TWO SIDES AND AN ANGLE ASS ARE GIVEN FOR A TRIANGLE ABC DETE

TWO SIDES AND AN ANGLE (ASS) ARE GIVEN FOR A TRIANGLE ABC. DETERMINE WHETHER THE GIVEN MEASUEMENTS PRODUCE ONE TRIANGLE, TWO TRIANGLES, OR NO TRIANGLES AT ALL SOLVING EACH TRIANGLE THAT RESULTS. ROUND LENGTHS TO NEAREST 10^TH, ANGLES TO NEAREST DEGREE a = 10, c = 8.9, A = 63 degree A = 35 degree, a = 12, b = 16 A = 75 degree, a = 51, b = 71

Solution

a. a=10,c=8.9,A=63

Using sine law

sin A/a=sinC/c

sin63/10=sinC/8.9

sin C= 8.9 sin 63/10

C=52.5

B= 180-A-C= 64.5 degree

SinA/a=SinB/b

sin63/10=sin64.5/b

b= 10sin64.5/sin63=10.1

b. A=35 degree, a=12,b=16

SinA/a=SinB/b

sin 35/12=sinB/16

sinB=16 sin35/12

B1=49.9 degree        B2= 130.1degree

C1=180-A-B=95.1      C2= 14.9 degree

Sin A/a=SinC1/c1

c1= a sinC1/sinA= 12 sin 95.1/ sin35= 20.8

c2=asinC2/sinA= 12sin 14.9/sin35=5.4

Here we have two such triangles

c. A=75 degree, a=51 , b=71

SinA/a=SinB/b

SinB= b SinA/a = 71 sin75/51=1.34

Here sin B=1.32 and sin theta always lies between [-1,1]. Hence no such triangle possible