Properties of Triangles 1 Question 16

16. There exists a ABC satisfying the conditions

(a) bsinA=a,A<π2

(b) bsinA>a,A>π2

(c) bsinA>a,A<π2

(d) bsinA<a,A<π2,b>a

Fill in the Blanks

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Answer:

Correct Answer: 16. (a, d)

Solution:

  1. The sine formula is

asinA=bsinBasinB=bsinA

(a) bsinA=aasinB=a

B=π2

Since, A<π2, therefore the triangle is possible.

(b) and (c) bsinA>a

asinB>asinB>1

ABC is not possible.

(d) bsinA<a

asinB<asinB<1B exists.

Now, b>aB>A

Since, A<π2

The triangle is possible.

Hence, (a) and (d) are the correct answers.



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