if given two angles, use A+B+C=180°
Then, if only one side, use law of sines
If three sides, use law of cosines to find an angle.
If only one angle, use law of sines to get a 2nd one. Then see step 1.
use the information to solve the triangle. If two solutions exist, find both solutions. How do you work this out?
1.A=36°, B=98°, c=16
2.a=4,b=8,c=10
3.A=35°,b=8,c=12
4.A=25°,b=28,a=18
5.B=130°,c=10.1,b=5.2
6.A=150°,b=4.8,a=9.4
2 answers
I assume you are familiar with the tradional way of identifying the information given about a triangle, such as ASA, SAS, SSS and SSA
use "cosine law" for SAS and SSS
use "sine law" for ASA and SSA
#1 - a case of ASA --> a side contained between two angles. We can find the third angle and then use the sine law
Angle C = 46°
a/sin36 = 16/sin46
a = 16sin36/sin45 = appr 13.1
find b in the same way
#2, you have all three sides, so SSS and cosine law.
I always find the smallest angle (opposite the smallest side). That way I don't get confused about the cosine of an obtuse angle being negative
4^2 = 8^2 + 10^2 - 2(8)(10)cosA
160cosA = 64+100-16
cosA = 148/160 = 37/40
angle A = appr 22.33°
Now use the sine law to find a second angle, then use the supplementary angle propery to find the third.
#3 -- cosine law to find a, then use the sine law to find a second angle
#4. a case of SSA, which could be trouble.
let's find angle B
sinB/28 = sin25/18
sinB = .6574..
B = 41.1° or B = 180-41.1 = 138.9°
Case1: Angle B=41.1, Angle A = 25°, so angleC= 113.9°
c/sin113.9 = 18/sin25
c = 38.9
case2: angle B = 138.9, A=25°, C = 16.1°
c/sin16.1 = 18/sin25
c = 11.8
#5 another SSA --- try the sine law
sinC/10.1 = sin130/9.4
sinC = .823..
C = 55.4° or 124.6°
case1: C = 55.4, B=130, which already is more than 180°, so this is not possible
case2: C=124.6, B = 130, way over 180 already.
So this triangle is not possible. (just because we can write down a bunch of data does not mean we can actually draw the triangle.
Try drawing a triangle with the original given data. )
#6
again make your sketch. case of SSA
sinB/4.8 = sin150/9.4
sinB = .253..
B = 14.8° or 165.2°
case1: A = 150, B = 14.8, then C = 15.2°
c/sin15.2 = 9.4/sin150
c = 4.9
case2: A = 150, B = 165.2 ---> not possible, over 180°
notice the problems arise when we have SSA
solving for the angle results in two cases, since the sine is positive in both the first and second quadrants.
use "cosine law" for SAS and SSS
use "sine law" for ASA and SSA
#1 - a case of ASA --> a side contained between two angles. We can find the third angle and then use the sine law
Angle C = 46°
a/sin36 = 16/sin46
a = 16sin36/sin45 = appr 13.1
find b in the same way
#2, you have all three sides, so SSS and cosine law.
I always find the smallest angle (opposite the smallest side). That way I don't get confused about the cosine of an obtuse angle being negative
4^2 = 8^2 + 10^2 - 2(8)(10)cosA
160cosA = 64+100-16
cosA = 148/160 = 37/40
angle A = appr 22.33°
Now use the sine law to find a second angle, then use the supplementary angle propery to find the third.
#3 -- cosine law to find a, then use the sine law to find a second angle
#4. a case of SSA, which could be trouble.
let's find angle B
sinB/28 = sin25/18
sinB = .6574..
B = 41.1° or B = 180-41.1 = 138.9°
Case1: Angle B=41.1, Angle A = 25°, so angleC= 113.9°
c/sin113.9 = 18/sin25
c = 38.9
case2: angle B = 138.9, A=25°, C = 16.1°
c/sin16.1 = 18/sin25
c = 11.8
#5 another SSA --- try the sine law
sinC/10.1 = sin130/9.4
sinC = .823..
C = 55.4° or 124.6°
case1: C = 55.4, B=130, which already is more than 180°, so this is not possible
case2: C=124.6, B = 130, way over 180 already.
So this triangle is not possible. (just because we can write down a bunch of data does not mean we can actually draw the triangle.
Try drawing a triangle with the original given data. )
#6
again make your sketch. case of SSA
sinB/4.8 = sin150/9.4
sinB = .253..
B = 14.8° or 165.2°
case1: A = 150, B = 14.8, then C = 15.2°
c/sin15.2 = 9.4/sin150
c = 4.9
case2: A = 150, B = 165.2 ---> not possible, over 180°
notice the problems arise when we have SSA
solving for the angle results in two cases, since the sine is positive in both the first and second quadrants.
1 answer