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JEE • MATHEMATICS • PROPERTIES OF TRIANGLES

Properties of triangles practice · Medium

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Showing 1-30 of 140 questions
1 Medium
1 of 140
Application +4 | -1
In Δ  ABC,b+ca\Delta \;ABC,\frac{{b + c}}{a} is equal to
View answer and explanation
Correct option: A
cos⁡12(B−C)sin⁡12  A\frac{{\cos \frac{1}{2}\left( {B - C} \right)}}{{\sin \frac{1}{2}\;A}}
Option A (cos⁡12(B−C)sin⁡12 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 b+ca\frac{{b + c}}{a} can be simplified to cos⁡12(B−C)sin⁡12 A\frac{{\cos \frac{1}{2}({B - C})}}{{\sin \frac{1}{2}\,A}}.
2 Medium
2 of 140
Application +4 | -1
If AA is the area and 2s2 s is the sum of the sides of a triangle, then
View answer and explanation
Correct option: A
A≤s233A \leq \frac{s^2}{3 \sqrt{3}}
Option A (A≤s233A \leq \frac{s^2}{3 \sqrt{3}}) is correct because it is derived using the AM-GM inequality on the sides of the triangle.
3 Medium
3 of 140
Analyzing +4 | -1
In an acute angled △ABC,r+r1=r2+r3\triangle A B C, r+r_1=r_2+r_3 and ∠B>π3\angle B>\frac{\pi}{3}, then
View answer and explanation
Correct option: D
b+3c<3a<3b+3cb+3 c<3 a<3 b+3 c
Option D (b+3c<3a<3b+3cb+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.
4 Medium
4 of 140
Application +4 | -1
In a triangle ABC\mathrm{ABC}, if
tan⁡(A2)=p,tan⁡(B2)=q,\tan \left(\frac{\mathrm{A}}{2}\right)=\mathrm{p}, \tan \left(\frac{\mathrm{B}}{2}\right)=\mathrm{q}, then

2(p+q)(1−pq)(1+p2)(1+q2)\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\sin C
Option C (sin⁡C\sin C) is correct because the given expression simplifies to sin⁡C\sin C using trigonometric identities.
5 Medium
5 of 140
Analyzing +4 | -1
sin⁡A,sin⁡B\sin A, \sin B and sin⁡C\sin C are in APA P for the △ABC\triangle A B C. Then,
View answer and explanation
Correct option: B
the altitudes are in HPH 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.
6 Medium
6 of 140
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
2:6:3+12:\sqrt 6 :\sqrt 3 + 1
Option A (2:6:3+12:\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.
7 Medium
7 of 140
Application +4 | -1
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: B
r/Rr/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.
8 Medium
8 of 140
Application +4 | -1
If A,B,CA, B, C are vertices of triangle, then cos⁡2A+cos⁡2B+cos⁡2C+2cos⁡Acos⁡Bcos⁡C\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
1
Option A (1) is correct because by using trigonometric identities and the fact that A+B+C = pi, the expression simplifies to 1.
9 Medium
9 of 140
Application +4 | -1
Let ABCA B C be a triangle such that ∠ACB=π6\angle A C B=\frac{\pi}{6} and a,ba, b and cc denote the lengths of the sides opposite to A,BA, B and CC, respectively. The value(s) of xx for which a=x2+x+1,b=x2−1a=x^2+x+1, b=x^2-1 and c=2x+1c=2 x+1 is(are)
View answer and explanation
Correct option: D
434 \sqrt{3}
The provided solution is incorrect. The correct answer should be 1+31+\sqrt{3}. By using the cosine rule and simplifying, we find that x=1+3x=1+\sqrt{3}.
10 Medium
10 of 140
Analyzing +4 | -1
If in a Δ ABC\Delta \,ABC,

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

∣ abcbcacab ∣=\left| {\,\begin{array}{cccccccccccccccccccc} a&b&c\\ b&c&a\\ c&a&b \end{array}\,} \right| =
View answer and explanation
Correct option: A
0\mathrm{0}
Option A (0\mathrm{0}) is correct because the given condition implies that A=B=C=60∘A = B = C = 60^\circ, making the determinant zero.
11 Medium
11 of 140
Application +4 | -1
In a △ABC\triangle A B C, if 3tan⁡A2tan⁡C2=13 \tan \frac{A}{2} \tan \frac{C}{2}=1, then sides a,ba, b and cc are in
View answer and explanation
Correct option: A
AP
Option A (AP) is correct because the given condition leads to the relationship a + c = 2b, which defines an arithmetic progression.
12 Medium
12 of 140
Application +4 | -1
The area (in square unit) of the triangle formed by the points with polar coordinates (1, 0), (2,π3)\left( {2,\frac{\pi }{3}} \right) and (3,2π3)\left( {3,\frac{{2\pi }}{3}} \right) is
View answer and explanation
Correct option: B
534\frac{{5\sqrt 3 }}{4}
Option B (534\frac{{5\sqrt 3 }}{4}) is correct because it is the area of the triangle calculated using the polar coordinates of the vertices.
13 Medium
13 of 140
Application +4 | -1
In a Δ ABC\Delta \,ABC, if a  cos⁡2  C2+c  cos⁡2  A2=3 b2a\,\,{\cos ^2}\,\,\frac{C}{2} + c\,\,{\cos ^2}\,\,\frac{A}{2} = \frac{{3\,b}}{2} then a,  b,  ca,\,\,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.
14 Medium
14 of 140
Application +4 | -1
The area of right-angled triangle in terms of rr and r1r_1, if ∠A=90∘\angle A=90^{\circ} (where r,r1r, r_1 have their usual meanings), is
View answer and explanation
Correct option: B
rr1r r_1
Option B (rr1r 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.
15 Medium
15 of 140
Analyzing +4 | -1
In △ABC\triangle A B C, Σ(b+c)tan⁡A2tan⁡(B−C2)\Sigma(b+c) \tan \frac{A}{2} \tan \left(\frac{B-C}{2}\right) is equal to
View answer and explanation
Correct option: D
0
Option D (0) is correct because the expression simplifies to b-c + c-a + a-b, which equals zero due to the tangent half-angle and difference identities.

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16 Medium
16 of 140
Application +4 | -1
In a Δ ABC\Delta \,ABC, if A=30∘,  b=2,  c=3+1A = 30^\circ ,\,\,b = 2,\,\,c = \sqrt 3 + 1, then C−B2=\frac{{C - B}}{2} =
View answer and explanation
Correct option: B
30∘30^\circ
Option B (30∘30^\circ) is correct because applying the tangent half-angle formula and solving for C−B2\frac{{C - B}}{2} yields 30∘30^\circ.
17 Medium
17 of 140
Application +4 | -1
In a △ABC,a=5,b=4\triangle A B C, a=5, b=4 and tan⁡C2=79\tan \frac{C}{2}=\sqrt{\frac{7}{9}}. The side cc is
View answer and explanation
Correct option: A
6
Option A (6) is correct because it is the solution obtained by applying the tangent half-angle formula and solving for side c.
18 Medium
18 of 140
Application +4 | -1
The area of a △ABC\triangle A B C is a2−(b−c)2a^2-(b-c)^2. Then, tan⁡A\tan A is equal to&nbsp;
View answer and explanation
Correct option: C
815\frac{8}{15}
Option C (815\frac{{8}}{{15}}) is correct because by using the area formula and trigonometric identities, we can find the value of tan A.
19 Medium
19 of 140
Application +4 | -1
If p1,p2,  p3{p_1},{p_2},\;{p_3} are respectively the perpendicular from the vertices of a triangle to the opposite sides, then p1p2p3{p_1}{p_2}{p_3} is equal to
View answer and explanation
Correct option: D
a2b2c28R2\frac{{{a^2}{b^2}{c^2}}}{{8{R^2}}}
Option D (a2b2c28R2\frac{{{a^2}{b^2}{c^2}}}{{8{R^2}}}) is correct because the product of the perpendiculars from the vertices to the opposite sides can be expressed in terms of the sides and the circumradius of the triangle.
20 Medium
20 of 140
Analyzing +4 | -1
The sides of a triangle are 3x+4y,  4x+3y3x + 4y,\,\,4x + 3y and 5x+5y5x + 5y, where, x, y>0x,\,y > 0 then the triangle is
View answer and explanation
Correct option: B
Obtuse angled
Option B (Obtuse angled) is correct because by using the cosine rule, we find that cos⁡C<0\cos C < 0, which means angle C is obtuse.
21 Medium
21 of 140
Application +4 | -1
In a Δ ABC\Delta \,ABC if a=5,  b=4a = 5,\,\,b = 4 and tan⁡C2=73\tan \frac{C}{2} = \frac{{\sqrt 7 }}{3}, then c=c =
View answer and explanation
Correct option: C
6
Option C (6) is correct because by using the Law of Cosines and the given value of tan⁡C2\tan \frac{C}{2}, we can find the value of c.
22 Medium
22 of 140
Analyzing +4 | -1
If in a Δ  ABC,r3=r1+r2+r\Delta \;ABC,{r_3} = {r_1} + {r_2} + r, then ∠A+∠B\angle A + \angle B is equal to
View answer and explanation
Correct option: C
90∘90^\circ
Option C (90∘90^\circ) is correct because the given condition r3=r1+r2+r{r_3} = {r_1} + {r_2} + r leads to C=90∘C = 90^\circ, and therefore A+B=90∘A + B = 90^\circ.
23 Medium
23 of 140
Analyzing +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 by using trigonometric identities and the definitions of inradius and circumradius, the expression can be simplified to 1r−12R\frac{1}{r}-\frac{1}{2 R}.
24 Medium
24 of 140
Application +4 | -1
If in a Δ  ABC,cos⁡A+2cos⁡B+cos⁡C=2\Delta \;ABC,\cos A + 2\cos B + \cos C = 2, then a,  b,  ca,\;b,\;c are in
View answer and explanation
Correct option: A
AP
Option A (AP) is correct because the given condition leads to the relationship sin⁡A+sin⁡C=2sin⁡B\sin A + \sin C = 2\sin B, which implies that a, b, and c are in arithmetic progression.
25 Medium
25 of 140
Application +4 | -1
In △ABC(a+b)(a−b)=c(b+c)\triangle A B C(a+b)(a-b)=c(b+c), the measure of angle AA, is-
View answer and explanation
Correct option: D
120∘120^{\circ}
Option D (120∘120^{\circ}) is correct because by applying the law of cosines and simplifying the given equation, we find that cos⁡A=−1/2\cos A = -1/2, which corresponds to an angle of 120∘120^{\circ}.
26 Medium
26 of 140
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.
27 Medium
27 of 140
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.
28 Medium
28 of 140
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}.
29 Medium
29 of 140
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.
30 Medium
30 of 140
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.
Revision notes and key takeaways

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  • ✓ This set contains 140 single-correct-answer MCQs on PROPERTIES OF TRIANGLES.
  • ✓ Try each question before opening its answer. Read the explanation where one is available.
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