Solve for A B and C A2C 5 9 5 6 9 Solution9Pythagoras theore

A2C 5 9 5 6 9

Solution

9).Pythagoras\' theorem

AB^2 = BC^2 + AC^2

AB^2 = 6^2 +15^2

AB^2 = 36+225

AB^2 = 261

AB = sqrt(261)

AB=16.2

we know all side lengths AB(c) =16.2 AC(b)=6 , BC(a) =15

from the triangle C=90 degree

we know the sine law

a/sinA =b/sinB =c/sinC

15/sinA = 16.2/sin90

15/sinA =16.2

sinA= 15/16.2

sinA =0.926

A = arc sin 0.926

A=67.8 = 68(approximatly)

we know sum of angles in a triangle is 180

A+B+C=180

68+B+90=180

158+B=180

B=180-158

B=22

A=68,B=22,C=90

11).

AB^2 = AC^2 +CB^2

AB^2= 2^2 + 5.9^2

AB^2 = 4+34.81

AB^2=38.81

AB= sqrt(38.81)

AB(c)=6.2

we know AB(c) =6.2 , BC(a) = 5.9 , AC(b) =2 , C=90 degree

again apply sine law

a/sinA=b/sinB=c/sinC

5.9/sinA= 6.2/sin90

5.9/sinA= 6.2/1

sinA= 5.9/6.2

sinA=0.952

A =arc sin 0.952

A=72 degree

sum of angles in a triangle =180

A+B+C=180

72+B+90=180

B+162=180

B=180-162

B=18

A=72,B=18C=90