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JEE • MATHEMATICS • PROPERTIES OF TRIANGLES

Properties of triangles practice · Mixed difficulty

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Practice format and chapter quiz

These sets use single-correct-answer MCQs. They do not cover every JEE Main or JEE Advanced question format and are not full-length mock papers.

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Showing 61-90 of 434 questions
61 Easy
61 of 434
Application +4 | -1
If the area of a △ABC\triangle A B C be λ\lambda, then a2sin⁡2B+b2sin⁡2Aa^2 \sin 2 B+b^2 \sin 2 A is equal to
View answer and explanation
Correct option: C
4λ4 \lambda
Option C (4λ4 \lambda) is correct because using trigonometric identities and the sine rule, the expression simplifies to four times the area of the triangle.
62 Easy
62 of 434
Application +4 | -1
In any ΔABC\Delta ABC under usual notation, a(bcos⁡C−ccos⁡B)a\left( {b\cos C - c\cos B} \right) is equal to
View answer and explanation
Correct option: A
b2−c2{b^2} - {c^2}
Option A (b2−c2{b^2} - {c^2}) is correct because using the cosine rule, the expression simplifies to b2−c2{b^2} - {c^2}.
63 Easy
63 of 434
Application +4 | -1
If the sides a,  ba,\,\,b and cc of a Δ ABC\Delta \,ABC are in A.P., then
(tan⁡  A2+tan⁡  C2):cot⁡  B2\left( {\tan \,\,\frac{A}{2} + \tan \,\,\frac{C}{2}} \right):\cot \,\,\frac{B}{2}
View answer and explanation
Correct option: D
None of these
Option D (None of these) is correct because the ratio simplifies to 2/3, which is not among the given options.
64 Medium
64 of 434
Analyzing +4 | -1
In any △ABC\triangle A B C, the least value of (sin⁡2A+sin⁡A+1sin⁡A)\left(\frac{\sin ^2 A+\sin A+1}{\sin A}\right) is
View answer and explanation
Correct option: A
3
Option A (3) is correct because the minimum value of the given expression occurs when sin⁡A=1\sin A = 1, and the expression evaluates to 3.
65 Easy
65 of 434
Application +4 | -1
A △ABC\triangle A B C is right angled at BB. Then, the diameter of the incircle of the triangle is
View answer and explanation
Correct option: C
c+a−bc+a-b
Option C (c+a−bc+a-b) is correct because it represents the diameter (2r) of the incircle of a right-angled triangle.
66 Medium
66 of 434
Analyzing +4 | -1
In △ABC,a=5,b=4\triangle A B C, a=5, b=4 and cos⁡(A−B)=3132\cos (A-B)=\frac{31}{32}, then side cc is
View answer and explanation
Correct option: A
6
Option A (6) is correct because by using the tangent half-angle formula and the cosine rule, the side c can be calculated to be 6.
67 Easy
67 of 434
Application +4 | -1
In Δ  ABC,1+cos⁡(A−B)cos⁡C1+cos⁡(A−C)cos⁡B\Delta \;ABC,\frac{{1 + \cos \left( {A - B} \right)\cos C}}{{1 + \cos \left( {A - C} \right)\cos B}} is equal to
View answer and explanation
Correct option: D
a2+b2a2+c2\frac{{{a^2} + {b^2}}}{{{a^2} + {c^2}}}
Option D ((a²+b²)/(a²+c²)) is correct because the given expression simplifies to (sin²A + sin²B)/(sin²A + sin²C), which is equal to (a²+b²)/(a²+c²) by the sine rule.
68 Easy
68 of 434
Analyzing +4 | -1
In Δ  ABC,ifsin⁡2A2,sin⁡2B2,sin⁡2C2\Delta \;ABC,\mathrm{if}{\sin ^2}\frac{A}{2},{\sin ^2}\frac{B}{2},{\sin ^2}\frac{C}{2} be in HP. Then, a,b,ca,b,c will be in
View answer and explanation
Correct option: C
HP
Option C (HP) is correct because the reciprocals of the given terms are in AP, leading to a, b, c being in HP.
69 Easy
69 of 434
Application +4 | -1
If in a ΔABC,  a=6\Delta ABC,\;a = 6 cm, b=8b = 8 cm, c=10c = 10 cm, then the value of sin⁡2A\sin 2A is
View answer and explanation
Correct option: D
24/2524/25
Option D (24/25) is correct because using the cosine rule to find cos A, then sin A, and finally sin 2A gives 24/25.
70 Easy
70 of 434
Application +4 | -1
The circumradius of the triangle whose sides are 13, 12 and 5, is
View answer and explanation
Correct option: B
132\frac{{13}}{2}
Option B (132\frac{{13}}{2}) is correct because the triangle is right-angled, and the circumradius is half the length of the hypotenuse.
71 Medium
71 of 434
Application +4 | -1
Two sides of a triangle are 3−1\sqrt{3}-1 and 3+1\sqrt{3}+1 units and their included angle is 60∘60^{\circ}.
View answer and explanation
Correct option: A
6\sqrt{6}
Option A (6\sqrt{6}) is correct because using the cosine rule with the given sides and included angle yields a side length of 6\sqrt{6}.
72 Easy
72 of 434
Application +4 | -1
In a Δ  ABC,  a(bcos⁡C−ccos⁡B)\Delta \;ABC,\;a\left( {b\cos C - c\cos B} \right) is equal to
View answer and explanation
Correct option: B
b2−c2{b^2} - {c^2}
Option B b2−c2{b^2} - {c^2} is correct because it is derived from applying the cosine rule and simplifying the expression.
73 Easy
73 of 434
Application +4 | -1
The greatest angle of a triangle whose sides are x2+x+1,2x+1x^2+x+1,2 x+1 and x2−1x^2-1, is
View answer and explanation
Correct option: C
120∘120^{\circ}
Option A (120∘120^{\circ}) is correct because the greatest angle is found using the cosine rule, with x2+x+1x^2+x+1 being the longest side.
74 Medium
74 of 434
Application +4 | -1
The sides of triangle are in the ratio 1:3:21:\sqrt 3 :2, then the angles of the triangle are in ratio
View answer and explanation
Correct option: D
1:2:31:2:3
Option D (1:2:31:2:3) is correct because the sides are in the ratio 1:3:21:\sqrt 3 :2, indicating a 30-60-90 triangle, and the angles are in the ratio 1:2:31:2:3.
75 Medium
75 of 434
Analyzing +4 | -1
Consider the following statements :

1. b2−c2asin⁡(B−C)=2R\frac{{{b^2} - {c^2}}}{{a\sin \left( {B - C} \right)}} = 2R

2. asin⁡(B−C)+bsin⁡(C−A)+csin⁡(A−B)=0a\sin \left( {B - C} \right) + b\sin \left( {C - A} \right) + c\sin \left( {A - B} \right) = 0

Which of these is/are correct?
View answer and explanation
Correct option: C
Both (1) and (2)
Option C (Both (1) and (2)) is correct because both statements are trigonometric identities derived from the sine rule and angle relationships in a triangle.
76 Easy
76 of 434
Application +4 | -1
If the angles of a triangle are in the ratio 1  :2  :31\;:2\;:3, the corresponding sides are in the ratio

View answer and explanation
Correct option: D
1  :3  :21\;:\sqrt 3 \;:2
Option D (1:3:21:\sqrt 3 :2) is correct because the ratio of sides is the same as the ratio of sines of the angles, and the angles are 30, 60, and 90 degrees.
77 Medium
77 of 434
Application +4 | -1
A tower subtends a right angle at the top of a pole and distance between tower and pole is 10 m,
then minimum height of the tower is
View answer and explanation
Correct option: C
20 m
Option D (20 m) is correct because the minimum height is achieved when the angle is 45 degrees, resulting in a height of 20 m.
78 Medium
78 of 434
Analyzing +4 | -1
If the cotangents of half the angles of a triangle are in AP, then its sides are in
View answer and explanation
Correct option: A
AP\mathrm{AP}
Option A (AP) is correct because if the cotangents of half the angles of a triangle are in AP, then the sides of the triangle are also in AP.
79 Easy
79 of 434
Application +4 | -1
In a ΔABC,acot⁡A+bcot⁡B+ccot⁡C\Delta ABC,a\cot A + b\cot B + c\cot C is equal to
View answer and explanation
Correct option: C
2(r+R)2\left( {r + R} \right)
Option C (2(r+R)2\left( {r + R} \right)) is correct because acot⁡A+bcot⁡B+ccot⁡Ca\cot A + b\cot B + c\cot C can be simplified to 2R(cos⁡A+cos⁡B+cos⁡C)2R(\cos A + \cos B + \cos C), which equals 2(r+R)2(r+R).
80 Easy
80 of 434
Application +4 | -1
In Δ  ABC,if(a+b+c)(a−b+c)=3ac\Delta \;ABC,\mathrm{if}\left( {a + b + c} \right)\left( {a - b + c} \right) = 3ac, then
View answer and explanation
Correct option: A
∠B=60∘\angle B = 60^\circ
Option A (∠B=60∘\angle B = 60^\circ) is correct because by simplifying the given equation and using the cosine rule, we find that angle B is 60 degrees.
81 Easy
81 of 434
Application +4 | -1
In a Δ  ABC\Delta \;ABC, if b2+c2=3a2{b^2} + {c^2} = 3{a^2}, then cot⁡B+cot⁡C−cot⁡A\cot B + \cot C - \cot A is equal to
View answer and explanation
Correct option: C
0
Option C (0) is correct because substituting the given condition b² + c² = 3a² into the expression and simplifying leads to 0.
82 Easy
82 of 434
Application +4 | -1
If in a △ABC,a2+b2+c2=8R2\triangle A B C, a^2+b^2+c^2=8 R^2, where R=R= circumradius, then the triangle is
View answer and explanation
Correct option: C
right angled
Option C (right angled) is correct because the given condition a2+b2+c2=8R2a^2 + b^2 + c^2 = 8R^2 leads to one angle being 90 degrees.
83 Medium
83 of 434
Analyzing +4 | -1
A tower subtends angles θ,2θ,3θ\theta, 2 \theta, 3 \theta at points A,BA, B and CC respectively all lying on a horizontal line through the foot of tower. If ABBC=1+2\frac{A B}{B C}=1+\sqrt{2}, then θ\theta is equal to
View answer and explanation
Correct option: A
π8\frac{\pi}{8}
Option A (π8\frac{\pi}{8}) is correct because using trigonometric identities and the given ratio, we can derive that θ=π8\theta = \frac{\pi}{8}.
84 Easy
84 of 434
Application +4 | -1
The vertices of a triangle are A(1,1),B(4,5)A(1,1), B(4,5) and C(6,13)C(6,13). Then, the value of cos⁡A\cos A is
View answer and explanation
Correct option: A
6365\frac{63}{65}
Option A (6365\frac{63}{65}) is correct because it is the cosine of angle A, calculated using the cosine rule with the given vertices of the triangle.
85 Easy
85 of 434
Analyzing +4 | -1
The ex-radii of a triangle r1,  r2,  r3{r_1},\,\,{r_2},\,\,{r_3} are in harmonic progression, then the sides a,  b,  ca,\,\,b,\,\,c are
View answer and explanation
Correct option: B
In A.P.
Option B (In A.P.) is correct because the harmonic progression of ex-radii implies an arithmetic progression of the sides.
86 Easy
86 of 434
Application +4 | -1
If x,y,zx,y,z are perpendicular drawn from the vertices of triangle having sides a,ba,b and cc, then the value of bxc+cya+azb\frac{{bx}}{c} + \frac{{cy}}{a} + \frac{{az}}{b} will be
View answer and explanation
Correct option: A
a2+b2+c22R\frac{{{a^2} + {b^2} + {c^2}}}{{2R}}
Option A (a2+b2+c22R\frac{{{a^2} + {b^2} + {c^2}}}{{2R}}) is correct because it is derived from the area of the triangle and the circumradius.
87 Hard
87 of 434
Application +4 | -1
In a △ABC,r1bc+r2ca+r3ab\triangle A B C, \frac{r_1}{b c}+\frac{r_2}{c a}+\frac{r_3}{a b} is equal to
View answer and explanation
Correct option: D
1r−12R\frac{1}{r}-\frac{1}{2 R}
Option D (1r−12R\frac{1}{r}-\frac{1}{2 R}) is correct because it is derived by using the formulas for exradii and the inradius of a triangle.
88 Easy
88 of 434
Application +4 | -1
In a △ABC,R=\triangle A B C, R= circumradius and r=r= inradius. The value of acos⁡A+bcos⁡B+ccos⁡Ca+b+c\frac{a \cos A+b \cos B+c \cos C}{a+b+c} is equal to
View answer and explanation
Correct option: C
rR\frac{r}{R}
Option C (r/Rr/R) is correct because by using trigonometric identities and the definitions of inradius and circumradius, the given expression simplifies to r/Rr/R.
89 Medium
89 of 434
Application +4 | -1
The angle of elevation of the top CC of a vertical tower CDC D of height hh from a point AA in the horizontal plane is 45∘45^{\circ} and from a point BB at a distance aa from AA on the line making an angle 30∘30^{\circ} with ADA D, it is 60∘60^{\circ}, then
View answer and explanation
Correct option: C
a=h(3−1)a=h(\sqrt{3}-1)
Option C (a=h(3−1)a=h(\sqrt{3}-1)) is correct because it is derived by using trigonometric relationships and the sine rule in the given geometric setup.
90 Easy
90 of 434
Application +4 | -1
In a Δ ABC\Delta \,ABC, if a,  c,  ba,\,\,c,\,\,b are in A.P., then the value of a  cos⁡  B−b  cos⁡  Aa−b\frac{{a\,\,\cos \,\,B - b\,\,\cos \,\,A}}{{a - b}}, is
View answer and explanation
Correct option: B
2\mathrm{2}
Option B (2) is correct because the value of the expression simplifies to 2 using the cosine rule and the given A.P. condition.
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