Solve the triangle completely and determine how many triangles there are:
Given: angle a equals 41.2 degrees side a equals 8.1 side b equals 10.6
I worked out this triangle and got:
---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
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<10.6 one 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
The issue is that angle b is supposed to be an acute angle but it is not because it is 59.5 degrees while angle a is 41.2 degrees. What did I do wrong?