Please help I just have a few little things I get stuck on with these problems but I am not sure what they are. Thank you so much for taking time to read this!
Tell how many triangles there are and solve each triangle completely
1 angle a equals 39.7 degrees angle c equals 91.6 degrees the length
of b equals 16.4
I solved it and got
---angle b=48.7 degrees
---a=13.9
---c=21.8
---# of triangles -------idk------------
This is a ASA triangle so I am unsure how to determine how many triangles there are. I was thinking I should use the SSA rules?
3 angle a equals 41.2 degrees side a equals 8.1 side b equals 10.6
---angle c=79.3 degrees
---angle b=59.5 degrees
---c=12.1
---# of triangles 2
the second triangle
---angle b=120.5 degrees
---angle c=18.3 degrees
---c=3.9
how i got the second triangle because for an SSA example having side a
8.1 side b 10.6 and angle a 41.2
1. find h=bsinangleA if h>a there is no triangle
h=10.6sin41.2=6.98 6.98 is not > 8.1 so there must be a triangle
2. if h=a there is one right triangle angle b is 90 degrees and side b
is hypotenuse solve using right angle trig
6.98 is not equal to 8.1
3. if h<a<b there are 2 triangles this one is correct 6.98<8.1<10.6one with angle b acute and one with
angle b obtuse. find the acute b using law of sines. law of sines sin41.2/8.1 = sinB/10.6 --- sinB=10.6sin41.2/8.1 angle B=59.5 59.5 cannot be the acute angle because angle a is 41.2 degrees so what am I doing wrong... subtract it from
180 degrees to get the obtuse angle b. in each of the two triangles
find angle y using angle y=180 degrees-angle a-angle b and side c using
law of sines
4 if a is greater than or equal to b there is only one triangle and
angle b is acute (angle a or angle y might be obtuse) find angle b
using law of sines then find angle y using y=180-angle a-angle b. find
c using law of sines
6.98 is not > or = 8.1
**************************ISSUE ANGLE B IS SUPPOSED TO BE ACUTE BUT IT
IS NOT IT IS 59.5 DEGREES WHILE ANGLE A IS 41.2 DEGREES WHAT DID I DO
WRONG**************************