One Angle, One Side

Section 6.1

How is this done? Do 34 in step by step fashion.

 

34.

given: A=60°degrees, a=10

find the value of b such that triangle has

a. one solution
b. two solutions
c. no solution

(a)

The triangle side is a=10 and angle is A=60°

The laws of sines :

a/sin(A)=b/sin(B)=c/sin(C)

The angle A=60° is acute angle.

Recall the relation from the ambiguous case:

h=b*sin(A)

If triangle has one solution and A is an acute angle , then a=h and a>=b

h=10 and b<=10

Substitute h=10, in A=60° in h=b*sin(A) .

10=b*sin(60°)

b=10/sin(60°)

If triangle has one solution, then values of b are:

b<=10 and b=10/sin(60°)


(b)

If triangle has two solutions and A is an acute angle , h< a < b.

Substitute the corresponding value in above formula

b*sin(A) < a/sin(A) < b/sin(A)

b < a/sin(A)

b < a/sin(60°)

Finally conclude that

10< b < 10/sin(60°)


(c)

If triangle has no solution and A is an acute angle, then a<h.

a<b*sin(A)

Divide each side by sin(A)

a/sin(A) <b

or

b >a/sin(A)

Substitute a=10 and A=60° in above expression:

b >10/sin(60°)

 

I don't there is a sample question for one angle, one side given. This is why I posted this thread. I will 33 and 35 when time allows.