Answer:
Step-by-step explanation:
You want to know how many triangles are specified by sides and angle of a = 8, c = 6, and C = 10°, and the solutions for each.
When two sides and an adjacent angle are specified, the number of triangles may be 0, 1, or 2. If the specified angle is opposite the longer of the two sides, then there will always be one triangle.
If the specified angle is opposite the shorter of the two sides, as here, the number of triangles depends on the relation between the given values:
short / long < sin(θ) ⇒ 0 triangles
short / long = sin(θ) ⇒ 1 right triangle
short / long > sin(θ) ⇒ 2 triangles
Here, we have 6/8 > sin(10°), so there will be two solutions.
The law of sines is used to find the angle opposite the other given side.
[tex]\dfrac{\sin(A)}{a}=\dfrac{\sin(C)}{c}\\\\A=\arcsin\left(\sin(C)\cdot\dfrac{a}{c}\right)\\\\A=\arcsin(\dfrac{4}{3}\sin(10^\circ))\approx13.4^\circ[/tex]
In this application, the angle whose sine is 4/3sin(10°) may be the supplement of this value as well.
A' = 180° -A = 166.6°
The third angle in the triangle will have the value that brings the total to 180°:
B = 180° -10° -A = 156.6°
B' = 180° -10° -A' = 3.4°
The law of sines is used again to find the length of the third side of the triangle.
[tex]\dfrac{b}{\sin(B)}=\dfrac{c}{\sin(C)}\\\\b=c\cdot\dfrac{\sin(B)}{\sin(C)}[/tex]
Then for the possible values of angle B, the possible values of side b are ...
[tex]b=6\cdot\dfrac{\sin(156.6^\circ)}{\sin(10^\circ)}\approx13.7\\\\b'=6\cdot\dfrac{\sin(3.4^\circ)}{\sin(10^\circ)}\approx2.04[/tex]
In summary, the solutions for the two possible triangles are ...
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Additional comment
We have given the side b=2.04 to two decimal places, because a side length of 2.0 would make that triangle look like a straight line segment.
