SOLUTION: A=99°, a=14.3​in., b=15.3in.
Can you please tell me the answer AND if there is: one solution, two solutions or if there are no solutions. Thank you!!!!
Algebra.Com
Question 1087771: A=99°, a=14.3in., b=15.3in.
Can you please tell me the answer AND if there is: one solution, two solutions or if there are no solutions. Thank you!!!!
Answer by jim_thompson5910(35256) (Show Source): You can put this solution on YOUR website!
Use the law of sines to solve for angle B.
sin(A)/a = sin(B)/b
sin(99)/14.3 = sin(B)/15.3
0.0690691147269327 = sin(B)/15.3 Use a calculator to compute the left side (make sure you are in degree mode)
15.3*0.0690691147269327 = 15.3*sin(B)/15.3 multiply both sides by 15.3
1.05675745532207 = sin(B)
sin(B) = 1.05675745532207
The right hand side value 1.05675745532207 is larger than 1. This value is outside the range of the sine function, which is -1 <= y <= 1.
The value of arcsin(1.05675745532207) is undefined.
This means that we cannot solve for B.
Therefore, No triangle is possible which is another way of saying that there are No solutions.
----------------------------------------------------
We can see this with a drawing
(Image generated by GeoGebra which is free graphing software)
The green angle is angle A = 99 degrees
The green segment is side a = 14.3 units
The red segment is side b = 15.3 units
The blue dashed segment (From point B1 to point B2) is the portion that doesn't exist. It's an empty gap. If it did exist, then we'd be able to make the triangle possible. However we come up short. This drawing is to scale and it visually confirms the answer.
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