Properties of Triangles JEE Main previous year questions with solutions

5 solved JEE Main 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.

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1. Q1JEE Main 2024 (06 Apr, Shift 2)
   In a triangle $\mathrm{ABC}, \mathrm{BC}=7, \mathrm{AC}=8, \mathrm{AB}=\alpha \in \mathrm{N}$ and $\cos \mathrm{A}=\frac{2}{3}$. If $49 \cos (3 \mathrm{C})+42=\frac{\mathrm{m}}{\mathrm{n}}$, where $\operatorname{gcd}(\mathrm{m}, \mathrm{n})=1$, then $\mathrm{m}+\mathrm{n}$ is equal to________
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   Answer: 39

   $\begin{aligned} & \cos \mathrm{A}=\frac{\mathrm{b}^2+\mathrm{c}^2-\mathrm{a}^2}{2 \mathrm{bc}} \\ & \frac{2}{3}=\frac{8^2+\mathrm{c}^2-7^2}{2 \times 8 \times \mathrm{c}} \\ & \mathrm{C}=9 \\ & \cos \mathrm{C}=\frac{7^2+8^2-9^2}{2 \times 7 \times 8}=\frac{2}{7} \\ & 49 \cos 3 \mathrm{C}+42 \\ & 49\left(4 \cos ^3 \mathrm{C}-3 \cos \mathrm{C}\right)+42 \\ & 49\left(4\left(\frac{2}{7}\right)^3-3\left(\frac{2}{7}\right)\right)+42 \\ & =\frac{32}{7} \\ & \mathrm{~m}+\mathrm{n}=32+7=39\end{aligned}$
2. Q2JEE Main 2015 (10 Apr)Measurement of Angles
   In a $\Delta ABC$, $\dfrac{a}{b}=2+\sqrt{3}$, and $\angle C=60^{\circ}.$ Then the ordered pair $(\angle A,\angle B)$ is equal to:
   1. A.$\left(105^{\circ},15^{\circ}\right)$
   2. B.$\left(15^{\circ},105^{\circ}\right)$
   3. C.$\left(45^{\circ},75^{\circ}\right)$
   4. D.$(75^{\circ},45^{\circ})$
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   Answer: (A)

   $\begin{matrix}\dfrac{a}{b}\gt 1\Rightarrow \angle A\gt \angle B\end{matrix}$ $A+B=120^{\circ}$ $\Rightarrow \dfrac{sin(120^{\circ}-B)}{sinB}=2+\sqrt{3}$ $\begin{matrix}\text{⇒}\dfrac{\sqrt{3}}{2}cotB+\dfrac{1}{2}=2+\sqrt{3}\end{matrix}$ $\begin{matrix}\Rightarrow cotB=\dfrac{2}{\sqrt{3}}\left(\dfrac{3}{2}+\sqrt{3}\right)\end{matrix}$ $\Rightarrow cotB=2+\sqrt{3}$ $\begin{matrix}\Rightarrow B=15^{\circ}\end{matrix}$ Hence, $(\angle A,\angle B)=(105^{\circ},15^{\circ})$
3. Q3JEE Main 2022 (25 Jun, Shift 1)
   Let $a,b$ and $c$ be the length of sides of a triangle $ABC$ such that $\dfrac{a+b}{7}=\dfrac{b+c}{8}=\dfrac{c+a}{9}$. If $r$ and $R$ are the radius of incircle and radius of circumcircle of the triangle $ABC$, respectively, then the value of $\dfrac{R}{r}$ is equal to
   1. A.$2$
   2. B.$\dfrac{3}{5}$
   3. C.$\dfrac{5}{2}$
   4. D.$1$
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   Answer: (C)

   Let $\dfrac{a+b}{7}=\dfrac{b+c}{8}=\dfrac{c+a}{9}=k$ so $a+b=7k,b+c=8k,c+a=9k$ Solving this we get $a=4k,b=3k,c=5k$ and calculating semi-perimeter $S=\dfrac{a+b+c}{2}$ $\Rightarrow S=6k$ We know that $\dfrac{r}{R}=4\sin \dfrac{A}{2}\sin \dfrac{B}{2}\sin \dfrac{C}{2}$ $\Rightarrow \dfrac{r}{R}=4\sqrt{\dfrac{\left(s-b\right)\left(s-c\right)}{bc}}\times \sqrt{\dfrac{\left(s-a\right)\left(s-c\right)}{ac}}\times \sqrt{\dfrac{\left(s-a\right)\left(s-b\right)}{ab}}$ $\Rightarrow \dfrac{r}{R}=4\cdot \dfrac{\left(s-a\right)\left(s-b\right)\left(s-c\right)}{abc}$ $\Rightarrow \dfrac{r}{R}=4\cdot \dfrac{\left(2k\right)\left(3k\right)\left(k\right)}{\left(4k\right)\left(3k\right)\left(5k\right)}$ $\Rightarrow \dfrac{R}{r}=\dfrac{5}{2}$
4. Q4JEE Main 2019 (08 Apr, Shift 2)
   If the lengths of the sides of a triangle are in A.P and the greatest angle is double the smallest, then a ratio of lengths of the sides of this triangle is:
   1. A.$3:4:5$
   2. B.$5:6:7$
   3. C.$5:9:13$
   4. D.$4:5:6$
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   Answer: (D)

   Let $a\lt b\lt c$ are three sides and $\theta$ is the smallest angle. Hence three angles are $\theta , \pi -3\theta , 2\theta$ Given, $2b=a+c\Rightarrow 2\sin ⁡B=\sin ⁡A+\sin ⁡C$ (using sine rule) $\Rightarrow 2sin 3\theta =sin \theta +sin 2\theta$ $\Rightarrow 2\left(3sin \theta -4{sin}^{3}\theta \right)=sin \theta \left(1+2cos \theta \right)$ $\Rightarrow 6-8\left(1-{cos}^{2}\theta \right)=1+2cos \theta$ $\Rightarrow cos \theta =\dfrac{3}{4}, -\dfrac{1}{2}$ ( $-\dfrac{1}{2}$ is rejected since $\theta$ is acute) $∴ a:b:c=sinA:sin B:sin C=sin \theta :sin 3\theta : sin 2\theta$ $=1:3-4 {sin}^{2}\theta :2cos \theta =1:4{cos}^{2}\theta -1:2 cos \theta$ $=4:5:6$
5. Q5JEE Main 2019 (11 Jan, Shift 2)
   Given $\frac{\mathrm{b}+\mathrm{c}}{11}=\frac{\mathrm{c}+\mathrm{a}}{12}=\frac{\mathrm{a}+\mathrm{b}}{13}$ for a $\Delta \mathrm{ABC}$ with usual notation. If $\frac{\cos \mathrm{A}}{\alpha}=\frac{\cos \mathrm{B}}{\beta}=\frac{\cos \mathrm{C}}{\gamma},$ then the ordered triad$(\alpha, \beta, \gamma)$ has a value
   1. A.(7,19,25)
   2. B.(3,4,5)
   3. C.(5,12,13)
   4. D.(19,7,25)
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   Answer: (A)

   Let $\frac{b+c}{11}=\frac{c+a}{12}=\frac{a+b}{13}=k$ (Say). $\therefore \quad b+c=11 k, c+a=12 k, a+b=13 k$ $\therefore \quad a+b+c=18 k$ $\therefore \quad a=7 k, b=6 k$ and $c=5 k$ $\therefore \quad \cos A=\frac{36 k^{2}+25 k^{2}-49 k^{2}}{2.30 k^{2}}=\frac{1}{5}$ and $\cos B=\frac{49 k^{2}+25 k^{2}-36 k^{2}}{2.35 k^{2}}=\frac{19}{35}$ and $\cos C=\frac{49 k^{2}+36 k^{2}-25 k^{2}}{2.42 k^{2}}=\frac{5}{7}$ $\therefore \quad \cos A: \cos B: \cos C=7: 19: 25$ $\therefore \quad \frac{\cos A}{7}=\frac{\cos B}{19}=\frac{\cos C}{25}$ Hence, required ordered triplet is (7,19,25)