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find the missing parts of ∆ABC when A=50 b=16in and c=11in.show your solution using law of sines

given:
A=?
B=50°
C=?
a=?
b=16in
c=11in

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Sagot :

Answer with Step-by-step explanation:

See answers in underlined bold fonts.

Law of Sine:

a/sinA = b/sinB = c/sinC

Given:

  • b=16 in;   B=50°
  • c=11 in;   C=?
  • a=? ;        A=?

Find C:

b/sinB = c/sinC

16/sin(50) = 11/sinC

Cross multiply:

sinC = [sin (50) (11)] / 16

sin C = [(0.766)(11)]/16

sin⁻¹ C= 0.526625

C = 31.779  

C = 32°

Find A:  The sum of interior angles is 180°.

A = 180° - (50 + 32)°

A = 180° - 82°

A = 98°

Find a:

a/sinA = 16/sin50

a = [(sin(98))(16)]/sin50

a = [(0.990)(16)]/0.766)

a = 20.67

a = 21 in

Please see attached photo for the illustration of the triangle with sides and angles measures.

View image Аноним