JEE Advanced 2020 Concept Recapitulation Test 1 Paper 2

(a) In a triangle ABC,A+B+C=π, so tan‌A+tan‌B+tan‌C=tan‌A‌tan‌B‌tan‌C. Since none of tan‌A,tan‌B,tan‌C can be zero, a triangle given in (a) is not possible.
(b) ‌

(sin‌A)

2

=‌

(sin‌B)

3

=‌

(sinC)

7

, so that by the laws of sines ‌

a

2

=‌

b

3

=‌

c

7

⇒‌

a+b

c

=‌

5

7

which is not possible, as the sum of two sides of a triangle is greater than the third.
(c) We have (a+b)2=c2+ab,
so that ‌

a2+b2−c2

2ab

=−‌

1

2

⇒cos‌C=−‌

1

2

⇒C=120∘ and √2(sin‌A+cos‌A)=√3
⇒‌

1

√2

‌sin‌A+‌

1

√2

‌cos‌A=‌

√3

2

⇒sin(A+‌

π

4

)=‌

√3

2

⇒A+‌

π

4

=‌

π

3

That is, A=‌

π

12

=15∘ and B=45∘. Sucha triangle is possible.
(d) Since cos‌A‌cos‌B=sin‌A‌sin‌B=‌

√3

4

, we get cos(A+B)=0, i.e. A+B=‌

π

2

,
Also, from cos(A−B)=‌

√3

2

,
we get A−B=‌

π

6

, so that A=‌

π

3

,B=‌

π

6

and sin‌A+sinB=‌

√3

2

+‌

1

2

=‌

√3+1

2

Hence, such a triangle is also possible.