangle ABC,
angle A= $\alpha$ = 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= $\alpha$ = 35 degrees, B' = 50 degrees, C = 95 degrees
a= 6 cm, b = 8 cm, c = AB' = 10.4 cm

**magentarita** · September 13th 2008, 09:15 PM

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?

Your post is well named

Alpha = side a

A for a, get it?

**magentarita** · September 13th 2008, 09:16 PM

Originally Posted by Shyam

here you see, a<b, so calculate, $b \sin \alpha$

$\alpha= A$ ,

$b \sin \alpha= 8\times \sin 35$

$=4.5<6$

$a > b \sin \alpha$

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'

$\frac{8}{\sin B'}= \frac{6}{\sin \alpha}$

$\frac{8}{\sin B'}= \frac{6}{\sin 35}$

$6 \sin B' = 8 \sin 35$

$sin B' = 0.764768581$

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,

$\frac {AB}{\sin 15}=\frac{6}{\sin 35}$

AB = 2.7 cm

Use sin law again in triangle ACB',

$\frac {AB'}{\sin 95}=\frac{6}{\sin 35}$

AB = 10.4 cm

Solution is :

For triangle ABC,
angle A= $\alpha$ = 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= $\alpha$ = 35 degrees, B' = 50 degrees, C = 95 degrees
a= 6 cm, b = 8 cm, c = AB' = 10.4 cm

Perfectly done!

**mr fantastic** · September 13th 2008, 10:57 PM

Originally Posted by magentarita

Alpha = side a

A for a, get it?

1. Clarity when asking your questions is not too much to ask for.

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?

**CaptainBlack** · September 13th 2008, 11:41 PM

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

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