Topic: Ambiguous case sine
June 27, 2019 / By Jami Question:
Given triangle ABC with sides of a = 6, b = 10 and measure of angle A = (in degrees, minutes, seconds) 31˚10', find the other sides (c) and other angle measures (B)(C)
This IS an ambiguous case problem.
10 points for correct answer.
Ephah | 4 days ago
use sine rule
31.10 = 31.17 degrees
a/sinA = b/sinB = c/sinC
6/sin(31.17) = 10/sin(B)
11.59 = 10/sinB
sinB = 10/11.59
sin(B) = 0.86
B = 59.3 degrees
so C = 180 - (A+B) = 180 -(31.17 + 59.3) = 89.53
now a/sinA = c/sinC
11.59 = c/sin(89.53)
c = 11.59*sin(89.53) = 11.59
It is almost right angled triangle with c as hypotenuse
a thank you to unravel those issues is by using making use of the regulation of Sines to unravel for the sine of the unknown perspective, ?, say. For any answer to be achievable, we could have sin ? ? a million, so anticipate that it is so. on account that ? and ? ? ? have the comparable sine, computing ? from sin ? will many times produce 2 strategies. If the sine is a million, i.e., ? = ?/2, there's a different answer. additionally, one answer could be impossible because it may produce a triangle the sum of whose angles exceeds ?. In the two of those 2 circumstances, the answer is unique; in any different case, it rather is the ambiguous case. So, the ambiguous case is whilst the unknown perspective (Z on your case) does have 2 achievable values. on your problem we've x / sin X = z / sin Z ?sin Z = z sin X / x = 7.5 sin fifty 8?/9.3 = 0.6839 ?Z = 40 3.15? or 136.80 5?. on account that 136.80 5? + fifty 8? > one hundred eighty?, the 2nd answer is impossible and Z = 40 3.15?. So, your answer is actual.
try using this online calculator just plug in the values it should work. http://www.csgnetwork.com/trianglessacalc.html
Originally Answered: Question on ambiguous case triangle.please explain?
Just adding on the previous answer:
Draw a triangle and make the angle A be the 60 degree angle, and the side opposite is 40cm. It doesn't matter how you label the other two sides, but let's make the side opposite the angle B be the 15cm side. now, using Law of Sines, you get that
sin(60)/40 = sin(B)/15
Solve for sin(B):
sin(B) = 15sin(60)/40 = 3sqrt(3)/16
Now, the angle B = arcsin(3sqrt(3)/16), in degrees = 18.951 degrees, or 19 if you wanna round. BUT Sine is positive in both Quadrants 1 and 2, so there are possible two answers. Angle B could be either 19 degrees or 180-19=161 degrees.
However we can clearly see that B can't be 161 as 161+60 = 221 which is clearly more than 180 degrees which is the sum of the angles in a triangle. So there is only one solution of 19 degrees.