Question 1138025
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Area of triangle = 0.5*side1*side2*sin(angle)
The "angle" is between side1 and side2
Your book may refer to this as the SAS, or side-angle-side, triangle area formula.


In this case,
area = 18
side1 = p = 6
side2 = q = 8
angle = R = unknown


Let's use this to solve for R
Area of triangle = 0.5*side1*side2*sin(angle)
Area of triangle = 0.5*p*q*sin(R)
18 = 0.5*p*q*sin(R)
18 = 0.5*6*8*sin(R)
18 = 24*sin(R)
24*sin(R) = 18
sin(R) = 18/24
sin(R) = 0.75
R = arcsin(0.75) or R = 180 - arcsin(0.75)
R = 48.590378 or R = 180 - 48.590378
R = 48.590378 or R = 131.409622
R = 48.6 or R = 131.4


The angle R could be either <font color=red>R = 48.6 degrees or R = 131.4 degrees</font>


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As a check, let's plug in p = 6, q = 8, R = 48.6
Area of triangle = 0.5*p*q*sin(R)
Area of triangle = 0.5*6*8*sin(48.6)
Area of triangle = 18.002666
which is off a bit, but keep in mind that rounding error has occurred. If we used a more accurate value of R (say something like R = 48.590378), then we would be much closer to area = 18. However, we're close enough to effectively verify this value of R.


Let's also check the other angle as well
Plug in p = 6, q = 8, R = 131.4
Area of triangle = 0.5*p*q*sin(R)
Area of triangle = 0.5*6*8*sin(131.4)
Area of triangle = 18.002666
The same issue pops up as before.
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