JEE Main Previous Year Trigonometry Questions and Solutions

Q 1 :

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 [2024]

396
408
414
312

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(2)

Let radius of inscribed circle be $r .$

$r = \frac{1}{3} A D$

$= \frac{1}{3} \sqrt{a^{2} - \frac{a^{2}}{4}}, a is side of triangle.$

$= \frac{a}{2 \sqrt{3}} = \frac{12}{2 \sqrt{3}} = 2 \sqrt{3}$

Area of square DEFG, $m = \frac{1}{2} \times {(diagonal)}^{2}$

$= \frac{1}{2} \times {(2 r)}^{2} = \frac{1}{2} \times 4 \times 12 = 24 sq. units$

Side of square $= 2 \sqrt{6}$

Perimeter of square, $n = 4 \times 2 \sqrt{6} = 8 \sqrt{6}$

$∴ m + n^{2} = 24 + 384 = 408$

Q 2 :

If P(6,1) be the orthocentre of the triangle whose vertices are A(5, -2), B(8, 3) and C(h, k), then the point C lies on the circle [2024]

$x^{2} + y^{2} - 65 = 0$
$x^{2} + y^{2} - 74 = 0$
$x^{2} + y^{2} - 52 = 0$
$x^{2} + y^{2} - 61 = 0$

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(1)

Slope of $A P = \frac{1 + 2}{6 - 5} = 3$

Slope of $B C = \frac{- 1}{Slope of A D}$

$= \frac{- 1}{Slope of A P}$

$= \frac{- 1}{3}$

Equation of line BC is given by $(y - 3) = \frac{- 1}{3} (x - 8)$

$\Rightarrow 3 y + x - 17 = 0$ ...(i)

Now, slope of $B P = \frac{1 - 3}{6 - 8} = 1$

Slope of $A C = - 1$

Equation of AC is $y + 2 = - 1 (x - 5)$

i.e., $y + x - 3 = 0$ ...(ii)

C is the point of intersection of AC and BC, then solving equation (i) and (ii), we get

$C = (- 4, 7), which lies on the circle x^{2} + y^{2} - 65 = 0 .$

Q 3 :

Let $(5, \frac{a}{4})$ be the circumcenter of a triangle with vertices $A (a, - 2),$ $B (a, 6)$ and $C (\frac{a}{4}, - 2)$ . Let $α$ denote the circumradius, $β$ denote the area, and $γ$ denote the perimeter of the triangle. Then $α + β + γ$ is [2024]

62
60
30
53

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(4)

The circumcentre of the triangle ABC is $O (5, \frac{a}{4})$ $∵ O A^{2} = O B^{2}$

$\Rightarrow {(5 - a)}^{2} + {(\frac{a}{4} + 2)}^{2} = {(5 - a)}^{2} + {(\frac{a}{4} - 6)}^{2}$

$\Rightarrow 4 a = 32 \Rightarrow a = 8$

$∴$ $A \equiv (8, - 2), B \equiv (8, 6)$ and $C \equiv (2, - 2)$

$A C = \sqrt{36 + 0} = 6, B C = \sqrt{36 + 64} = 10, A B = \sqrt{64} = 8$

Perimeter, $γ = 6 + 10 + 8 = 24$

$Circumradius, α = O A = \sqrt{9 + 16} = 5$

$Area, β = \frac{1}{2} \times 6 \times 8 = 24$

$∴$ $α + β + γ = 5 + 24 + 24 = 53$

Q 4 :

In a triangle ABC, BC = 7, AC = 8, $A B = α \in N$ and $\cos A = \frac{2}{3} .$ If $49 \cos (3 C) + 42 = \frac{m}{n},$ where $\gcd (m, n) = 1,$ then $m + n$ is equal to ________ . [2024]

0%

(39)

$\cos A = \frac{b^{2} + c^{2} - a^{2}}{2 b c} = \frac{8^{2} + α^{2} - 7^{2}}{2 \cdot 8 \cdot α}$

$\Rightarrow \frac{2}{3} = \frac{α^{2} + 15}{16 α}$

$\Rightarrow 32 α = 3 α^{2} + 45$

$\Rightarrow 3 α^{2} - 32 α + 45 = 0 \Rightarrow α = \frac{5}{3}, 9$

$\Rightarrow α = 9 [∵ α \in N]$

$Now, \cos C = \frac{7^{2} + 8^{2} - 9^{2}}{2 \times 7 \times 8} \Rightarrow \cos C = \frac{2}{7}$

$Now, \cos 3 C = 4 \cos^{3} C - 3 \cos C = \frac{4 \times 8}{7^{3}} - \frac{6}{7}$

$So, 49 \cos 3 C + 42 = 49 (\frac{4 \times 8}{7^{3}} - \frac{6}{7}) + 42 = \frac{32}{7} - 42 + 42 = \frac{m}{n}$

$\Rightarrow m = 32, n = 7; m + n = 32 + 7 = 39$

Q 5 :

If the orthocentre of the triangle formed by the lines $2 x + 3 y - 1 = 0, x + 2 y - 1 = 0$ and $a x + b y - 1 = 0,$ is the centroid of another triangle, whose circumcentre and orthocentre respectively are (3, 4) and (-6, -8), then the value of $| a - b |$ is _________ . [2024]

0%

(16)

Let $∆ A B C$ be the given triangle with sides $2 x + 3 y = 1, x + 2 y = 1 and a x + b y = 1$

Let $∆ P Q R$ be another triangle whose circumcentre, orthocentre and centroid be $C_{1}, H_{1}$ and $G_{1}$ respectively. We know centroid divides circumcentre and orthocentre in the ratio 1 : 2.

$∴$ $Coordinates of G_{1} = (\frac{6 - 6}{3}, \frac{8 - 8}{3}) = (0, 0)$

$∴$ $Orthocentre of ∆ A B C, H_{2} = (0, 0)$

$Slope of A H_{2} = \frac{- 1}{Slope of B C} \Rightarrow \frac{1 - 0}{- 1 - 0} = \frac{- 1}{\frac{- a}{b}} \Rightarrow - a = b$

$Slope of A B = \frac{- 2}{3}$

Now, $slope of C H_{2} = \frac{3}{2}$

$∴$ $Equation of C H_{2} is given by y = \frac{3}{2} x$

Now, C is the point of intersection of $y = \frac{3}{2} x$ and $x + 2 y - 1 = 0$

$∴$ $C (\frac{1}{4}, \frac{3}{8}) are the coordinates of C which will also satisfy the equation of B C .$

$\Rightarrow \frac{a}{4} - \frac{3}{8} a - 1 = 0 \Rightarrow a = - 8, b = 8$

$\Rightarrow | a - b | = | - 8 - 8 | = 16$

Q 6 :

In a triangle ABC, if $\cos A + 2 \cos B + \cos C = 2$ and the lengths of the sides opposite to angles A and C are 3 and 7 respectively, then $\cos A - \cos C$ is equal to [2023]

$\frac{5}{7}$
$\frac{10}{7}$
$\frac{3}{7}$
$\frac{9}{7}$

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(2)

$We have, \cos A + 2 \cos B + \cos C = 2$

$∴$ $[\frac{b^{2} + c^{2} - a^{2}}{2 b c}] + 2 [\frac{c^{2} + a^{2} - b^{2}}{2 a c}] + [\frac{a^{2} + b^{2} - c^{2}}{2 a b}] = 2$

$\Rightarrow [\frac{b^{2} + 49 - 9}{14 b}] + 2 [\frac{49 + 9 - b^{2}}{42}] + [\frac{9 + b^{2} - 49}{6 b}] = 2$ $[∵ a = 3, c = 7]$

$\Rightarrow \frac{b^{2} + 40}{14 b} + \frac{58 - b^{2}}{21} + \frac{- 40 + b^{2}}{6 b} = 2$

$\Rightarrow b^{3} - 5 b^{2} - 16 b + 80 = 0$

$\Rightarrow (b - 4) (b + 4) (b - 5) = 0$

$\Rightarrow b = - 4$ or $4$ or $5$

$b \neq - 4$ $(∵ b cannot be - 4)$

$For b = 4, triangle cannot be constructed$

$∴$ $b = 5$

$Now, \cos A - \cos C = \frac{b^{2} + c^{2} - a^{2}}{2 b c} - \frac{a^{2} + b^{2} - c^{2}}{2 a b}$

$= \frac{49 + 25 - 9}{2 \times 7 \times 5} - \frac{9 + 25 - 49}{2 \times 3 \times 5} = \frac{13}{14} + \frac{1}{2} = \frac{10}{7}$

Q 7 :

For a triangle ABC, the value of $\cos 2 A + \cos 2 B + \cos 2 C$ is least. If its inradius is 3 and incentre is M, then which of the following is NOT correct? [2023]

area of $∆ A B C$ is $\frac{27 \sqrt{3}}{2}$
$\sin 2 A + \sin 2 B + \sin 2 C = \sin A + \sin B + \sin C$
perimeter of $∆ A B C$ is $18 \sqrt{3}$
$\vec{M A} \cdot \vec{M B} = - 18$

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(1)

$Here, \cos 2 A + \cos 2 B + \cos 2 C = - \frac{3}{2}$

$\Rightarrow 2 A = 2 B = 2 C = 120 ° \Rightarrow A = B = C = 60 °$

$Now, inradius = \frac{Area of ∆ A B C}{Semiperimeter}$

$3 = \frac{\frac{\sqrt{3}}{4} a^{2}}{\frac{3 a}{2}}$ $[Where a = side of ∆]$

$\Rightarrow a = \frac{18}{\sqrt{3}} = 6 \sqrt{3}$

$Perimeter of ∆ A B C = 18 \sqrt{3}$

$Area of ∆ A B C = \frac{\sqrt{3}}{4} \times {(6 \sqrt{3})}^{2} = 27 \sqrt{3}$

$A = 60 \Rightarrow 2 A = 120$

$\sin 2 A = \sin A$

$\sum \sin 2 A = \sum \sin A$

$| \vec{M A} |$ $=$ $| \vec{M B} |$ $\Rightarrow 6 = 2 r$

$\vec{M A} \cdot \vec{M B} = 36 \cos 120 ° = - 18$

Q 8 :

In the figure, $θ_{1} + θ_{2} = \frac{π}{2}$ and $\sqrt{3} (B E) = 4 (A B) .$ If the area of $∆ C A B$ is $2 \sqrt{3} - 3$ ${unit}^{2}$ , when $\frac{θ_{2}}{θ_{1}}$ is the largest, then the perimeter (in unit) of $∆ C E D$ is equal to ______ . [2023]

0%

(6)

$Given, \sqrt{3} (B E) = 4 (A B)$

$ar (∆ C A B) = 2 \sqrt{3} - 3$

$So, \frac{1}{2} x^{2} \tan θ_{1} = 2 \sqrt{3} - 3$

$B E = B D + D E$

$= x (\tan θ_{1} + \tan θ_{2})$

$B E = A B (\tan θ_{1} + \cot θ_{1})$

$\frac{4}{\sqrt{3}} = \tan θ_{1} + \cot θ_{1} \Rightarrow \tan θ_{1} = \sqrt{3}, \frac{1}{\sqrt{3}}$

$θ_{1} = \frac{π}{6}, θ_{2} = \frac{π}{3};$ $θ_{1} = \frac{π}{3}, θ_{2} = \frac{π}{6}$

$as \frac{θ_{2}}{θ_{1}} is largest$

$∴$ $θ_{1} = \frac{π}{6}, θ_{2} = \frac{π}{3}$

$∴$ $x^{2} = \frac{(2 \sqrt{3} - 3) \times 2}{\tan θ_{1}} = \frac{\sqrt{3} (2 - \sqrt{3}) \times 2}{\tan \frac{π}{6}}$

$\Rightarrow x^{2} = 12 - 6 \sqrt{3} = {(3 - \sqrt{3})}^{2} \Rightarrow x = 3 - \sqrt{3}$

$Perimeter of ∆ C E D = C D + D E + C E$

$= 3 - \sqrt{3} + (3 - \sqrt{3}) \sqrt{3} + (3 - \sqrt{3}) \times 2 = 6 units$

Topic Question Set