+0  
 
0
201
2
avatar

Solve each triangle described below.Round measures of sides and angles to the nearest tenth. 

1. B=71° , c=8 , b=16

2. A=40° , B=14° , a=52

 Oct 1, 2019
 #1
avatar+11714 
+1

Solve each triangle described below.Round measures of sides and angles to the nearest tenth. 

laugh

 Oct 1, 2019
 #2
avatar+109450 
+2

1.  B =71°, c  = 8, b  = 16

We have a SSA situation.....so.....we will have either 1 triangle, 2 triangles or 0 triangles

Using the Law of Sines

sin 71 / 16  = sin C / 8

sin C  = .5sin71

arcsin (.5sin 71)  = C ≈  28.2°  or   180 -  28.2  = 151.8°

Note that the second angle value isn't possible since  71 + 151.8  > 180°

So....only one triangle exists

A  = 180 - 71 - 28.2  = 80.8°

And

a/ sin (80.8)  = 16/sin 71

a  = 16sin (80.8)/sin 71  ≈ 16.7

 

A = 80.8°   a  = 16.7

B = 71°      b  = 16

C = 28.2°   c =  8

 

cool cool cool

 Oct 1, 2019

21 Online Users

avatar
avatar
avatar