In ΔABC, given that a=14.2cm and angle A=38˚ and angle B=75˚, find the length of side b.?

Question

In ΔABC, given that a=14.2cm and angle A=38˚ and angle B=75˚, find the length of side b.

Answer 1

a/sinA=b/sinB
14.2/sin38*=b/sin75*
b=14.2(sin75*/sin38*)

Answer 2

on the grounds which you recognize that a triangle has one hundred eighty tiers, perspective B = one hundred eighty - (38 + seventy 5) = sixty seven. you are able to then use the regulation of sines to compute the fee. sin(perspective B) / part b = sin(perspective A) / part a sin(sixty seven) / b = sin(38) / 14.2 .9205/ b = .6156 / 14.2 .9205 / b = .04335 / a million Now go multiply. .04335b = .9205 b= 21.23 approximately.

Answer 3

By applying sine rule,

a/sin A = b/sin B
14.2/sin 38° = b/sin75°
b = 14.2sin75° / sin38°
b = 22.279 cm

Answer 4

The most important rule for the law of sines are that sin(α)/a=sin(β)/b=sin(γ)/c

sin(38)/14.2=sin(75)/b
b=(sin(75)*14.2)/sin(38)
b=22.3 cm.

Check:
sin(38)/14.2=.0433
sin(75)/22.3=.0433

The guy above me also has the right answer even though he did the problem differently.

Answer 5

14.2/sin(38) = b/sin(75)
b = (14.2sin(75))/(sin(38))
b = about 22.28cm

Using Law of Signs