(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=