Properties of Triangles JEE Advanced previous year questions with solutions

5 solved JEE Advanced questions on Properties of Triangles, free to read — no sign-in needed. The full chapter has 25 questions; sign in to attempt the remaining 20 in the exam simulator.

1. Q1JEE Advanced Adv 2021 (Paper 1)
   In a triangle $ABC$, let $AB=\sqrt{23},BC=3$ and $CA=4$. Then the value of $\dfrac{\cot A+\cot C}{\cot B}$ is
   Show answer & solution

   Answer: 2

   $\dfrac{\cot A+\cot C}{\cot B}=\dfrac{\dfrac{\cos A}{\sin A}+\dfrac{\cos C}{\sin C}}{\dfrac{\cos B}{\sin B}}$ $=\dfrac{\cos A\sin C+\cos C\sin A}{\sin A\sin C}\cdot \dfrac{\sin B}{\cos B}$ $=\dfrac{\sin (A+C)}{\sin A\sin C}\cdot \dfrac{\sin B}{\cos B}$ $=\dfrac{\sin (\pi -B)}{\sin A\sin C}\cdot \dfrac{\sin B}{\cos B}$ $=\dfrac{{\sin }^{2}B}{\sin A\sin C\cos B}$ Let, $AB=c,BC=a,CA=b$ By using sine rule, we get $=\dfrac{{b}^{2}}{ac\cos B}$ $=\dfrac{{b}^{2}}{ac\left(\dfrac{{a}^{2}+{c}^{2}-{b}^{2}}{2ac}\right)}$ $=\dfrac{2{b}^{2}}{{a}^{2}+{c}^{2}-{b}^{2}}$ $=\dfrac{32}{9+23-16}$ $=\dfrac{32}{16}=2$
2. Q2JEE Advanced Adv 2020 (Paper 1)
   Let $x,y$ and $z$ be positive real numbers. Suppose $x,y$ and $z$ are the lengths of the sides of a triangle opposite to its angles $X,Y$ and $Z$ respectively. If $\tan \dfrac{X}{2}+\tan \dfrac{Z}{2}=\dfrac{2y}{x+y+z}$, then which of the following statements is/are TRUE?
   1. A.$2Y=X+Z$
   2. B.$Y=X+Z$
   3. C.$\tan \dfrac{X}{2}=\dfrac{x}{y+z}$
   4. D.${x}^{2}+{z}^{2}-{y}^{2}=xz$
   Show answer & solution

   Answer: B,C

   We have, $\tan \dfrac{X}{2}+\tan \dfrac{Z}{2}=\dfrac{2y}{x+y+z}\&s=x+y+z=$perimeter of a triangle Apply half-angle formula and $x+y+z=s$ then, $\Rightarrow \dfrac{∆}{s\left(s-x\right)}+\dfrac{∆}{s\left(s-z\right)}=\dfrac{y}{s}\Rightarrow ∆=\left(s-x\right)\left(s-z\right)$ $\Rightarrow {∆}^{2}=s\left(s-x\right)\left(s-y\right)\left(s-z\right)={\left(s-x\right)}^{2}{\left(s-z\right)}^{2}$ $\Rightarrow {y}^{2}={x}^{2}+{z}^{2}\Rightarrow \angle Y=90^{\circ}$ $\Rightarrow \angle Y=\angle X+\angle Z$ An option $\left(B\right)$ is correct. Now, $\dfrac{x}{y+z}=\dfrac{\sin X}{\sin Y+\sin Z}=\dfrac{\sin X}{1+\sin \left(\dfrac{\pi }{2}-X\right)}=\dfrac{2\sin \dfrac{X}{2}\cos \dfrac{X}{2}}{2{\cos }^{2}\dfrac{X}{2}}=\tan \dfrac{X}{2}$ An Option $\left(C\right)$ is correct.
3. Q3JEE Advanced Adv 2017 (Paper 2)
   Paragraph: Let $O$ be the origin, and $\overrightarrow{O X}, \overrightarrow{O Y}, \overrightarrow{O Z}$ be three unit vectors in the directions of the sides $\overrightarrow{Q R}, \overrightarrow{R P}$, $\overrightarrow{P Q}$, respectively, of a triangle $P Q R$. Question: If the triangle $P Q R$ varies, then the minimum value of $\cos (P+Q)+\cos (Q+R)+\cos (R+P)$ is
   1. A.$-\dfrac{3}{2}$
   2. B.$\dfrac{3}{2}$
   3. C.$\dfrac{5}{3}$
   4. D.$-\dfrac{5}{3}$
   Show answer & solution

   Answer: (A)

   $\cos \left(P+Q\right)+\cos \left(Q+R\right)+\cos \left(R+P\right)$ $=\cos \left(180-R\right)+\cos \left(180-P\right)+\cos \left(180-Q\right)$ $=-\left[\cos P+\cos Q+\cos R\right]$ In any $\Delta PQR$, $\cos P+\cos Q+\cos R\leq \dfrac{3}{2}$ $\Rightarrow -\left(\cos P+\cos Q+\cos R\right)\geq -\dfrac{3}{2}$
4. Q4JEE Advanced Adv 2014 (Paper 2)
   In a triangle the sum of two sides is x and the product of the same two sides is y. If ${x}^{2}-{c}^{2}=y,$ (where c is the third side of the triangle) then the ratio of the inradius to the circumradius of the triangle is
   1. A.$\dfrac{3y}{2x\left(x+c\right)}$
   2. B.$\dfrac{3y}{2c\left(x+c\right)}$
   3. C.$\dfrac{3y}{4x\left(x+c\right)}$
   4. D.$\dfrac{3y}{4c\left(x+c\right)}$
   Show answer & solution

   Answer: (B)

   Let $x=a+b$ $y=ab$ ${x}^{2}-{c}^{2}=y$ $\Rightarrow \dfrac{{a}^{2}+{b}^{2}-{c}^{2}}{2ab}=-\dfrac{1}{2}=\cos ⁡\left({120}^{o}\right)$ $\Rightarrow \angle C=\dfrac{2\pi }{3}$ $\Rightarrow R=\dfrac{abc}{4∆},r=\dfrac{∆}{s}$ $\Rightarrow \dfrac{r}{R}=\dfrac{4{∆}^{2}}{s\left(abc\right)}=\dfrac{4{\left[\dfrac{1}{2}ab\sin ⁡\left(\dfrac{2\pi }{3}\right)\right]}^{2}}{\dfrac{x+c}{2}y.c}$ $\dfrac{r}{R}=\dfrac{3y}{2c\left(x+c\right)}$
5. Q5JEE Advanced Adv 2012 (Paper 2)
   Let $P Q R$ be a triangle of area $\Delta$ with $a=2, b=\frac{7}{2}$ and $c=\frac{5}{2} ;$ where $a, b$, and $c$ are the lengths of the sides of the triangle opposite to the angles at $P, Q$ and $R$ respectively. Then $\frac{2 \sin P-\sin 2 P}{2 \sin P+\sin 2 P}$ equals.
   1. A.$\frac{3}{4 \Delta}$
   2. B.$\frac{45}{4 \Delta}$
   3. C.$\left(\frac{3}{4 \Delta}\right)^{2}$
   4. D.$\left(\frac{45}{4 \Lambda}\right)^{2}$
   Show answer & solution

   Answer: (C)

   $\frac{2 \sin P-\sin 2 P}{2 \sin P+\sin 2 P}=\frac{2 \sin P-2 \sin P \cos P}{2 \sin P+2 \sin P \cos P}=\frac{1-\cos P}{1-\sin P}$ $=\frac{2 \sin ^{2} \frac{P}{2}}{2 \cos ^{2} \frac{P}{2}}=\tan ^{2} \frac{P}{2}=\frac{(s-b)(s-c)}{s(s-a)}$ where $s=\frac{a+b+c}{2}$ $=\frac{(s-b)^{2}(s-c)^{2}}{s(s-a)(s-b)(s-c)}=\frac{(a+c-b)^{2}(a+b-c)^{2}}{16 . \Delta^{2}}$ $=\frac{\left(2+\frac{5}{2}-\frac{7}{2}\right)^{2}\left(2+\frac{7}{2}-\frac{5}{2}\right)^{2}}{16 \Delta^{2}}=\frac{1 \times 9}{16 \Delta^{2}}=\left(\frac{3}{4 \Delta}\right)^{2}$