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Section 6.1
How is this done? Do 34 in step by step fashion.
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°)