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Given a=10 and B=150° in Triangle ABC, determine the values of b for which A has

A) exactly one value

B) two possible values

C) no value​

1 Answer

7 votes

Answer:

A) b > 10

B) none

C) b ≤ 10

Explanation:

For the purpose of this answer, we assume ABC is a triangle if and only if all angles and side lengths have measures greater than 0, and the sum of angles is exactly 180°.

The fact that B = 150° means A+C = 30°. That is, 0 < A < 30°. Taking the sine of A, we have ...

0 < sin(A) < 1/2

A, B)

The Law of Sines tells us ...

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

sin(A) = a·sin(B)/b = 10·(1/2)/b = 5/b

Substituting this into the inequality for sin(A), we get ...

0 < 5/b < 1/2

0 < 10 < b . . . . . multiply by 2b (which is > 0)

In order to form a triangle, the value of b must be greater than 10. In that case, there will always be only one value for A:

A = arcsin(5/b) . . . . b > 10

__

C)

For b = 10, the "triangle" is a straight line, a disallowed condition. That is A will have no allowed value.

For b < 10, A > 30°, which forces C < 0°. This is also a disallowed condition. So we can say ...

b ≤ 10, A has no allowed value

_____

The attached diagram shows triangle ABC. Point A can be anywhere on ray BA. As segment c gets longer, angle C increases from near 0 to near 30°. At the same time, angle A decreases from near 30° to near 0°. When b=10, c=0, and the figure is not a triangle. Nor is it a triangle for b < 10.

Given a=10 and B=150° in Triangle ABC, determine the values of b for which A has A-example-1
User Phil Cook
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