Suppose you wish to apply SSA to a triangle, in order to find an angle measure. Also suppose the given side opposite the given angle is less than the other given side. In addition, the ratio of the longer side to the shorter side, multiplied by the sine of the angle opposite the shorter side, is less than 1. Which of the following statements is true?

A. There will be infinitely many solutions for the angle.
B. There will be zero solutions for the angle.
C. There will be one solution for the angle.
D. There will be two solutions for the angle.

It says that the ratio of the longer side to the shorter side, multiplied by the sine of the angle opposite the shorter side, is less than 1. It means

[tex] \frac{BC}{AB} [/tex] · sin(C) < 1

We know about Law of Sines of Triangle is given by :-

[tex] \frac{AB}{sin(C)} =\frac{BC}{sin(A)} =\frac{AC}{sin(B)} \\\\
Solving \;\;\frac{AB}{sin(C)} =\frac{BC}{sin(A)} \\\\
Cross\;\; Multiplying\\\\
AB\; sin(A)=BC\; sin(C)\\\\
sin(A)=\frac{BC}{AB}\; sin(C)\\\\
sin(A) <1 [/tex]\\

We got sin(A) < 1, and BC > AB.

Therefore, angle ∡A could be either acute or obtuse angle.

So, There will be two solutions for the angle ∡A.

⇒ option D is the final answer.

boffeemadrid boffeemadrid · Answer 2 · 2019-03-13T07:56:53+01:00

Answer:

(D)

Step-by-step explanation:

Consider a triangle ΔABC, in which sides AB, BC and angle ∠C are given.

Now, according to the question, Angle C is the angle opposite of side AB.

And, the given side opposite the given angle is less than the other given side. that is AB < BC.

Also, it is given that the ratio of the longer side to the shorter side, multiplied by the sine of the angle opposite the shorter side, is less than 1, that is

[tex](\frac{BC}{AB})SinC<1[/tex] (1)

Now, Law of Sines of Triangle, we get

[tex]\frac{AB}{sinC}=\frac{BC}{sinA}=\frac{AC}{sinB}[/tex]

Taking the first and second equality, we have

[tex]\frac{AB}{sinC}=\frac{BC}{sinA}[/tex]

⇒[tex]ABsinA=BCsinC[/tex]

⇒[tex]sinA=\frac{BC}{AB}sinC[/tex]

Using equation (1),we get

[tex]sinA<1[/tex]

Therefore, BC > AB

Hence, angle A could be either acute or obtuse angle.

Therefore, There will be two solutions for the angle A making option D correct.