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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 421-434 of 434 questions

421 Medium

421 of 434

Application +4 | -1

In a

\triangle \mathrm{ABC}

\angle \mathrm{A}=60^{\circ}, \frac{\mathrm{b}}{\mathrm{c}}=\frac{\sqrt{3}+1}{2}

then

\angle \mathrm{B}-\angle \mathrm{C}

has value equal to

View answer and explanation

Correct option: B

30^{\circ}

Option B (

30^{\circ}

) is correct because it is the result of applying the tangent half-angle formula and simplifying.

422 Easy

422 of 434

Application +4 | -1

In a

\Delta \,ABC

, the HM of the ex-radii is equal to

View answer and explanation

Correct option: A

3\,r

Option A (

3r

) is correct because the harmonic mean of the ex-radii is equal to

3r

, where

r

is the inradius of the triangle.

423 Medium

423 of 434

Application +4 | -1

In a

\Delta \,ABC

, if

C = 60^\circ

, then

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

View answer and explanation

Correct option: B

\mathrm{1}

Option A (b/a) is correct because by using trigonometric identities and subtracting the equations for α and β, we can derive the value of tan((α + β)/2).

424 Easy

424 of 434

Analyzing +4 | -1

If in a

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

, then

View answer and explanation

Correct option: B

2 \sin A \cos B=\sin C

Option D (None of the above) is incorrect. Option B (

2 \sin A \cos B = \sin C

) is the correct answer because it is derived from the given condition using trigonometric identities and the sine rule.

425 Easy

425 of 434

Application +4 | -1

\Delta \;ABC

, if

\frac{1}{{b + c}} + \frac{1}{{c + a}} = \frac{3}{{a + b + c}}

, then

C

is equal to

View answer and explanation

Correct option: B

60^\circ

Option B (

60^\circ

) is correct because the given condition leads to the equation

a^2 + b^2 - c^2 = ab

, which implies

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

426 Medium

426 of 434

Application +4 | -1

In a triangle, if

{r_1} + {r_3} = kco{s^2}\frac{B}{2}

, then

k

is equal to

View answer and explanation

Correct option: D

4R

Option D (

4R

) is correct because, after simplification using trigonometric identities and the area of a triangle formula, k is found to be equal to 4R.

427 Easy

427 of 434

Application +4 | -1

\Delta \;ABC,\left( {b + c} \right)\cos A + \left( {c + a} \right)\cos B + \left( {a + b} \right)\cos C

is equal to

View answer and explanation

Correct option: C

a + b + c

Option C

a + b + c

is correct because it is derived by expanding the expression and using the projection formula.

428 Easy

428 of 434

Application +4 | -1

In a

\Delta \;ABC,\left( {b + c - a} \right)\tan \frac{A}{2}

is equal to

View answer and explanation

Correct option: A

\frac{{2\Delta }}{s}

Option A (

\frac{{2\Delta }}{s}

) is correct because

\left( {b + c - a} \right)\tan \frac{A}{2}

can be simplified using trigonometric identities and the area formula to equal

\frac{{2\Delta }}{s}

429 Easy

429 of 434

Application +4 | -1

If angles

A, B

and

C

are in AP, then

\frac{a+c}{b}

is equal to

View answer and explanation

Correct option: B

\cos \left(\frac{A-C}{2}\right)

Option B (

\cos \left(\frac{A-C}{2}\right)

) is correct because using the properties of AP and trigonometric identities, the expression simplifies to this result.

430 Medium

430 of 434

Application +4 | -1

In a

\triangle A B C, A=\frac{2 \pi}{3}, b-c=3 \sqrt{3} \mathrm{~cm}

and

\operatorname{ar}(\triangle A B C)=\frac{9 \sqrt{3}}{2} \mathrm{~cm}^2

. Then,

a

View answer and explanation

Correct option: B

9 \mathrm{~cm}

Option B (

9 \text{{ cm}}

) is correct because using the area and angle A, we can find bc, and with b-c, we can find a using the cosine rule.

431 Easy

431 of 434

Analyzing +4 | -1

{p_1},\,\,{p_2},\,\,{p_3}

are altitude of a triangle

ABC

from the vertices

A,\,\,B,\,\,C

and

\Delta

, the area of the triangle, then

\frac{1}{{p_1^2}} + \frac{1}{{p_2^2}} + \frac{1}{{p_3^2}} =

View answer and explanation

Correct option: A

\frac{{\cot \,\,A + \cos \,\,B + \cot \,\,C}}{\Delta }

Option A is correct because it correctly relates the altitudes of a triangle to its area and cotangents of its angles.

432 Easy

432 of 434

Application +4 | -1

In a triangle

A B C, 2 a c \sin \frac{1}{2}(A-B+C)=

View answer and explanation

Correct option: B

c^2+a^2-b^2

Option B (

c^2+a^2-b^2

) is correct because using the given condition and trigonometric identities, the expression simplifies to this form.

433 Easy

433 of 434

Application +4 | -1

If the sides of the triangle are

p,q,\sqrt {{p^2} + {q^2} + pq}

, then the greatest angle is

View answer and explanation

Correct option: C

\frac{{2\pi }}{3}

Option C (

\frac{{2\pi }}{3}

) is correct because the greatest angle is found using the cosine rule.

434 Medium

434 of 434

Application +4 | -1

The sides of a triangle are

8 \mathrm{~cm}

10 \mathrm{~cm}

and

12 \mathrm{~cm}

. Then, the greatest angle is

View answer and explanation

Correct option: B

double of the smallest angle

Option B (double of the smallest angle) is correct because by using the cosine rule, we find that cos(C) = 1/8 and cos(A) = 3/4, and cos(2A) = 1/8, implying C = 2A.

Revision notes and key takeaways

Review PROPERTIES OF TRIANGLES in your textbook or class notes, then use this set to check your understanding.

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