Type I Sine, Cosine, Projection Formula
1. In a triangle $ABC$, if $\cos A + 2 \cos B + \cos C = 2$ and the lengths of the sides opposite to the angle $A$ and $C$ are $3$ and $7$ respectively, then $\cos A - \cos C$ is equal to
(a) $\frac{3}{7}$
(b) $\frac{9}{7}$
(c) $\frac{10}{7}$
(d) $\frac{5}{7}$
JEE Main-12.04.2023, Shift-I
Assam CEE-2022
Ans. (c) : Given that,
$$\cos A + 2\cos B + \cos C = 2$$Sides of triangle are $3, 7, 5$
$$\cos A + \cos C = 2(1 - \cos B)$$ $$2 \cos \left(\frac{A+C}{2}\right) \cos \left(\frac{A-C}{2}\right) = 4 \sin^2 \left(\frac{B}{2}\right)$$$\because \cos\left(\frac{A+C}{2}\right) = \sin\left(\frac{B}{2}\right)$
$$2 \sin\left(\frac{B}{2}\right) \cos\left(\frac{A-C}{2}\right) = 4 \sin^2\left(\frac{B}{2}\right)$$ $$\cos\left(\frac{A-C}{2}\right) = 2 \sin\left(\frac{B}{2}\right)$$Multiplying both sides by $2 \cos\left(\frac{B}{2}\right)$:
$$2 \cos\left(\frac{B}{2}\right) \times \cos\left(\frac{A-C}{2}\right) = 4 \sin\left(\frac{B}{2}\right) \cos\left(\frac{B}{2}\right)$$ $$2 \cos\left(\frac{B}{2}\right) \times \cos\left(\frac{A-C}{2}\right) = 2 \sin B$$$\because A + B + C = \pi \implies A + C = \pi - B$
$$\sin\left(\frac{A+C}{2}\right) = \sin\left(\frac{\pi - B}{2}\right) \implies \sin\left(\frac{A+C}{2}\right) = \cos\left(\frac{B}{2}\right)$$Then, $2 \sin\left(\frac{A+C}{2}\right) \cos\left(\frac{A-C}{2}\right) = 2 \sin B$
$$\sin A + \sin C = 2 \sin B$$ $$a + c = 2b \implies a = 3, c = 7, b = 5$$ $$\cos A - \cos C = \frac{b^2 + c^2 - a^2}{2bc} - \frac{a^2 + b^2 - c^2}{2ab}$$ $$= \frac{25 + 49 - 9}{70} - \frac{9 + 25 - 49}{30}$$ $$= \frac{65}{70} - \frac{-15}{30} = \frac{13}{14} + \frac{1}{2} = \frac{20}{14} = \frac{10}{7}$$2. Let $\frac{\sin A}{\sin B} = \frac{\sin(A - C)}{\sin(C - B)}$, where $A, B$ and $C$ are angles of a $\Delta ABC$. If the lengths of the sides opposite these angles are $a, b$ and $c$ respectively, then
(a) $b^2 - a^2 = a^2 + c^2$
(b) $b^2, c^2$ and $a^2$ are in AP
(c) $c^2, a^2$ and $b^2$ are in AP
(d) $a^2, b^2$ and $c^2$ are in AP
JEE Main 27.08.2021, Shift-I
Ans. (b) : Given that,
$$\frac{\sin A}{\sin B} = \frac{\sin(A - C)}{\sin(C - B)}$$The sum of $\angle A + \angle B + \angle C = 180^\circ$ in triangle $ABC$.
$$\therefore \frac{\sin[\pi - (B + C)]}{\sin[\pi - (C + A)]} = \frac{\sin(A - C)}{\sin(C - B)}$$ $$\frac{\sin(B + C)}{\sin(C + A)} = \frac{\sin(A - C)}{\sin(C - B)}$$ $$\sin(B + C) \sin(C - B) = \sin(C + A) \sin(A - C)$$ $$\sin^2 C - \sin^2 B = \sin^2 A - \sin^2 C$$ $$2 \sin^2 C = \sin^2 A + \sin^2 B$$ $$\sin^2 C = \frac{\sin^2 A + \sin^2 B}{2}$$Hence, $b^2, c^2, a^2$ are in A.P.
3. In a triangle $ABC$, $BC = 7$, $AC = 8$, $AB = \alpha \in \mathbb{N}$ and $\cos A = \frac{2}{3}$. If $49 \cos(3C) + 42 = \frac{m}{n}$, where $\gcd(m,n) = 1$, then $m + n$ is equal to_______.
JEE Mains 06/04/2024 Shift-II
Ans. (39) :
$$\because \cos A = \frac{b^2 + c^2 - a^2}{2bc}$$ $$\frac{2}{3} = \frac{8^2 + \alpha^2 - 7^2}{2 \times 8 \times \alpha}$$ $$\frac{2}{3} = \frac{15 + \alpha^2}{16\alpha}$$ $$32\alpha = 3\alpha^2 + 45$$ $$3\alpha^2 - 32\alpha + 45 = 0$$Now, solving this quadratic equation:
$$\alpha = 9 \quad \text{or} \quad \alpha = \frac{5}{3}$$ $$\implies \alpha = 9 \quad (\because \alpha \in \mathbb{N})$$Now, $\cos C = \frac{a^2 + b^2 - c^2}{2ab}$
$$\cos C = \frac{7^2 + 8^2 - 9^2}{2 \times 7 \times 8} = \frac{32}{112} = \frac{2}{7}$$The value of $49 \cos 3C + 42$:
$$= 49 (4 \cos^3 C - 3 \cos C) + 42$$ $$= 49 \left( 4 \left(\frac{2}{7}\right)^3 - 3 \left(\frac{2}{7}\right) \right) + 42$$ $$= 49 \left( \frac{32}{343} - \frac{6}{7} \right) + 42$$ $$= 49 \left( \frac{32 - 294}{343} \right) + 42$$ $$= \frac{-262}{7} + 42 = \frac{-262 + 294}{7} = \frac{32}{7}$$ $$\Rightarrow \frac{m}{n} = \frac{32}{7}$$So the value of $m + n = 32 + 7 = 39$
4. If the line $x = y = z$ intersects the line $x \sin A + y \sin B + z \sin C - 18 = 0 = x \sin 2A + y \sin 2B + z \sin 2C - 9$, where $A, B, C$ are the angles of a triangle $ABC$, then $80 \left( \sin \frac{A}{2} \sin \frac{B}{2} \sin \frac{C}{2} \right)$ is equal to ____
JEE Mains 15/04/2023 Shift-I
Ans. (5) : Given,
$$x \sin A + x \sin B + x \sin C = 18 \quad (\because x = y = z)$$ $$\sin A + \sin B + \sin C = \frac{18}{x} \quad \text{---(i)}$$and $x \sin 2A + x \sin 2B + x \sin 2C = 9 \quad (\because x = y = z)$
$$\sin 2A + \sin 2B + \sin 2C = \frac{9}{x} \quad \text{---(ii)}$$On dividing equation (i) by equation (ii), we get:
$$\frac{\sin A + \sin B + \sin C}{\sin 2A + \sin 2B + \sin 2C} = \frac{2}{1}$$ $$\frac{4 \cos \frac{A}{2} \cos \frac{B}{2} \cos \frac{C}{2}}{4 \sin A \sin B \sin C} = 2$$ $$\frac{4 \cos \frac{A}{2} \cos \frac{B}{2} \cos \frac{C}{2}}{32 \sin \frac{A}{2} \cos \frac{A}{2} \sin \frac{B}{2} \cos \frac{B}{2} \sin \frac{C}{2} \cos \frac{C}{2}} = 2$$ $$\frac{1}{8 \sin \frac{A}{2} \sin \frac{B}{2} \sin \frac{C}{2}} = 2$$ $$\sin \frac{A}{2} \sin \frac{B}{2} \sin \frac{C}{2} = \frac{1}{16}$$Hence, $80 \left( \sin \frac{A}{2} \sin \frac{B}{2} \sin \frac{C}{2} \right) = 80 \left( \frac{1}{16} \right) = 5$
5. In a triangle $ABC$, if $\cos A + 2 \cos B + \cos C = 2$ and the lengths of the sides opposite to the angles $A$ and $C$ are $3$ and $7$ respectively, then $\cos A - \cos C$ is equal to
(a) $\frac{5}{7}$
(b) $\frac{9}{7}$
(c) $\frac{3}{7}$
(d) $\frac{10}{7}$
JEE Mains 12/04/2023 Shift-I
Ans. (d) : Given,
$$\cos A + 2 \cos B + \cos C = 2, \quad a = 3, \, c = 7$$ $$\frac{b^2 + c^2 - a^2}{2bc} + 2 \left( \frac{a^2 + c^2 - b^2}{2ac} \right) + \frac{a^2 + b^2 - c^2}{2ab} = 2$$ $$\frac{b^2 + 49 - 9}{14b} + \frac{9 + 49 - b^2}{21} + \frac{9 + b^2 - 49}{6b} = 2$$ $$\frac{b^2 + 40}{14b} + \frac{58 - b^2}{21} + \frac{b^2 - 40}{6b} = 2$$Solving for $b$ gives $b = 5$.
Now,
$$\cos A - \cos C = \frac{b^2 + c^2 - a^2}{2bc} - \frac{a^2 + b^2 - c^2}{2ab}$$ $$= \frac{25 + 49 - 9}{70} - \frac{9 + 25 - 49}{30}$$ $$= \frac{65}{70} + \frac{15}{30} = \frac{13}{14} + \frac{1}{2} = \frac{20}{14} = \frac{10}{7}$$6. Let $ABC$ be an isosceles triangle in which $A$ is at $(-1, 0)$, $\angle A = \frac{2\pi}{3}$, $AB = AC$ and $B$ is on the positive $x$-axis. If $BC = 4\sqrt{3}$ and the line $BC$ intersects the line $y = x + 3$ at $(\alpha, \beta)$, then $\frac{\beta^4}{\alpha^2}$ is_________.
JEE Mains 01/02/2024 Shift-II
Ans. = 36: Given,
$$\angle A = \frac{2\pi}{3} \quad \text{and} \quad AB = AC$$According to question, by sine rule:
$$\frac{a}{\sin A} = \frac{b}{\sin B} = \frac{c}{\sin C}$$ $$\frac{c}{\sin \frac{\pi}{6}} = \frac{4\sqrt{3}}{\sin \frac{2\pi}{3}}$$ $$\frac{c}{\frac{1}{2}} = \frac{4\sqrt{3}}{\frac{\sqrt{3}}{2}}$$ $$2c = 8 \implies c = 4$$ $$AB = b = 1 + 3 = 4 \quad (B \text{ is at } (3, 0))$$$b = 3$, $m_{AB} = 0$
$$m_{BC} = -\frac{1}{\sqrt{3}}$$ $$\text{Line } BC: y - 0 = -\frac{1}{\sqrt{3}}(x - 3)$$ $$\sqrt{3}y + x = 3$$Now, finding the point of intersection of $y = x + 3$ and $\sqrt{3}y + x = 3$:
$$\sqrt{3}(x + 3) + x = 3 \implies (\sqrt{3} + 1)x = 3 - 3\sqrt{3}$$ $$\alpha = \frac{3(1 - \sqrt{3})}{\sqrt{3} + 1} = \frac{3(1 - \sqrt{3})(\sqrt{3} - 1)}{2} = \frac{-3(4 - 2\sqrt{3})}{2} = 3\sqrt{3} - 6$$ $$\beta = \alpha + 3 = 3\sqrt{3} - 3$$ $$\therefore \frac{\beta^4}{\alpha^2} = 36$$Type II Area of Triangle m-n Rule
7. For a triangle $ABC$, the value of $\cos 2A + \cos 2B + \cos 2C$ is least. If its inradius is $3$ and incentre is $M$, then which of the following is NOT correct ?
(a) area of $\Delta ABC$ is $\frac{27\sqrt{3}}{2}$
(b) $\sin 2A + \sin 2B + \sin 2C = \sin A + \sin B + \sin C$
(c) perimeter of $ABC$ is $18\sqrt{3}$
(d) $MA \cdot MB = -18$
JEE Main-01.02.2023, Shift-I
Ans. (a) : Given that,
$$\text{Radius, } r = 3$$If $\cos 2A + \cos 2B + \cos 2C$ is minimum, then $A = B = C = 60^\circ$ [because radii to all sides are equal].
So, $\Delta ABC$ is equilateral.
In $\Delta MBD$, we have:
$$\frac{MD}{BD} = \tan 30^\circ$$ $$\frac{r}{a/2} = \tan 30^\circ$$ $$\frac{3}{a/2} = \frac{1}{\sqrt{3}} \implies a = 6\sqrt{3}$$ $$\text{Perimeter of } \Delta ABC = 3a = 18\sqrt{3}$$ $$\text{Area of } \Delta ABC = \frac{\sqrt{3}}{4} a^2 = \frac{\sqrt{3}}{4} (6\sqrt{3})^2 = 27\sqrt{3}$$Thus option (a) is NOT correct (as it gives $\frac{27\sqrt{3}}{2}$).
8. If in a triangle $ABC$, $AB = 5$ units, $\angle B = \cos^{-1}\left(\frac{3}{5}\right)$ and radius of circumcircle of $ABC$ is $5$ units, then the area (in square units) of $\Delta ABC$ is
(a) $10 + 6\sqrt{2}$
(b) $8 + 2\sqrt{2}$
(c) $6 + 8\sqrt{3}$
(d) $4 + 2\sqrt{3}$
JEE Main 20.07.2021, Shift-I
Ans. (c) : Given that in $\Delta ABC$,
$$\cos B = \frac{3}{5}, \quad R = 5$$ $$\therefore \sin B = \frac{4}{5}$$We know that,
$$\frac{c}{\sin C} = \frac{a}{\sin A} = \frac{b}{\sin B} = 2R$$ $$\therefore \frac{5}{\sin C} = \frac{a}{\sin A} = \frac{b}{4/5} = 10$$ $$\sin C = \frac{1}{2} \implies C = 30^\circ \quad \text{and} \quad b = 8$$ $$\because \sin A = \sin(B + C) = \sin B \cos C + \cos B \sin C$$ $$\sin A = \frac{4}{5} \cdot \frac{\sqrt{3}}{2} + \frac{3}{5} \cdot \frac{1}{2} = \frac{4\sqrt{3} + 3}{10}$$ $$\Delta = \frac{1}{2} bc \sin A = \frac{1}{2} \times 8 \times 5 \times \left( \frac{4\sqrt{3} + 3}{10} \right) = 6 + 8\sqrt{3}$$9. Let $ABC$ be an equilateral triangle. A new triangle is formed by joining the middle points of all sides of the triangle $ABC$ and the same process is repeated infinitely many times. If $P$ is the sum of perimeters and $Q$ is the sum of areas of all the triangles formed in this process, then:
(a) $P^2 = 36\sqrt{3} Q$
(b) $P^2 = 6\sqrt{3} Q$
(c) $P^2 = 36 \cdot 3 Q$
(d) $P^2 = 72\sqrt{3} Q$
JEE Mains 06/04/2024 Shift-II
Ans. (a) : We know that,
$$\text{Area of equilateral triangle} = \frac{\sqrt{3} a^2}{4}$$ $$\text{Area of 1st triangle} = \frac{\sqrt{3} a^2}{4}$$ $$\text{Area of 2nd triangle} = \frac{\sqrt{3}}{4} \left(\frac{a}{2}\right)^2 = \frac{\sqrt{3} a^2}{16}$$ $$\text{Area of 3rd triangle} = \frac{\sqrt{3} a^2}{64}$$ $$\text{Sum of all areas } (Q) = \frac{\sqrt{3} a^2}{4} \left( 1 + \frac{1}{4} + \frac{1}{16} + \frac{1}{64} + \dots \right)$$ $$Q = \frac{\sqrt{3} a^2}{4} \left( \frac{1}{1 - 1/4} \right) = \frac{\sqrt{3} a^2}{3}$$ $$\text{Perimeter of 1st triangle} = 3a$$ $$\text{Perimeter of 2nd triangle} = \frac{3a}{2}$$ $$\text{Perimeter of 3rd triangle} = \frac{3a}{4}$$ $$P = 3a \left( 1 + \frac{1}{2} + \frac{1}{4} + \dots \right) = 3a \left( \frac{1}{1 - 1/2} \right) = 6a$$ $$a = \frac{P}{6}$$Substituting $a = \frac{P}{6}$ into $Q$:
$$Q = \frac{\sqrt{3} \left(\frac{P}{6}\right)^2}{3} = \frac{\sqrt{3} P^2}{108}$$ $$P^2 = \frac{108}{\sqrt{3}} Q = 36\sqrt{3} Q$$10. For a triangle $ABC$, the value of $\cos 2A + \cos 2B + \cos 2C$ is least. If its inradius is $3$ and incentre is $M$, then which of the following is NOT correct?
(a) $\sin 2A + \sin 2B + \sin 2C = \sin A + \sin B + \sin C$
(b) perimeter of $\Delta ABC$ is $18\sqrt{3}$
(c) $MA \cdot MB = -18$
(d) Area of $\Delta ABC$ is $\frac{27\sqrt{3}}{2}$
JEE Mains 01/02/2023 Shift-I
Ans. (d) : Given, Radius $r = 3$, Centre $= M$
$$s = \frac{a + b + c}{2} = \frac{3a}{2}$$We know that inradius $r = \frac{\Delta}{s}$
$$r \cdot s = \Delta$$ $$3 \times \frac{3a}{2} = \frac{\sqrt{3}}{4} a^2$$ $$\frac{9a}{2} = \frac{\sqrt{3}}{4} a^2 \implies a = 6\sqrt{3}$$ $$\text{Perimeter of } \Delta ABC = 3a = 3 \times 6\sqrt{3} = 18\sqrt{3}$$ $$\text{Area of triangle} = \frac{\sqrt{3}}{4} a^2 = \frac{\sqrt{3}}{4} (6\sqrt{3})^2 = 27\sqrt{3}$$So, option (d) is correct as the false statement.
11. A circle is inscribed in an equilateral triangle of side of length $12$. If the area and perimeter of any square inscribed in this circle are $m$ and $n$, respectively, then $m + n^2$ is equal to
(a) 396
(b) 408
(c) 312
(d) 414
JEE Mains 06/04/2024 Shift-I
Ans. (b) : We know that $s = \frac{a + b + c}{2}$.
For equilateral triangle, $s = \frac{3a}{2}$ and area of equilateral triangle $(\Delta) = \frac{\sqrt{3} a^2}{4}$.
$$r = \frac{\Delta}{s} = \frac{\frac{\sqrt{3} a^2}{4}}{\frac{3a}{2}} = \frac{a}{2\sqrt{3}} = \frac{12}{2\sqrt{3}} = 2\sqrt{3}$$Let side of square be $b$.
$$b^2 + b^2 = (2r)^2 \implies 2b^2 = 4r^2 \implies b^2 = 2r^2$$ $$b^2 = 2(2\sqrt{3})^2 = 24 \implies b = 2\sqrt{6}$$ $$\text{Area of square } (m) = b^2 = 24$$ $$\text{Perimeter of square } (n) = 4b = 8\sqrt{6}$$ $$\text{So, the value of } m + n^2 = 24 + (8\sqrt{6})^2 = 24 + 384 = 408$$12. In the figure, $\theta_1 + \theta_2 = \frac{\pi}{3}$ and $3 (BE) = 4 (AB)$. If the area of $\Delta CAB$ is $2\sqrt{3} - 3 \text{ unit}^2$, when $\frac{\theta_2}{\theta_1}$ is the largest, then the perimeter (in unit) of $\Delta CED$ is equal to _____.
JEE Mains 10/04/2023 Shift-II
Ans. (6) :
$$3 BE = 4 AB$$ $$\text{Area of } \Delta CAB = 2\sqrt{3} - 3$$Let $AB = CD = x$
$$\Rightarrow \frac{1}{2} x^2 \tan \theta_1 = 2\sqrt{3} - 3 \quad \text{---(i)}$$Now, $BE = BD + DE \implies BE = x(\tan \theta_1 + \tan \theta_2)$
$$\frac{BE}{AB} = \tan \theta_1 + \tan \theta_2 = \frac{4}{3}$$ $$\tan \theta_1 + \tan \theta_2 = \frac{4}{3}$$Given $\theta_1 + \theta_2 = \frac{\pi}{3}$, for $\frac{\theta_2}{\theta_1}$ to be largest, $\theta_1 = \frac{\pi}{6}$ and $\theta_2 = \frac{\pi}{3}$.
From (i):
$$\frac{1}{2} x^2 \tan \frac{\pi}{6} = 2\sqrt{3} - 3$$ $$\frac{1}{2} x^2 \frac{1}{\sqrt{3}} = 2\sqrt{3} - 3 \implies x^2 = 2\sqrt{3}(2\sqrt{3} - 3) = 12 - 6\sqrt{3}$$ $$\text{Perimeter of } \Delta CED = CD + DE + CE = 6$$Type III Radius of Circumcircle, Incircle, Ex-circle
13. Let $a, b$ and $c$ be the length of sides of a triangle $ABC$ such that $\frac{a+b}{7} = \frac{b+c}{8} = \frac{c+a}{9}$. If $r$ and $R$ are the radius of incircle and radius of circumcircle of the triangle $ABC$, respectively, then the value of $\frac{R}{r}$ is equal to
(a) $\frac{5}{2}$
(b) 2
(c) $\frac{3}{2}$
(d) 1
JEE Main-25.06.2022, Shift-I
Ans. (a) : Given that,
$$\frac{a+b}{7} = \frac{b+c}{8} = \frac{c+a}{9} = \lambda$$ $$a + b = 7\lambda$$ $$b + c = 8\lambda$$ $$a + c = 9\lambda$$Adding all these equations above:
$$2(a + b + c) = 24\lambda \implies a + b + c = 12\lambda$$ $$\text{Now } a = 4\lambda, \, b = 3\lambda, \, c = 5\lambda$$ $$\because c^2 = a^2 + b^2 \implies \angle C = 90^\circ$$ $$\Delta = \frac{1}{2} ab \sin C = \frac{1}{2} ab = \frac{1}{2} (4\lambda)(3\lambda) = 6\lambda^2$$ $$\because 2s = 12\lambda \implies s = 6\lambda$$ $$R = \frac{c}{2\sin C} = \frac{5\lambda}{2(1)} = \frac{5\lambda}{2}$$ $$r = \frac{\Delta}{s} = \frac{6\lambda^2}{6\lambda} = \lambda$$ $$\frac{R}{r} = \frac{\frac{5\lambda}{2}}{\lambda} = \frac{5}{2}$$14. In a triangle, the sum of lengths of two sides is $x$ and the product of the lengths of the same two sides is $y$. If $x^2 - c^2 = y$, where $c$ is the length of the third side of the triangle, then the circumradius of the triangle is
(a) $\frac{c}{3}$
(b) $\frac{c}{\sqrt{3}}$
(c) $\frac{3y}{2}$
(d) $\frac{y}{\sqrt{3}}$
JEE Main 11.01.2019, Shift-I
Ans. (b) : Let the sides of the triangle $\Delta ABC$ be $a, b, c$.
$$a + b = x \quad \text{and} \quad ab = y$$ $$\because x^2 - c^2 = y$$Putting the values, we get:
$$(a + b)^2 - c^2 = ab$$ $$a^2 + b^2 + 2ab - c^2 = ab$$ $$a^2 + b^2 - c^2 = -ab$$ $$\frac{a^2 + b^2 - c^2}{2ab} = -\frac{1}{2}$$We know that $\cos C = \frac{a^2 + b^2 - c^2}{2ab}$:
$$\cos C = -\frac{1}{2} \implies C = 120^\circ$$ $$2R = \frac{c}{\sin C} \implies R = \frac{c}{2 \sin 120^\circ} = \frac{c}{2 \times \frac{\sqrt{3}}{2}} = \frac{c}{\sqrt{3}}$$15. If $5, 5r, 5r^2$ are the lengths of the sides of a triangle, then $r$ cannot be equal to
(a) $\frac{5}{4}$
(b) $\frac{7}{4}$
(c) $\frac{3}{2}$
(d) $\frac{3}{4}$
JEE Main 10.01.2019, Shift-I
Ans. (b) :
The triangle is possible only if the sum of any two sides is always greater than the third side.
$$5 + 5r > 5r^2 \quad \text{---(i)}$$ $$5r + 5r^2 > 5 \quad \text{---(ii)}$$ $$5r^2 + 5 > 5r \quad \text{---(iii)}$$From equation (i):
$$5 + 5r > 5r^2 \implies r^2 - r - 1 < 0$$ $$\frac{1 - \sqrt{5}}{2} < r < \frac{1 + \sqrt{5}}{2}$$Since $r > 0$ (length of side must be positive):
$$0 < r < \frac{\sqrt{5} + 1}{2} \approx 1.618 \quad \text{---(iv)}$$From equation (ii):
$$r^2 + r - 1 > 0 \implies r > \frac{\sqrt{5} - 1}{2} \approx 0.618 \quad \text{---(v)}$$From equation (iii):
$$r^2 - r + 1 > 0 \quad (\text{True for all } r \in \mathbb{R})$$Combining (iv) and (v):
$$\frac{\sqrt{5} - 1}{2} < r < \frac{\sqrt{5} + 1}{2}$$Hence, $r$ cannot be equal to $\frac{7}{4} = 1.75$.
16. In $\Delta ABC$, the lengths of sides $AC$ and $AB$ are $12\text{ cm}$ and $5\text{ cm}$, respectively. If the area of $\Delta ABC$ is $30\text{ cm}^2$ and $R$ and $r$ are respectively the radii of circumcircle and incircle of $\Delta ABC$, then the value of $2R + r$ (in cm) is equal to...... .
JEE Main 16.03.2021, Shift-II
Ans. (15) : Given that,
Length of sides of triangle $b = AC = 12\text{ cm}$, $c = AB = 5\text{ cm}$.
$$\text{Area of triangle} = 30\text{ cm}^2 = \frac{1}{2} bc \sin \theta$$ $$30 = \frac{1}{2} \times 5 \times 12 \sin \theta$$ $$30 = 30 \sin \theta \implies \sin \theta = 1 \implies \theta = 90^\circ$$This triangle is a right-angled triangle, with hypotenuse $a = BC = \sqrt{12^2 + 5^2} = 13\text{ cm}$.
$$2R = a = 13$$ $$r = \frac{\Delta}{s} = \frac{30}{\frac{5 + 12 + 13}{2}} = \frac{30}{15} = 2$$ $$2R + r = 13 + 2 = 15$$17. A pole stands vertically inside a triangular park $ABC$. Let the angle of elevation of the top of the pole from each corner of the park be $\frac{\pi}{3}$. If the radius of the circumcircle of $\Delta ABC$ is $2$, then the height of the pole is equal to
(a) $\frac{2\sqrt{3}}{3}$
(b) $2\sqrt{3}$
(c) $\sqrt{3}$
(d) $\frac{1}{\sqrt{3}}$
JEE Main 18.03.2021, Shift-II
Ans. (b) : Given that,
$$\text{Angle of elevation} = \frac{\pi}{3}$$ $$\text{Radius of circumcircle } R = 2$$Since the angle of elevation of the top of the pole from each vertex of $\Delta ABC$ is equal, the foot of the pole $D$ must be the circumcentre of $\Delta ABC$.
$$\tan \frac{\pi}{3} = \frac{PD}{CD} = \frac{h}{R}$$ $$h = R \tan \frac{\pi}{3} = R \tan 60^\circ = R \sqrt{3}$$ $$h = 2\sqrt{3}$$18. Let $P$ and $Q$ be the points on the line $\frac{x+3}{8} = \frac{y-4}{2} = \frac{z+1}{2}$ which are at a distance of $6$ units from the point $R(1,2,3)$. If the centroid of the triangle $PQR$ is $(\alpha, \beta, \gamma)$, then $\alpha^2 + \beta^2 + \gamma^2$ is:
(a) 18
(b) 24
(c) 26
(d) 36
JEE Mains 01/02/2024 Shift-II
Ans. (a) : According to question,
$$\text{Centroid of } \Delta PQR = \left( \frac{x_1 + x_2 + x_3}{3}, \frac{y_1 + y_2 + y_3}{3}, \frac{z_1 + z_2 + z_3}{3} \right)$$ $$\frac{x+3}{8} = \frac{y-4}{2} = \frac{z+1}{2} = k$$Point on line: $(8k - 3, 2k + 4, 2k - 1)$ and $R(1, 2, 3)$.
$$PR = 6 \implies PR^2 = 36$$ $$(8k - 4)^2 + (2k + 2)^2 + (2k - 4)^2 = 36$$ $$4(4k - 2)^2 + 4(k + 1)^2 + 4(k - 2)^2 = 36$$ $$(4k - 2)^2 + (k + 1)^2 + (k - 2)^2 = 9$$ $$(16k^2 - 16k + 4) + (k^2 + 2k + 1) + (k^2 - 4k + 4) = 9$$ $$18k^2 - 18k + 9 = 9$$ $$18k(k - 1) = 0 \implies k = 0 \quad \text{and} \quad k = 1$$Therefore, $P(-3, 4, -1)$, $Q(5, 6, 1)$ and $R(1, 2, 3)$.
$$\text{Centroid of } \Delta PQR = \left( \frac{-3 + 5 + 1}{3}, \frac{4 + 6 + 2}{3}, \frac{-1 + 1 + 3}{3} \right) = (1, 4, 1)$$ $$\therefore (\alpha, \beta, \gamma) = (1, 4, 1)$$ $$\alpha = 1, \quad \beta = 4, \quad \gamma = 1$$ $$\text{Hence, } \alpha^2 + \beta^2 + \gamma^2 = (1)^2 + (4)^2 + (1)^2 = 1 + 16 + 1 = 18$$
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