Question

Trig and triangles. Is this possible?

One sec, I am typing it up.

Answer 1

Ok, it stats that \( \large a\geq b\space and \space a>h =b * sin A \) there is one triangle

Well I have a situation where a is greater than b and a is greater than h so there should be one triangle Well, using Sine Law, I get that there can be two angles for sine A, which should mean there are two triangles correct?

Answer 2

o-o some one would have to be a pro to figure that out...

Answer 3

It is trig Pre Cal

Answer 4

what grade you in?

Answer 5

College

Answer 6

Yes.
Are there choices? I can give you the answer.

Answer 7

oh wow

Answer 8

No choices. Let me see if I can make one up close to it.

Answer 9

a =40, b =10 , B=5 degrees a and b are sides.

Ok, If we do

\( \large \frac{sin5}{10}=\frac{sin A}{40} = 20.40 \)

Now the law states\( \large a\geq b\space and \space a>h =b * sin A \)

But clearly we have another angle we can use for A. We can use 159.6, which should mean we have two triangles correct?

Answer 10

h = 3.48

Answer 11

@Astrophysics @math1234

Answer 12

@dan815

Answer 13

@Loser66

Answer 14

what is \(h\)

Answer 15

h is the height

Answer 16

This has to do with Sine Law

Answer 17

It has to do with Sine Law and figuring out if there is one or two triangles

Answer 18

height drawn from vertex A to its opposite base?

Answer 19

The rule stats that \( \huge \large a\geq b\space and \space a>h =b * sin A \) is one triange but Sine Law says different because I can use another angle for A.

Answer 20

I am drawing a picture

Answer 21

\(a\geq b\) and \(h = b sinA\)
since BH is the height, we have sin A = h/c
hence \(sinA = \dfrac{h}{c}=\dfrac{bsinA}{c}\), hence b =c

Answer 22

But I am not sure what you want me to do. hehehe....

Answer 23

a =40, b =10 , B=5 degrees

Well, the rule states \( \large a\geq b\space and \space a>h =b * sin A \)
a is greater than b and a is greater than h so we should have one triangle. Ok, but now Sine Law goes

\(\large A=\frac{sin5}{10}=\frac{sin A}{40} = 20.40 \)
h = 10* sin(20.40) =3.42

So a is greater than h

but we can use another angle since 180-20.40 = 159.6 and B only = 5 so 159.6+ 5 = 169.6 so C can equal 10.4 if we use angle 159.6. But this shows we have two triangles using sine law correct but the other rule must not hold true?

Answer 24

15 owlbucks to help me understand. To me it seems the rule(\( \large (a\geq b\space and \space a>h =b * sin A ) \) = one triangle) does not hold true.

Answer 25

In my studies this is call the Ambiguous Case of (SSA) Side Side Angle

Answer 26

I think they are using the rule to differentiate between these two triangles: