Question

Match the following by appropriately matching the lists based on the information given in $\text{Column I}$ and $\text{Column II}$.

$\begin{array}{cc}ColumnI& ColumnII\left(Typeof△ABC\right)\\ \text{a.}\cot \frac{A}{2}=\frac{b+c}{a}& \text{p. always right angled}\\ \text{b.}a\tan A+b\tan B=(a+b)\tan \frac{A+B}{2}& \text{q. always isosceles}\\ \text{c.}a\cos A=b\cos B& \text{r. may be right angled}\\ \text{d.}\cos A=\frac{\sin B}{2\sin C}& \text{s. may be right angled isosceles}\end{array}$

• A. $a-p,q;b-q,r;c-p,s;d-q,r$
• B. $a-p,q;b-q,s;c-p,s;d-q,s$
• C. $a-p,s;b-q,r;c-p,s;d-q,r$
• D. $a-p,s;b-q,r,s;c-r,s;d-q,r,s$

Answer 1

The correct option is D $a-p,s;b-q,r,s;c-r,s;d-q,r,s$

$\cot \frac{A}{2}=\frac{b+c}{a}\Rightarrow \frac{\cos \left(A/2\right)}{\sin \left(A/2\right)}=\frac{\sin B+\sin C}{\sin A}\Rightarrow \frac{\cos \left(A/2\right)}{\sin \left(A/2\right)}=\frac{2\sin \frac{B+C}{2}\cos \frac{B-C}{2}}{2\sin \frac{A}{2}\cos \frac{A}{2}}\Rightarrow \cos \frac{A}{2}=\cos \frac{B-C}{2}A=B-C\Rightarrow A+C=BA+B+C=2B=\pi \Rightarrow B=\frac{\pi }{2}$
So, its a right angled triangle

B.$a\tan A+b\tan B=(a+b)\tan \frac{A+B}{2}$
It can be written as
$a[\tan A-\tan \frac{1}{2}(A+B)]=b[\tan \frac{1}{2}(A+B)-\tan B]$
$\Rightarrow \frac{a\sin (A-\frac{A+B}{2})}{\cos A\cos \frac{A+B}{2}}=\frac{b\sin (\frac{A+B}{2}-B)}{\cos \frac{A+B}{2}\cos B}\Rightarrow \frac{2R\sin A\sin \frac{A-B}{2}}{\cos A}=\frac{2R\sin B\sin \frac{A-B}{2}}{\cos B}\Rightarrow \sin \frac{A-B}{2}\times (\tan A-\tan B)=0\Rightarrow A=B$
hence triangle is an isosceles triangle and can be right angled.

C.
$a\cos A=b\cos B2R\sin A\cos A=2R\sin B\cos B\sin 2A=\sin 2BA=BorA+B=\frac{\pi }{2}$
Hence triangle is isosceles or right angled.

D.
$\cos A=\frac{\sin B}{2\sin C}\Rightarrow \frac{b^2+c^2-a^2}{2bc}=\frac{b}{2c}{b}^2+c^2-a^2=b^2\Rightarrow c^2=a^2,c=a$
Hence the triangle is isosceles