In $\Delta \;ABC,\frac{{b + c}}{a}$ is equal to

$\frac{{\sin \frac{1}{2}\left( {B - C} \right)}}{{\cos \frac{1}{2}\;A}}$

$\frac{{\cos \frac{1}{2}\left( {B + C} \right)}}{{\cos \frac{1}{2}\;A}}$

If in a $\Delta \,ABC$, ${\sin ^3}A + {\sin ^3}B + {\sin ^3}C = 3\,\,\sin A\,\,\sin \,\,B\,\,\sin \,\,C$, then $\left| {\,\begin{array}{*{20}{c}} a&b&c\\ b&c&a\\ c&a&b \end{array}\,} \right| = $

${\left( {a + b + c} \right)^3}$

$\left( {a + b + c} \right)\,\,\left( {ab + bc + ca} \right)$

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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 1-30 of 434 questions

1 Easy

1 of 434

Application +4 | -1

{p_1},{p_2},{p_3}

are altitudes of a

\Delta \;ABC

drawn from the vertices

A,B,C

and

\Delta

the area of the triangle, then

p_1^{ - 2} + p_2^{ - 2} + p_3^{ - 2}

is equal to

View answer and explanation

Correct option: B

\frac{{{a^2} + {b^2} + {c^2}}}{{4{\Delta ^2}}}

Option B

\frac{{{a^2} + {b^2} + {c^2}}}{{4{\Delta ^2}}}

is correct because it is derived from the relationship between altitudes and the area of the triangle.

2 Easy

2 of 434

Application +4 | -1

If the sides of the triangles are

5k,\;6k,\;5k

and radius of incircle is 6, the value of

k

is equal to

View answer and explanation

Correct option: A

Option A (4) is correct because by using Heron's formula to find the area of the triangle and then using the formula for the inradius, we can solve for

k

3 Easy

3 of 434

Application +4 | -1

In a

\Delta \,ABC

, if

a = \left( {b - c} \right)\sec \theta

, then

\frac{{2\,\sqrt {bc} }}{{b - c}}\,\,\sin \,\,\frac{A}{2} =

View answer and explanation

Correct option: C

\tan \,\,\theta

Option C (

\tan \,\,\theta

) is correct because by substituting

a = (b - c)\sec \theta

into the cosine rule and simplifying, we find that

\frac{{2\,\sqrt {bc} }}{{b - c}}\,\,\sin \,\,\frac{A}{2} = \tan \,\,\theta

4 Medium

4 of 434

Application +4 | -1

\Delta \;ABC,\frac{{b + c}}{a}

is equal to

View answer and explanation

Correct option: A

\frac{{\cos \frac{1}{2}\left( {B - C} \right)}}{{\sin \frac{1}{2}\;A}}

Option A (

\frac{{\cos \frac{1}{2}({B - C})}}{{\sin \frac{1}{2}\,A}}

) is correct because by using the sine rule and sum-to-product trigonometric identities, the expression

\frac{{b + c}}{a}

can be simplified to

\frac{{\cos \frac{1}{2}({B - C})}}{{\sin \frac{1}{2}\,A}}

5 Easy

5 of 434

Application +4 | -1

\Delta = {a^2} - {\left( {b - c} \right)^2}

, where

\Delta

is the area of

\Delta \;ABC

, then

\tan A

is equal to

View answer and explanation

Correct option: C

\frac{8}{{15}}

Option C (

\frac{8}{{15}}

) is correct because by using the given condition

\Delta = {a^2} - {\left( {b - c} \right)^2}

and trigonometric identities, we can find that

\tan A = \frac{8}{{15}}

6 Medium

6 of 434

Application +4 | -1

A

is the area and

2 s

is the sum of the sides of a triangle, then

View answer and explanation

Correct option: A

A \leq \frac{s^2}{3 \sqrt{3}}

Option A (

A \leq \frac{s^2}{3 \sqrt{3}}

) is correct because it is derived using the AM-GM inequality on the sides of the triangle.

7 Medium

7 of 434

Analyzing +4 | -1

In an acute angled

\triangle A B C, r+r_1=r_2+r_3

and

\angle B>\frac{\pi}{3}

, then

View answer and explanation

Correct option: D

b+3 c<3 a<3 b+3 c

Option D (

b+3 c<3 a<3 b+3 c

) is correct because by using the given condition and trigonometric identities, the relationship between a, b, and c can be established.

8 Easy

8 of 434

Application +4 | -1

In a triangle

A B C

, angles are in AP and

b: c=\sqrt{3}: \sqrt{2}

. Then the angle

A

View answer and explanation

Correct option: B

75^{\circ}

Option B (

75^{\circ}

) is correct because using the given ratio of sides and the fact that angles are in AP allows us to determine angle C as 45 degrees, leading to angle A being 75 degrees.

9 Medium

9 of 434

Application +4 | -1

In a triangle

\mathrm{ABC}

, if

\tan \left(\frac{\mathrm{A}}{2}\right)=\mathrm{p}, \tan \left(\frac{\mathrm{B}}{2}\right)=\mathrm{q},

then

\frac{2(p+q)(1-p q)}{\left(1+p^2\right)\left(1+q^2\right)}

then

View answer and explanation

Correct option: C

\sin C

Option C (

\sin C

) is correct because the given expression simplifies to

\sin C

using trigonometric identities.

10 Easy

10 of 434

Application +4 | -1

If in a

\Delta \;ABC,\frac{1}{{a + c}} + \frac{1}{{b + c}} = \frac{3}{{a + b + c}}

, then the value of the angle

C

View answer and explanation

Correct option: A

60^\circ

Option A (

60^\circ

) is correct because the given equation leads to

\cos C = \frac{1}{2}

, which implies

C = 60^\circ

11 Medium

11 of 434

Analyzing +4 | -1

\sin A, \sin B

and

\sin C

are in

A P

for the

\triangle A B C

. Then,

View answer and explanation

Correct option: B

the altitudes are in

H P

Option B (the altitudes are in HP) is correct because if the sines of the angles are in AP, then the sides are in AP, and therefore the altitudes are in HP.

12 Medium

12 of 434

Application +4 | -1

If the angles of a triangle are in the ratio 3:4:5, then the sides are in the ratio

View answer and explanation

Correct option: A

2:\sqrt 6 :\sqrt 3 + 1

Option A (

2:\sqrt 6 :\sqrt 3 + 1

) is correct because the angles are 45, 60, and 75 degrees, and the ratio of their sines gives the ratio of the sides.

13 Easy

13 of 434

Application +4 | -1

PQ

and

PR

are the two sides of a triangle, then the angle between them which gives maximum area of the triangle, is

View answer and explanation

Correct option: D

\frac{\pi }{2}

Option D (

2a

) is correct because the length of the subnormal of the parabola

{y^2} = 4ax

is constant and equal to

2a

14 Easy

14 of 434

Application +4 | -1

In a triangle,

1-\tan \frac{A}{2} \tan \frac{B}{2}=

View answer and explanation

Correct option: B

\frac{2 c}{a+b+c}

Option B (

\frac{2 c}{a+b+c}

) is correct because it is the simplified form of the given expression using the properties of a triangle.

15 Easy

15 of 434

Application +4 | -1

If the sides of a triangle are

7\;\mathrm{cm},\,\,4\,\sqrt 3 \mathrm{\;cm}

and

\sqrt {13} \,\,\;\mathrm{cm}

, then the smallest angle of the triangle is

View answer and explanation

Correct option: C

30^\circ

Option C (

30^\circ

) is correct because it is the smallest angle calculated using the Law of Cosines with the given side lengths.

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16 Easy

16 of 434

Application +4 | -1

In a

\Delta \;ABC,\frac{{b - c\cos A}}{{c - b\cos A}}

is equal to

View answer and explanation

Correct option: B

\frac{{\cos C}}{{\cos B}}

Option B (

\frac{{\cos C}}{{\cos B}}

) is correct because the expression can be simplified using the Law of Cosines and the Law of Sines.

17 Easy

17 of 434

Application +4 | -1

In a

\triangle A B C, 2 s=

perimeter and

R=

circumradius. Then,

\frac{S}{R}

is equal to

View answer and explanation

Correct option: A

\sin A+\sin B+\sin C

Option A (

\sin A + \sin B + \sin C

) is correct because using the sine rule,

s/R

can be expressed as the sum of sines of the angles.

18 Medium

18 of 434

Application +4 | -1

\frac{{a\cos A + b\cos B + c\cos C}}{{a + b + c}}

is equal to

View answer and explanation

Correct option: B

r/R

Option B (r/R) is correct because the expression simplifies to r/R using trigonometric identities and the relationships between the sides, angles, and radii of the triangle.

19 Easy

19 of 434

Application +4 | -1

If two angles of

\Delta \;ABC

are

45^\circ

and

60^\circ

, then the ratio of the smallest and the greatest sides are

View answer and explanation

Correct option: A

\left( {\sqrt 3 - 1} \right):1

Option A (

\left( {\sqrt 3 - 1} \right):1

) is correct because it represents the ratio of the smallest to the greatest side, calculated using the sine rule.

20 Easy

20 of 434

Analyzing +4 | -1

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

repectively, then

View answer and explanation

Correct option: B

{b^2},{c^2}\,\,\,\mathrm{and}\,\,\,{a^2}

are in AP

Option C (

{b^2},{c^2}\,\,\,\mathrm{and}\,\,\,{a^2}

are in AP) is correct because the given condition leads to the relationship

2c^2 = a^2 + b^2

, which defines an arithmetic progression.

21 Easy

21 of 434

Application +4 | -1

If in a triangle

ABC,\,\,\frac{{\cos \,\,A}}{a}\,\, = \frac{{\cos \,\,B}}{b} = \frac{{\cos \,\,C}}{c}

, then the triangle is

View answer and explanation

Correct option: C

Equilateral

Option C (Equilateral) is correct because if

\frac{{\cos \,A}}{a}\, = \frac{{\cos \,B}}{b} = \frac{{\cos \,C}}{c}

, then A = B = C, which implies the triangle is equilateral.

22 Medium

22 of 434

Application +4 | -1

A, B, C

are vertices of triangle, then

\cos ^2 A+\cos ^2 B+\cos ^2 C+2 \cos A \cos B \cos C

is equal to

View answer and explanation

Correct option: A

Option A (1) is correct because by using trigonometric identities and the fact that A+B+C = pi, the expression simplifies to 1.

23 Medium

23 of 434

Application +4 | -1

Let

A B C

be a triangle such that

\angle A C B=\frac{\pi}{6}

and

a, b

and

c

denote the lengths of the sides opposite to

A, B

and

C

, respectively. The value(s) of

x

for which

a=x^2+x+1, b=x^2-1

and

c=2 x+1

is(are)

View answer and explanation

Correct option: D

4 \sqrt{3}

The provided solution is incorrect. The correct answer should be

1+\sqrt{3}

. By using the cosine rule and simplifying, we find that

x=1+\sqrt{3}

24 Easy

24 of 434

Application +4 | -1

Two sides of a triangle are

2\,\sqrt 2 \;\mathrm{cm}

and

2\,\sqrt 3 \;\,\mathrm{cm}

and the angle opposite to the shorter side of the two is

\frac{\pi }{4}

. The largest possible length of the third side is

View answer and explanation

Correct option: A

\left( {\sqrt 6 + \sqrt 2 } \right)\;\mathrm{cm}

The correct answer is None of these because the expression simplifies to

\cos \,11^\circ

, which is not among the given options.

25 Medium

25 of 434

Analyzing +4 | -1

If in a

\Delta \,ABC

{\sin ^3}A + {\sin ^3}B + {\sin ^3}C = 3\,\,\sin A\,\,\sin \,\,B\,\,\sin \,\,C

, then

\left| {\,\begin{array}{cccccccccccccccccccc} a&b&c\\ b&c&a\\ c&a&b \end{array}\,} \right| =

View answer and explanation

Correct option: A

\mathrm{0}

Option A (

\mathrm{0}

) is correct because the given condition implies that

A = B = C = 60^\circ

, making the determinant zero.

26 Medium

26 of 434

Application +4 | -1

In a

\triangle A B C

, if

3 \tan \frac{A}{2} \tan \frac{C}{2}=1

, then sides

a, b

and

c

are in

View answer and explanation

Correct option: A

Option A (AP) is correct because the given condition leads to the relationship a + c = 2b, which defines an arithmetic progression.

27 Easy

27 of 434

Application +4 | -1

a = 2,\,\,b = 3,\,\,c = 5

\Delta \,ABC

, then

C =

View answer and explanation

Correct option: D

None of these

The correct answer is D (None of these) because the calculated value of C is \

\pi

, which is impossible in a triangle.

28 Medium

28 of 434

Application +4 | -1

The area (in square unit) of the triangle formed by the points with polar coordinates (1, 0),

\left( {2,\frac{\pi }{3}} \right)

and

\left( {3,\frac{{2\pi }}{3}} \right)

View answer and explanation

Correct option: B

\frac{{5\sqrt 3 }}{4}

Option B (

\frac{{5\sqrt 3 }}{4}

) is correct because it is the area of the triangle calculated using the polar coordinates of the vertices.

29 Medium

29 of 434

Application +4 | -1

In a

\Delta \,ABC

, if

a\,\,{\cos ^2}\,\,\frac{C}{2} + c\,\,{\cos ^2}\,\,\frac{A}{2} = \frac{{3\,b}}{2}

then

a,\,\,b,\,\,c

are in

View answer and explanation

Correct option: A

A.P.

Option A (A.P.) is correct because the given condition implies that a, b, and c are in arithmetic progression.

30 Medium

30 of 434

Application +4 | -1

The area of right-angled triangle in terms of

r

and

r_1

, if

\angle A=90^{\circ}

(where

r, r_1

have their usual meanings), is

View answer and explanation

Correct option: B

r r_1

Option B (

r r_1

) is correct because the area of the right-angled triangle is equal to the product of the inradius and the exradius opposite to the right angle.

Revision notes and key takeaways

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Key Takeaways

✓ This set contains 434 single-correct-answer MCQs on PROPERTIES OF TRIANGLES.
✓ Try each question before opening its answer. Read the explanation where one is available.
✓ Sign in to save submitted answers. Revisit incorrect answers alongside your textbook or class notes.

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