Properties of Triangle
55 Questions
Numerical
2025
JEE Mains
Numerical
JEE Main 2025 (Online) 22nd January Evening Shift
Let $\mathrm{A}(6,8), \mathrm{B}(10 \cos \alpha,-10 \sin \alpha)$ and $\mathrm{C}(-10 \sin \alpha, 10 \cos \alpha)$, be the vertices of a triangle. If $L(a, 9)$ and $G(h, k)$ be its orthocenter and centroid respectively, then $(5 a-3 h+6 k+100 \sin 2 \alpha)$ is equal to ___________.
Correct Answer: 145
Explanation:
All the three points $A, B, C$ lie on the circle $x^2+y^2=100$ so circumcentre is $(0,0)$
$\begin{aligned} & \frac{a+0}{3}=h \Rightarrow a=3 h \\ & \text { and } \frac{9+0}{3}=k \Rightarrow k=3 \\ & \text { also centroid } \frac{6+10 \cos \alpha-10 \sin \alpha}{3}=h \\ & \Rightarrow 10(\cos \alpha-\sin \alpha)=3 h-6\quad\text{.... (i)} \end{aligned}$
$\begin{aligned} &\begin{aligned} & \text { and } \frac{8+10 \cos \alpha-10 \sin \alpha}{3}=\mathrm{k} \\ & \Rightarrow 10(\cos \alpha-\sin \alpha)=3 \mathrm{k}-8=9-8=1 \quad\text{.... (ii)}\\ & \text { on squaring } 100(1-\sin 2 \alpha)=1 \\ & \Rightarrow 100 \sin 2 \alpha=99 \end{aligned}\\ &\text { from equ. (i) and (ii) we get } \mathrm{h}=\frac{7}{3}\\ &\begin{aligned} & \text { Now } 5 a-3 h+6 k+100 \sin 2 \alpha \\ & =15 h-3 h+6 k+100 \sin 2 \alpha \\ & =12 \times \frac{7}{3}+18+99 \\ & =145 \end{aligned} \end{aligned}$
2024
JEE Mains
Numerical
JEE Main 2024 (Online) 6th April Evening Shift
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 $\operatorname{gcd}(m, n)=1$, then $m+n$ is equal to _________.
Correct Answer: 39
Explanation:
$\begin{aligned} & \cos A=\frac{2}{3}=\frac{\alpha^2+8^2-7^2}{2 \cdot \alpha \cdot 8} \\ & \Rightarrow\left(\alpha^2+15\right) 3=32 \alpha \\ & 3 \alpha^2-32 \alpha+45=0 \\ & \Rightarrow \alpha=\frac{5}{3}, 9 \\ & \because \alpha \in N \Rightarrow \alpha=9 \\ & \cos C=\frac{7^2+8^2-9^2}{2 \cdot 7 \cdot 8}=\frac{2}{7} \end{aligned}$
$ \begin{aligned} \cos 3 C & =4 \cos ^3 C-3 \cos C \\ = & \frac{4 \times 8}{7^3}-\frac{6}{7} \\ 49 \cos 3 C & =\frac{32}{7}-42 \\ \Rightarrow \quad & 49 \cos 3 C+42=\frac{32}{7} \\ \Rightarrow \quad & m+n =39 \end{aligned} $
2024
JEE Mains
Numerical
JEE Main 2024 (Online) 4th April Evening Shift
Consider a triangle $\mathrm{ABC}$ having the vertices $\mathrm{A}(1,2), \mathrm{B}(\alpha, \beta)$ and $\mathrm{C}(\gamma, \delta)$ and angles $\angle A B C=\frac{\pi}{6}$ and $\angle B A C=\frac{2 \pi}{3}$. If the points $\mathrm{B}$ and $\mathrm{C}$ lie on the line $y=x+4$, then $\alpha^2+\gamma^2$ is equal to _______.
Correct Answer: 14
Explanation:
$\begin{aligned} & P=\frac{|2-1-4|}{\sqrt{1^2+1^2}}=\frac{3}{\sqrt{2}} \\ & \sin \left(\frac{\pi}{6}\right)=\frac{3 / \sqrt{2}}{A B}=\frac{1}{2} \Rightarrow A B=\frac{6}{\sqrt{2}} \\ & \Rightarrow(\alpha-1)^2+(\alpha+4-2)^2=18 \end{aligned}$
$\Rightarrow 2 \alpha^2+2 \alpha-13=0 \rightarrow \alpha$ and $\gamma$ satisfy same equation
$\begin{aligned} & \Rightarrow \alpha^2+\gamma^2=(\alpha+\gamma)^2-2 \alpha \gamma \\ & =(-1)^2-2\left(\frac{-13}{2}\right)=1+13=14 \end{aligned}$
2023
JEE Mains
Numerical
JEE Main 2023 (Online) 15th April Morning Shift
If the line $x=y=z$ intersects the line
$x \sin A+y \sin B+z \sin C-18=0=x \sin 2 A+y \sin 2 B+z \sin 2 C-9$,
where $A, B, C$ are the angles of a triangle $A B C$, then $80\left(\sin \frac{A}{2} \sin \frac{B}{2} \sin \frac{C}{2}\right)$
is equal to ______________.
$x \sin A+y \sin B+z \sin C-18=0=x \sin 2 A+y \sin 2 B+z \sin 2 C-9$,
where $A, B, C$ are the angles of a triangle $A B C$, then $80\left(\sin \frac{A}{2} \sin \frac{B}{2} \sin \frac{C}{2}\right)$
is equal to ______________.
Correct Answer: 5
Explanation:
$
\begin{aligned}
&x= y=z=k(\text { let }) \\\\
&\therefore k(\sin A+\sin B+\sin C)=18 \\\\
&\Rightarrow k\left(4 \cos \frac{A}{2} \cdot \cos \frac{B}{2} \cdot \cos \frac{C}{2}\right)=18 \\\\
& k(\sin 2 A+\sin 2 B+\sin 2 C)=9 \\\\
&\Rightarrow k(4 \sin A \cdot \sin B \cdot \sin C)=9 \ldots \text { (ii) } \\\\
& \text {(ii)} /\text {(i)} \\\\
& 8 \sin \frac{A}{2} \cdot \sin \frac{B}{2} \cdot \sin \frac{C}{2}=\frac{9}{18} \\\\
&\Rightarrow 80 \sin \frac{A}{2} \cdot \sin \frac{B}{2} \cdot \sin \frac{C}{2}=5
\end{aligned}
$
2023
JEE Mains
Numerical
JEE Main 2023 (Online) 10th April Evening Shift
In the figure, $\theta_{1}+\theta_{2}=\frac{\pi}{2}$ and $\sqrt{3}(\mathrm{BE})=4(\mathrm{AB})$. If the area of $\triangle \mathrm{CAB}$ is $2 \sqrt{3}-3$ unit ${ }^{2}$, when $\frac{\theta_{2}}{\theta_{1}}$ is the largest, then the perimeter (in unit) of $\triangle \mathrm{CED}$ is equal to _________.
Correct Answer: 6
Explanation:
We have, $\theta_1+\theta_2=\frac{\pi}{2}$ and $\sqrt{3}(B E)=4 A B$
Let $A B=x$ unit
$ \begin{aligned} & A C=x \tan \theta_1 \\\\ & E D=x \tan \theta_2 \\\\ & B E=B D+D E \end{aligned} $
$ \begin{array}{rlrl} & \Rightarrow \frac{4}{\sqrt{3}} x =x\left(\tan \theta_1+\tan \theta_2\right) {[\because \sqrt{3} B E=4 A B]} \\\\ & \Rightarrow \frac{4}{\sqrt{3}}=\tan \theta_1+\tan \left(\frac{\pi}{2}-\theta_1\right) {\left[\because \theta_1+\theta_2=\frac{\pi}{2}\right]} \end{array} $
$ \begin{aligned} & \Rightarrow \tan \theta_1+\cot \theta_1=\frac{4}{\sqrt{3}}=\sqrt{3}+\frac{1}{\sqrt{3}} \\\\ & \Rightarrow \tan \theta_1=\sqrt{3} \text { or } \theta_1=\frac{\pi}{3} \text { and } \theta_2=\frac{\pi}{6} \\\\ & \operatorname{or} \theta_1=\frac{\pi}{6} \text { and } \theta_2=\frac{\pi}{3} \end{aligned} $
$\because \frac{\theta_2}{\theta_1}$ is largest
$ \therefore \theta_1=\frac{\pi}{6} \text { and } \theta_2=\frac{\pi}{3} $
$ \begin{aligned} & \text { Area of } \triangle C A B=\frac{1}{2} \times x \times x \tan \theta_1 \\\\ & \Rightarrow \frac{x^2 \tan \theta_1}{2}=2 \sqrt{3}-3 \\\\ & \Rightarrow x^2=\frac{2(2 \sqrt{3}-3)}{\tan \frac{\pi}{6}}=12-6 \sqrt{3} \\\\\ & \Rightarrow x=3-\sqrt{3} \end{aligned} $
$ \text { Also, } C E=\sqrt{x^2+x^2 \tan ^2 \frac{\pi}{3}}=(3-\sqrt{3}) \times 2=6-2 \sqrt{3} $
Perimeter of $\triangle C E D$
$ \begin{aligned} & =C D+D E+C E \\\\ & =(3-\sqrt{3})+(3-\sqrt{3}) \tan \frac{\pi}{3}+6-2 \sqrt{3} \\\\ & =3-\sqrt{3}+3 \sqrt{3}-3+6-2 \sqrt{3}=6 \end{aligned} $
2023
JEE Advanced
Numerical
JEE Advanced 2023 Paper 2 Online
$
\text { Let } a \text { be the area of the triangle } A B C \text {. Then the value of }(64 a)^2 \text { is }
$ :
Correct Answer: 1008
Explanation:
Area of $\triangle A B C=\frac{A B \cdot B C \cdot A C}{4 R}$
$ \begin{aligned} \Rightarrow a & =\frac{8 \sin A \cdot \sin B \sin C}{4} \\\\ & =2 \sin \left(\frac{\pi}{2}-2 C\right) \sin \left(\frac{\pi}{2}+C\right) \sin C \\\\ & =2 \cos 2 C \cdot \cos C \cdot \sin C \\\\ & =\cos 2 C \cdot \sin 2 C \\\\ & =\sqrt{1-\sin ^2 2 C} \cdot \sin 2 C \\\\ & =\sqrt{1-\frac{9}{6}} \cdot \times \frac{3}{4} \\\\ &\Rightarrow a =\frac{3 \sqrt{7}}{16} \\\\ & \therefore (64 a)^2 =1008 \end{aligned} $
$ \begin{aligned} \Rightarrow a & =\frac{8 \sin A \cdot \sin B \sin C}{4} \\\\ & =2 \sin \left(\frac{\pi}{2}-2 C\right) \sin \left(\frac{\pi}{2}+C\right) \sin C \\\\ & =2 \cos 2 C \cdot \cos C \cdot \sin C \\\\ & =\cos 2 C \cdot \sin 2 C \\\\ & =\sqrt{1-\sin ^2 2 C} \cdot \sin 2 C \\\\ & =\sqrt{1-\frac{9}{6}} \cdot \times \frac{3}{4} \\\\ &\Rightarrow a =\frac{3 \sqrt{7}}{16} \\\\ & \therefore (64 a)^2 =1008 \end{aligned} $
2021
JEE Mains
Numerical
JEE Main 2021 (Online) 25th July Evening Shift
If a rectangle is inscribed in an equilateral triangle of side length $2\sqrt 2 $ as shown in the figure, then the square of the largest area of such a rectangle is _____________.
Correct Answer: 3
Explanation:
In $\Delta$DBF
$\tan 60^\circ = {{2b} \over {2\sqrt 2 - l}} \Rightarrow b = {{\sqrt 3 \left( {2\sqrt 2 - l} \right)} \over 2}$
A = Area of rectangle = l $\times$ b
$A = l \times {{\sqrt 3 } \over 2}\left( {2\sqrt 2 - l} \right)$
${{dA} \over {dl}} = {{\sqrt 3 } \over 2}\left( {2\sqrt 2 - l} \right) - {{l.\sqrt 3 } \over 2} = 0$
$l = \sqrt 2 $
$A = l \times b = \sqrt 2 \times {{\sqrt 3 } \over 2}\left( {\sqrt 2 } \right) = \sqrt 3 $
$\Rightarrow$ A2 = 3
2021
JEE Mains
Numerical
JEE Main 2021 (Online) 16th March Evening Shift
In $\Delta$ABC, the lengths of sides AC and AB are 12 cm and 5 cm, respectively. If the area of $\Delta$ABC is 30 cm2 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 ___________.
Correct Answer: 15
Explanation:
Area = ${1 \over 2}(5)(12)\sin \theta = 30$
$\sin \theta = 1 \Rightarrow \theta = {\pi \over 2}$
$\Delta$ is right angle $\Delta$
$r = (s - a)\tan {A \over 2}$
$r = (s - a)$$\tan {{90} \over 2}$
$r = (s - a)$
$2R + r = s$, (As $a = 2R$)
$2R + r = {{5 + 12 + 13} \over 2} = 15$
2021
JEE Mains
Numerical
JEE Main 2021 (Online) 16th March Morning Shift
Let ABCD be a square of side of unit length. Let a circle C1 centered at A with unit radius is drawn. Another circle C2 which touches C1 and the lines AD and AB are tangent to it, is also drawn. Let a tangent line from the point C to the circle C2 meet the side AB at E. If the length of EB is $\alpha$ + ${\sqrt 3 }$ $\beta$, where $\alpha$, $\beta$ are integers, then $\alpha$ + $\beta$ is equal to ____________.
Correct Answer: 1
Explanation:
(i) $\sqrt 2 r + r = 1$
$r = {1 \over {\sqrt 2 + 1}}$
$r = \sqrt 2 - 1$
(ii) $C{C_2} = 2\sqrt 2 - 2 = 2\left( {\sqrt 2 - 1} \right)$
From $\Delta C{C_2}N = \sin \phi = {{\sqrt 2 - 1} \over {2\left( {\sqrt 2 - 1} \right)}}$
$\phi = 30^\circ $
(iii) In $\Delta$ACE apply sine law
${{AE} \over {\sin \phi }} = {{AC} \over {\sin 105^\circ }}$
$AE = {1 \over 2} \times {{\sqrt 2 } \over {\sqrt 3 + 1}}.2\sqrt 2 $
$AE = {2 \over {\sqrt 3 + 1}} = \sqrt 3 - 1$
$ \therefore $ $EB = 1 - \left( {\sqrt 3 - 1} \right)$
= $2 - \sqrt 3 $
$ \therefore $ $\alpha$ = 2, $\beta$ = $-$1 $ \Rightarrow $ $\alpha$ + $\beta$ = 1
2021
JEE Advanced
Numerical
JEE Advanced 2021 Paper 1 Online
In a triangle ABC, let AB = $\sqrt {23} $, BC = 3 and CA = 4. Then the value of ${{\cot A + \cot C} \over {\cot B}}$ is _________.
Correct Answer: 2
Explanation:
Given,
$AB = \sqrt {23} = c$
$BC = 3 = a$
$CA = 4 = b$
Now,
${{\cot A + \cot C} \over {\cot B}} = {{{{\cos A} \over {\sin A}} + {{\cos C} \over {\sin C}}} \over {{{\cos B} \over {\sin B}}}}$
$ = {{{{{b^2} + {c^2} - {a^2}} \over {2bc(\sin A)}} + {{{a^2} + {b^2} - {c^2}} \over {2ab(\sin C)}}} \over {{{{c^2} + {a^2} - {b^2}} \over {2ac(\sin B)}}}}$
$ = {{{{{b^2} + {c^2} - {a^2}} \over {4\Delta }} + {{{a^2} + {b^2} - {c^2}} \over {4\Delta }}} \over {{{{c^2} + {a^2} - {b^2}} \over {4\Delta }}}}$
$ = {{{b^2} + {c^2} - {a^2} + {a^2} + {b^2} - {c^2}} \over {{c^2} + {a^2} - {b^2}}}$
$ = {{2{b^2}} \over {{a^2} + {c^2} - {b^2}}} = {{2 \times 16} \over {9 + 23 - 16}} = {{32} \over {16}} = 2$
$AB = \sqrt {23} = c$
$BC = 3 = a$
$CA = 4 = b$
Now,
${{\cot A + \cot C} \over {\cot B}} = {{{{\cos A} \over {\sin A}} + {{\cos C} \over {\sin C}}} \over {{{\cos B} \over {\sin B}}}}$
$ = {{{{{b^2} + {c^2} - {a^2}} \over {2bc(\sin A)}} + {{{a^2} + {b^2} - {c^2}} \over {2ab(\sin C)}}} \over {{{{c^2} + {a^2} - {b^2}} \over {2ac(\sin B)}}}}$
$ = {{{{{b^2} + {c^2} - {a^2}} \over {4\Delta }} + {{{a^2} + {b^2} - {c^2}} \over {4\Delta }}} \over {{{{c^2} + {a^2} - {b^2}} \over {4\Delta }}}}$
$ = {{{b^2} + {c^2} - {a^2} + {a^2} + {b^2} - {c^2}} \over {{c^2} + {a^2} - {b^2}}}$
$ = {{2{b^2}} \over {{a^2} + {c^2} - {b^2}}} = {{2 \times 16} \over {9 + 23 - 16}} = {{32} \over {16}} = 2$
2020
JEE Advanced
Numerical
JEE Advanced 2020 Paper 1 Offline
In a triangle PQR, let a = QR, b = RP, and c = PQ. If |a| = 3, |b| = 4
and ${{a\,.(\,c - \,b)} \over {c\,.\,(a - \,b)}} = {{|a|} \over {|a| + |b|}}$, then the value of |a $ \times $ b|2 is ......
and ${{a\,.(\,c - \,b)} \over {c\,.\,(a - \,b)}} = {{|a|} \over {|a| + |b|}}$, then the value of |a $ \times $ b|2 is ......
Correct Answer: 108
Explanation:
It is given that, in a $\Delta $PQR
PQ = c, QR = a and RP = b, so a + b + c = 0
$ \therefore $ |a|2 + a . b + a . c = 0 ....(i)
and a . b + |b|2 + c . b = 0 ....(ii)
and, also given that |a| = 3, |b| = 4
and ${{a\,.\,c - a\,.\,b} \over {c\,.\,a - c\,.\,b}} = {{|a|} \over {|a| + |b|}} = {3 \over 7}$
$ \Rightarrow 7[(a\,.\,c) - (a\,.\,b)] = 3[(c\,.\,a) - (c\,.\,b)]$
from Eqs. (i) and (ii), on putting the values of a . c and c . b, we get,
$ \Rightarrow 7[ - 9 - (a\,.\,b) - (a\,.\,b)] = 3[ - 9 - (a\,.\,b) + 16 + (a\,.\,b)]$
$ \Rightarrow 7[ - 9 - 2(a\,.\,b)] = 3 \times 7$
$ \Rightarrow a\,.\,b = - 6$
$ \because $ $|a \times b{|^2} = {(|a||b|)^2} - {(a\,.\,b)^2}$
= $144 - 36 = 108$
PQ = c, QR = a and RP = b, so a + b + c = 0
$ \therefore $ |a|2 + a . b + a . c = 0 ....(i)
and a . b + |b|2 + c . b = 0 ....(ii)
and, also given that |a| = 3, |b| = 4
and ${{a\,.\,c - a\,.\,b} \over {c\,.\,a - c\,.\,b}} = {{|a|} \over {|a| + |b|}} = {3 \over 7}$
$ \Rightarrow 7[(a\,.\,c) - (a\,.\,b)] = 3[(c\,.\,a) - (c\,.\,b)]$
from Eqs. (i) and (ii), on putting the values of a . c and c . b, we get,
$ \Rightarrow 7[ - 9 - (a\,.\,b) - (a\,.\,b)] = 3[ - 9 - (a\,.\,b) + 16 + (a\,.\,b)]$
$ \Rightarrow 7[ - 9 - 2(a\,.\,b)] = 3 \times 7$
$ \Rightarrow a\,.\,b = - 6$
$ \because $ $|a \times b{|^2} = {(|a||b|)^2} - {(a\,.\,b)^2}$
= $144 - 36 = 108$
2010
JEE Advanced
Numerical
IIT-JEE 2010 Paper 2 Offline
Consider a triangle $ABC$ and let $a, b$ and $c$ denote the lengths of the sides opposit to vertices $A, B$ and $C$ respectively. Suppose $a = 6,b = 10$ and the area of the triangle is $15\sqrt 3 $, if $\angle ACB$ is obtuse and if $r$ denotes the radius of the incircle of the triangle, then r2 is equal to :
Correct Answer: 3
Explanation:
$\sin C = {{\sqrt 3 } \over 2}$ and C is given to be obtuse
$ \Rightarrow C = {{2\pi } \over 3} = \sqrt {{a^2} + {b^2} - 2ab\cos C} $
$ = \sqrt {{6^2} + {{10}^2} - 2 \times 6 \times 10 \times \cos {{2\pi } \over 3}} = 14$
$\therefore$ $r = {\Delta \over s} \Rightarrow {r^2} = {{225 \times 3} \over {{{\left( {{{6 + 10 + 14} \over 2}} \right)}^2}}} = 3$
2009
JEE Advanced
Numerical
IIT-JEE 2009 Paper 2 Offline
Let ABC and ABC' be two non-congruent triangles with sides AB = 4, AC = AC' = 2$\sqrt2$ and angle B = 30$^\circ$. The absolute value of the difference between the areas of these triangles is ___________.
Correct Answer: 4
Explanation:
We have,
$\cos \beta = {{{a^2} + 16 - 8} \over {2 \times a \times 4}}$
$ \Rightarrow {{\sqrt 3 } \over 2} = {{{a^2} + 8} \over {8a}}$
$ \Rightarrow {a^2} - 4\sqrt 3 a + 8 = 0$
$ \Rightarrow {a_1} + {a_2} = 4\sqrt 3 ,{a_1}{a_2} = 8$
$ \Rightarrow |{a_1} - {a_2}| = 4$
$ \Rightarrow |{\Delta _1} - {\Delta _2}| = {1 \over 2} \times 4\sin 30^\circ \times 4 = 4$
2003
JEE Advanced
Numerical
IIT-JEE 2003
If ${I_n}$ is the area of $n$ sided regular polygon inscribed in a circle of unit radius and ${O_n}$ be the area of the polygon circumscribing the given circle, prove that
$${I_n} = {{{O_n}} \over 2}\left( {1 + \sqrt {1 - {{\left( {{{2{I_n}} \over n}} \right)}^2}} } \right)$$
Correct Answer: Solve it.
2001
JEE Advanced
Numerical
IIT-JEE 2001
If $\Delta $ is the area of a triangle with side lengths $a, b, c, $ then show that $\Delta \le {1 \over 4}\sqrt {\left( {a + b + c} \right)abc} $. Also show that the equality occurs in the above inequality if and only if $a=b=c$.
Correct Answer: Solve it.
2000
JEE Advanced
Numerical
IIT-JEE 2000
Let $ABC$ be a triangle with incentre $I$ and inradius $r$. Let $D,E,F$ be the feet of the perpendiculars from $I$ to the sides $BC$, $CA$ and $AB$ respectively. If ${r_1},{r_2}$ and ${r_3}$ are the radii of circles inscribed in the quadrilaterals $AFIE$, $BDIF$ and $CEID$ respectively, prove that
$${{{r_1}} \over {r - {r_1}}} + {{{r_2}} \over {r - {r_2}}} + {{{r_3}} \over {r - {r_3}}} = {{{r_1}{r_2}{r_3}} \over {\left( {e - {r_1}} \right)\left( {r - {r_2}} \right)\left( {r - {r_3}} \right)}}$$
Correct Answer: Solve it.
1999
JEE Advanced
Numerical
IIT-JEE 1999
Let $ABC$ be a triangle having $O$ and $I$ as its circumcenter and in centre respectively. If $R$ and $r$ are the circumradius and the inradius, respectively, then prove that ${\left( {IO} \right)^2} = {R^2} - 2{\mathop{\rm Rr}\nolimits} $. Further show that the triangle BIO is a right-angled triangle if and only if $b$ is arithmetic mean of $a$ and $c$.
Correct Answer: Solve it.
1998
JEE Advanced
Numerical
IIT-JEE 1998
A bird flies in a circle on a horizontal plane. An observer stands at a point on the ground. Suppose ${60^ \circ }$ and ${30^ \circ }$ are the maximum and the minimum angles of elevation of the bird and that they occur when the bird is at the points $P$ and $Q$ respectively on its path. Let $\theta $ be the angle of elevation of the bird when it is a point on the are of the circle exactly midway between $P$ and $Q$. Find the numerical value of ${\tan ^2}\theta $. (Assume that the observer is not inside the vertical projection of the path of the bird.)
Correct Answer: $${3 \over 5}$$
1998
JEE Advanced
Numerical
IIT-JEE 1998
Prove that a triangle $ABC$ is equilateral if and only if $\tan A + \tan B + \tan C = 3\sqrt 3 $.
Correct Answer: Solve it.
1996
JEE Advanced
Numerical
IIT-JEE 1996
In a triangle $ABC$, $a:b:c=4:5:6$. The ratio of the radius of the circumcircle to that of the incircle is ............
Correct Answer: $$16:7$$
1994
JEE Advanced
Numerical
IIT-JEE 1994
A tower $AB$ leans towards west making an angle $\alpha $ with the vertical. The angular elevation of $B$, the topmost point of the tower is $\beta $ as observed from a point $C$ due west of $A$ at a distance $d$ from $A$. If the angular elevation of $B$ from a point $D$ due east of $C$ at a distance $2d$ from $C$ is $\gamma $, then prove that $2$ tan $\alpha = - \cot \beta + \cot \gamma $.
Correct Answer: Solve it.
1994
JEE Advanced
Numerical
IIT-JEE 1994
Consider the following statements connecting a triangle $ABC$
(i) The sides $a, b, c$ and area $\Delta $ are rational.
(ii) $a,\tan {B \over 2},\tan {c \over 2}$ are rational.
(iii) $a,\sin A,\sin B,\sin C$ are rational.
Prove that $\left( i \right) \Rightarrow \left( {ii} \right) \Rightarrow \left( {iii} \right) \Rightarrow \left( i \right)$
Correct Answer: Solve it.
1994
JEE Advanced
Numerical
IIT-JEE 1994
Let ${A_1},{A_2},........,{A_n}$ be the vertices of an $n$-sided regular polygon such that ${1 \over {{A_1}{A_2}}} = {1 \over {{A_1}{A_3}}} + {1 \over {{A_1}{A_4}}}$, Find the value of $n$.
Correct Answer: $$7$$
1994
JEE Advanced
Numerical
IIT-JEE 1994
A circle is inscribed in an equilateral triangle of side $a$. The area of any square inscribed in this circle is ..............
Correct Answer: $${{{a^2}} \over 6}$$ sq. unit
1994
JEE Advanced
Numerical
IIT-JEE 1994
In a triangle $ABC$, $AD$ is the altitude from $A$. Given $b>c$, $\angle C = {23^ \circ }$ and $AD = {{abc} \over {{b^2} - {c^2}}}$ then $\angle B = $.................
Correct Answer: $${113^ \circ }$$
1993
JEE Advanced
Numerical
IIT-JEE 1993
An observer at $O$ notices that the angle of elevation of the top of a tower is ${30^ \circ }$. The line joining $O$ to the base of the tower makes an angle of ${\tan ^{ - 1}}\left( {1/\sqrt 2 } \right)$ with the North and is inclined Eastwards. The observer travels a distance of $300$ meters towards the North to a point A and finds the tower to his East. The angle of elevation of the top of the tower at $A$ is $\phi $, Find $\phi $ and the height of the tower.
Correct Answer: $${45^ \circ }$$, $${150\sqrt 2 }$$ $$m$$
1993
JEE Advanced
Numerical
IIT-JEE 1993
If in a triangle $ABC$, ${{2\cos A} \over a} + {{\cos B} \over b} + {{2\cos C} \over c} = {a \over {bc}} + {b \over {ca}},$ then the value of the angle $A$ is .................... degrees.
Correct Answer: $${90^ \circ }$$
1992
JEE Advanced
Numerical
IIT-JEE 1992
Three circles touch the one another externally. The tangent at their point of contact meet at a point whose distance from a point of contact is $4$. Find the ratio of the product of the radii to the sum of the radii of the circles.
Correct Answer: $$16:1$$
1991
JEE Advanced
Numerical
IIT-JEE 1991
A man notices two objects in a straight line due west. After walking a distance $c$ due north he observes that the objects subtend an angle $\alpha $ at his eye; and, after walking a further distance $2c$ due north, an angle $\beta $. Show that the distance between the objects is ${{8c} \over {3\cot \beta - \cot \alpha }}$; the height of the man is being ignored.
Correct Answer: Solve it.
1991
JEE Advanced
Numerical
IIT-JEE 1991
In a triangle of base a the ratio of the other two sides is $r\left( { < 1} \right)$. Show that the altitude of the triangle is less than of equal to ${{ar} \over {1 - {r^2}}}$
Correct Answer: Solve it.
1991
JEE Advanced
Numerical
IIT-JEE 1991
The sides of a triangle are three consecutive natural numbers and its largest angle is twice the smallest one. Determine the sides of the triangle.
Correct Answer: $$4, 5, 6$$
1990
JEE Advanced
Numerical
IIT-JEE 1990
A vertical tower $PQ$ stands at a point $P$. Points $A$ and $B$ are located to the South and East of $P$ respectively. $M$ is the mid point of $AB$. $PAM$ is an equilateral triangle; and $N$ is the foot of the perpendicular from $P$ and $AB$. Let $AN$$=20$ mrtres and the angle of elevation of the top of the tower at $N$ is ${\tan ^{ - 1}}\left( 2 \right)$. Determine the height of the tower and the angles of elevation of the top of the tower at $A$ and $B$.
Correct Answer: $$40\sqrt 3 \,\,\,m,\,\,\,{60^ \circ },\,\,\,{45^ \circ }$$
1989
JEE Advanced
Numerical
IIT-JEE 1989
$ABC$ is a triangular park with $AB=AC=100$ $m$. A television tower stands at the midpoint of $BC$. The angles of elevetion of the top of the tower at $A, B, C$ are 45$^ \circ $, 60$^ \circ $, 60$^ \circ $, respectively. Find the height of the tower.
Correct Answer: $$50\sqrt 3 $$ $$m$$
1988
JEE Advanced
Numerical
IIT-JEE 1988
A sign -post in the form of an isosceles triangle $ABC$ is mounted on a pole of height $h$ fixed to the ground. The base $BC$ of the triangle is parallel to the ground. A man standing on the ground at a distance $d$ from the sign-post finds that the top vertex $A$ of the triangle subtends an angle $\beta $ and either of the other two vertices subtends the same angle $\alpha $ at his feet. Find the area of the triangle.
Correct Answer: $$\left( {d\tan \beta - h} \right)\sqrt {{h^2}{{\cot }^2}\alpha - {d^2}} $$
1988
JEE Advanced
Numerical
IIT-JEE 1988
If the angles of a triangle are ${30^ \circ }$ and ${45^ \circ }$ and the included side is $\left( {\sqrt 3 + 1} \right)$ cms, then the area of the triangle is ...............
Correct Answer: $${{\sqrt 3 + 1} \over 2}$$ sq. unit
1987
JEE Advanced
Numerical
IIT-JEE 1987
A polygon of nine sides, each of length $2$, is inscribed in a circle. The radius of the circle is .................
Correct Answer: $$\cos ec\,{\raise0.5ex\hbox{$\scriptstyle \pi $}
\kern-0.1em/\kern-0.15em
\lower0.25ex\hbox{$\scriptstyle 9$}}$$
1986
JEE Advanced
Numerical
IIT-JEE 1986
If in a triangle $ABC$, $\cos A\cos B + \sin A\sin B\sin C = 1,$ Show that $a:b:c = 1:1:\sqrt 2 $
Correct Answer: Solve it.
1985
JEE Advanced
Numerical
IIT-JEE 1985
In a triangle $ABC$, the median to the side $BC$ is of length
$${1 \over {\sqrt {11 - 6\sqrt 3 } }}$$ and it divides the angle $A$ into angles ${30^ \circ }$ and ${45^ \circ }$. Find the length of the side $BC$.
Correct Answer: $$2$$ units.
1985
JEE Advanced
Numerical
IIT-JEE 1985
A ladder rests against a wall at an angle $\alpha $ to the horizintal. Its foot is pulled away from the wall through a distance $a$, so that it slides $a$ distance $b$ down the wall making an angle $\beta $ with the horizontal. Show that $a = b\tan {1 \over 2}\left( {\alpha + \beta } \right)$
Correct Answer: Solve it.
1985
JEE Advanced
Numerical
IIT-JEE 1985
In a triangle $ABC$, if cot $A$, cot $B$, cot $C$ are in A.P., then ${a^2},{b^2},{c^2}$, are in ............... progression.
Correct Answer: Arithmetic
1985
JEE Advanced
Numerical
IIT-JEE 1985
The set of all real numbers $a$ such that ${a^2} + 2a,2a + 3$ and ${a^2} + 3a + 8$ are the sides of a triangle is ...........
Correct Answer: $$\left( {5,x} \right)$$
1984
JEE Advanced
Numerical
IIT-JEE 1984
With usual notation, if in a triangle $ABC$;
${{b + c} \over {11}} = {{c + a} \over {12}} = {{a + b} \over {13}}$ then prove that ${{\cos A} \over 7} = {{\cos B} \over {19}} = {{\cos C} \over {25}}$.
${{b + c} \over {11}} = {{c + a} \over {12}} = {{a + b} \over {13}}$ then prove that ${{\cos A} \over 7} = {{\cos B} \over {19}} = {{\cos C} \over {25}}$.
Correct Answer: Solve it.
1984
JEE Advanced
Numerical
IIT-JEE 1984
For a triangle $ABC$ it is given that $\cos A + \cos B + \cos C = {3 \over 2}$. Prove that the triangle is equilateral.
Correct Answer: Solve it.
1983
JEE Advanced
Numerical
IIT-JEE 1983
The ex-radii ${r_1},{r_2},{r_3}$ of $\Delta $$ABC$ are H.P. Show that its sides $a, b, c$ are in A.P.
Correct Answer: Solve it.
1982
JEE Advanced
Numerical
IIT-JEE 1982
A vertical pole stands at a point $Q$ on a horizontal ground. $A$ and $B$ are points on the ground, $d$ meters apart. The pole subtends angles $\alpha $ and $\beta $ at $A$ and $B$ respectively. $AB$ subtends an angle $\gamma $ and $Q$. Find the height of the pole.
Correct Answer: $${d \over {\sqrt {{{\cot }^2}\alpha + {{\cot }^2}\beta - \cot \alpha \cot \beta \cot \gamma } }}$$
1981
JEE Advanced
Numerical
IIT-JEE 1981
Let the angles $A, B, C$ of a triangle $ABC$ be in A.P. and let $b:c = \sqrt 3 :\sqrt 2 $. Find the angle $A$.
Correct Answer: $${75^ \circ }$$
1980
JEE Advanced
Numerical
IIT-JEE 1980
$ABC$ is a triangle with $AB=AC$. $D$ is any point on the side $BC$. $E$ and $F$ are points on the side $AB$ and $AC$, respectively, such that $DE$ is parallel to $AC$, and $DF$ is parallel to $AB$. Prove that
$$DF + FA + AE + ED = AB + AC$$
Correct Answer: Solve it.
1980
JEE Advanced
Numerical
IIT-JEE 1980
(i) $PQ$ is a vertical tower. $P$ is the foot and $Q$ is the top of the tower. $A, B, C$ are three points in the horizontal plane through $P$. The angles of elevation of $Q$ from $A$, $B$, $C$ are equal, and each is equal to $\theta $. The sides of the triangle $ABC$ are $a, b, c$; and the area of the triangle $ABC$ is $\Delta $. Show that the height of the tower is ${{abc\tan \theta } \over {4\Delta }}$.
(ii) $AB$ is vertical pole. The end $A$ is on the level ground. $C$ is the middle point of $AB$. $P$ is a point on the level ground. The portion $CB$ subtends an angle $\beta $ at $P$. If $AP = n\,AB,$ then show that tan$\beta $ $ = {n \over {2{n^2} + 1}}$
Correct Answer: (i) Solve it.
<p>(ii) Solve it.</p>
1980
JEE Advanced
Numerical
IIT-JEE 1980
$ABC$ is a triangle. $D$ is the middle point of $BC$. If $AD$ is perpendicular to $AC$, then prove that
$$\cos A\,\cos C = {{2\left( {{c^2} - {a^2}} \right)} \over {3ac}}$$
Correct Answer: Solve it.
1980
JEE Advanced
Numerical
IIT-JEE 1980
$ABC$ is a triangle, $P$ is a point on $AB$, and $Q$ is point on $AC$ such that $\angle AQP = \angle ABC$. Complete the relation
$${{area\,\,of\,\,\Delta APQ} \over {area\,\,of\,\,\Delta ABC}} = {{\left( {...} \right)} \over {A{C^2}}}$$
Correct Answer: $$A{P^2}$$