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

Question

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

Answer 1

Since you know two angles you can find the third: B = 180 - 38 - 75 = 67

Then use law of sines:

sin A/a = sin B/b
sin 38/14.2 = sin 67/b
b = sin67/sin38 (14.2) = 21.2 cm

Answer 2

Since you know that a triangle has 180 degrees, angle B = 180 - (38 + 75) = 67. You can then use the law of sines to compute the value. sin(angle B) / side b = sin(angle A) / side a
sin(67) / b = sin(38) / 14.2
.9205/ b = .6156 / 14.2
.9205 / b = .04335 / 1
Now cross multiply.
.04335b = .9205
b= 21.23 approximately.

Answer 3

the main mandatory rule for the regulation of sines are that sin(?)/a=sin(?)/b=sin(?)/c sin(38)/14.2=sin(seventy 5)/b b=(sin(seventy 5)*14.2)/sin(38) b=22.3 cm. verify: sin(38)/14.2=.0.33 sin(seventy 5)/22.3=.0.33 the guy above me additionally has the main suitable answer even nonetheless he did the project in a diverse way.

Answer 4

angle B =67 deg.

Side b =(sin 67deg)*14.2/(sin 38 deg)= 21.1 cm
aprxoximately

Answer 5

mLB=113

I think..