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

1

2

3

4

(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]

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

1

2

3

4

(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

1

2

3

4

(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 $x + 2 y - 1 = 0$

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

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

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

Topic Question Set

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]

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]

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]

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]

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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]

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Topic Question Set

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+n2 is equal to [2024]

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]

Q 3 : Let (5,a4) be the circumcenter of a triangle with vertices A(a,-2), B(a,6) and C(a4,-2). Let α denote the circumradius, β denote the area, and γ denote the perimeter of the triangle. Then α+β+γ is [2024]

Q 4 : In a triangle ABC, BC = 7, AC = 8, AB=α∈N and cosA=23. If 49cos(3C)+42=mn, where gcd(m,n)=1, then m+n is equal to ________ . [2024]

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Q 5 : If the orthocentre of the triangle formed by the lines 2x+3y-1=0, x+2y-1=0 and ax+by-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]

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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]

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]

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]

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]

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]