Find the missing angles and sides of the following triangles

Find the missing angles and sides of the following triangles: m B = 73, a = 18, b = 9 In Delta ABC, a = 14 cm, b = 9 cm, c = 6 cm Divide 6x^3 - 19x^2 + 16x - 4 by x - 2 and determine the zeros of a polynomial f(x) = 6x^3 - 19x^2 + 16x - 4 Simplify: (2 + 7i) + (Squareroot 2- i) (2 + 11i)(3-2i) Divide and write answer in standard form: 2 + 3i/5 - 2i, (2 + 7i) (3 - 5i)

Solution

a. <B=73, a=18,b=9

using sine law

sinB/b= SInA/a

Sin 73/9=sinA/18

sinA=18sin73/9= 1.91

Here the value of sin A is more then 1.In such case no such triangle possible

b. a=14, b=9,c=6

using cosine law

a^2= b^2 + c^2 - 2ac cos A

14^2= 9^2 +6^2 - 2(9)(6) cos A

108 cos A= -79

cos A= -79/108137 degree

A=137 degree

Using sine law

sin A/a=sinB/b

sin 137/14=sinB/9

sin B=9sin137/14

B= 26 degree

C=180-A-B= 17 degree