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JEE • MATHEMATICS • PROPERTIES OF TRIANGLES

Properties of triangles practice · Easy

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91 Easy
91 of 287
Application +4 | -1
In triangle ABC,a=2,b=3ABC,a = 2,b = 3 and sin⁡A=23\sin A = \frac{2}{3} then BB is equal to
View answer and explanation
Correct option: C
90∘90^\circ
Option C (90∘90^\circ) is correct because using the sine rule, we find that sin B = 1, which implies B = 90∘90^\circ.
92 Easy
92 of 287
Analyzing +4 | -1
If in a triangle ABCABC,

sin⁡  A  :sin⁡  C=sin⁡  (A−B)  :sin⁡  (B−C)\sin \,\,A\;:\sin \,\,C = \sin \,\,\left( {A - B} \right)\;:\sin \,\,\left( {B - C} \right) then, a2  :b2  :c2{a^2}\;:{b^2}\;:{c^2} are in

View answer and explanation
Correct option: A
A.P.
The correct answer is A (A.P.) because simplifying the given ratio of sines leads to the condition a2,b2,c2a^2, b^2, c^2 being in Arithmetic Progression.
93 Easy
93 of 287
Application +4 | -1
If the angle of a triangle are in A.P. with common difference equal 13\frac{1}{3} of the least angle, then the sides are in the ratio
View answer and explanation
Correct option: D
2 2  :2 3  :6+22\,\sqrt 2 \;:2\,\sqrt 3 \;:\sqrt 6 + \sqrt 2
Option D (22;:;23;:6+22\sqrt 2 \\;:\\;2\sqrt 3 \\;:\sqrt 6 + \sqrt 2) is correct because the sides are in the ratio of sines of the angles, which are in A.P.
94 Easy
94 of 287
Application +4 | -1
If in △ABC,2b2=a2+c2\triangle A B C, 2 b^2=a^2+c^2, then sin⁡3Bsin⁡B=\frac{\sin 3 B}{\sin B}=
View answer and explanation
Correct option: D
(c2−a22ca)2\left(\frac{c^2-a^2}{2 c a}\right)^2
Option D ((c2−a22ca)2\left(\frac{c^2-a^2}{2 c a}\right)^2) is correct because it is the simplified form of the expression sin⁡3Bsin⁡B\frac{\sin 3B}{\sin B} using the given condition and trigonometric identities.
95 Easy
95 of 287
Application +4 | -1
If two angles of a triangle are 45∘45^\circ and tan⁡−1(2){\tan ^{ - 1}}\left( 2 \right), then the third angle is
View answer and explanation
Correct option: C
tan⁡−13{\tan ^{ - 1}}3
Option C (tan⁡−13{\tan ^{ - 1}}3) is correct because, in a triangle, the sum of all angles is π\pi radians, and using trigonometric identities, the third angle can be found to be tan⁡−13{\tan ^{ - 1}}3.
96 Easy
96 of 287
Analyzing +4 | -1
If in a triangle ABCABC

2  cos⁡  Aa+cos⁡  Bb+2cos⁡  Cc=abc+bca2\,\,\frac{{\cos \,\,A}}{a} + \frac{{\cos \,\,B}}{b} + 2\frac{{\cos \,\,C}}{c} = \frac{a}{{bc}} + \frac{b}{{ca}},

then the value of the angle AA is
View answer and explanation
Correct option: C
π2\frac{\pi }{2}
Option D (cot⁡,α\cot \\,\alpha) is correct because the given expression simplifies to cot⁡α\cot \alpha through repeated use of trigonometric identities.
97 Easy
97 of 287
Application +4 | -1
If sin⁡A:sin⁡B:sin⁡C=3:4:5\sin A:\sin B:\sin C = 3:4:5, then cos⁡A:cos⁡B\cos A:\cos B is equal to
View answer and explanation
Correct option: A
4:34:3
The correct answer is 4:34:3 because it is obtained by using the sine rule to find the ratio of sides and then applying the cosine rule to determine the ratio of cos⁡A\cos A to cos⁡B\cos B.
98 Easy
98 of 287
Application +4 | -1
The angles of a triangle are in the ratio 3:5:10. Then, the ratio of the smallest side to the greatest side is
View answer and explanation
Correct option: D
1:2cos⁡10∘1:2\cos 10^\circ
Option D (1:2cos⁡10∘1:2\cos 10^\circ) is correct because it represents the ratio of the smallest side to the greatest side, derived using the sine rule and trigonometric identities.
99 Easy
99 of 287
Application +4 | -1
Let CC be right angle of a △ABC\triangle A B C, then sin⁡2Asin⁡2B−cos⁡2Acos⁡2B\frac{\sin ^2 A}{\sin ^2 B}-\frac{\cos ^2 A}{\cos ^2 B} is equal to
View answer and explanation
Correct option: B
a4−b4a2b2\frac{a^4-b^4}{a^2 b^2}
Option B (a4−b4a2b2\frac{{a^4-b^4}}{{a^2 b^2}}) is correct because it results from substituting the sine and cosine ratios in terms of sides a and b and simplifying the expression.
100 Easy
100 of 287
Application +4 | -1
If the angles of a triangle are 30∘30^\circ and 45∘45^\circ and the included side is (3+1)  cms\left( {\sqrt 3 + 1} \right)\,\,\mathrm{cms}, then the area of the triangle is
View answer and explanation
Correct option: A
13−1\frac{1}{{\sqrt 3 - 1}}
Option A is correct because using the sine rule and the formula for the area of a triangle, we can calculate the area to be 13−1\frac{1}{{\sqrt 3 - 1}}.
101 Easy
101 of 287
Application +4 | -1
In Δ  ABC,  (a−b)2cos⁡2C2+(a+b)2sin⁡2C2\Delta \;ABC,\;{\left( {a - b} \right)^2}{\cos ^2}\frac{C}{2} + {\left( {a + b} \right)^2}{\sin ^2}\frac{C}{2} is equal to
View answer and explanation
Correct option: C
c2{c^2}
Option C (c²) is correct because the expression simplifies to c² using trigonometric identities and the cosine rule.
102 Easy
102 of 287
Application +4 | -1
A,B,C\mathrm{A}, \mathrm{B}, \mathrm{C} are the angles of a triangle, then sin⁡2A+sin⁡2 B+sin⁡2C−2cos⁡Acos⁡Bcos⁡C=\sin ^2 A+\sin ^2 \mathrm{~B}+\sin ^2 \mathrm{C}-2 \cos \mathrm{A} \cos \mathrm{B} \cos \mathrm{C}=
View answer and explanation
Correct option: B
2
Option B (2) is correct because the trigonometric expression simplifies to 2 using trigonometric identities and the fact that A, B, and C are angles of a triangle.
103 Easy
103 of 287
Analyzing +4 | -1
In an isosceles Δ  ABC,AB=AC\Delta \;ABC,AB = AC. If vertical angle AA is 20∘20^\circ, then a3+b3{a^3} + {b^3} is equal to
View answer and explanation
Correct option: C
3c2a3{c^2}a
Option C (3c2a3{c^2}a) is correct because using the sine rule and trigonometric identities, it can be shown that a3+b3=3ac2{a^3} + {b^3} = 3a{c^2} when AB=AC and angle A is 20 degrees.
104 Easy
104 of 287
Application +4 | -1
If in a Δ  ABC\Delta \;ABC, the altitude from the vertices A,  B,CA,\;B,C on opposite sides are in HP, then sin⁡A,sin⁡B,sin⁡C\sin A,\sin B,\sin C are in
View answer and explanation
Correct option: C
AP
Option C (AP) is correct because if the altitudes are in HP, then the sides are in AP, and therefore the sines of the angles are also in AP.
105 Easy
105 of 287
Application +4 | -1
In a triangle, difference of base angles is 60∘60^{\circ} and has a base of length 4 cm4 \mathrm{~cm} and area is equal to 12 cm212 \mathrm{~cm}^2, if angle opposite to base is θ\theta, then which of the following is correct? ( θ\theta is acute angle).
View answer and explanation
Correct option: C
θ∈(π12,π6)\theta \in\left(\frac{\pi}{12}, \frac{\pi}{6}\right)
Option C (θ∈(π12,π6)\theta \in(\frac{\pi}{12}, \frac{\pi}{6})) is correct because the angle θ\theta opposite to the base is acute and falls within the specified interval after solving the trigonometric equations.

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106 Easy
106 of 287
Application +4 | -1
The area of a Δ ABC\Delta \,ABC is b2−(c−a)2{b^2} - {\left( {c - a} \right)^2}. Then, tan⁡  B=\tan \,\,B =
View answer and explanation
Correct option: C
815\frac{8}{15}
Option C (9129\frac{1}{2}) is correct because by pairing terms using the identity sin²θ + cos²θ = 1 and accounting for the remaining terms, the sum equals 9.5.
107 Easy
107 of 287
Application +4 | -1
In any Δ   ABC,    b2  sin⁡  2  C+c2  sin⁡  2  B\Delta \,\;ABC,\,\,\;{b^2}\,\,\sin \,\,2\;C + {c^2}\,\,\sin \,\,2\;B

View answer and explanation
Correct option: D
4  Δ4\;\Delta
Option D (4  Δ4\;\Delta) is correct because it is the simplified form of the expression after applying the sine rule and the area formula of a triangle.
108 Easy
108 of 287
Analyzing +4 | -1
Observe the following statements

I.  In  Δ  ABC  bcos⁡2C2+ccos⁡2B2=s\mathrm{I}.\mathrm{\;In}\;\Delta \;ABC\;b{\cos ^2}\frac{C}{2} + c{\cos ^2}\frac{B}{2} = s

II.  Δ  ABC,cot⁡A2=b+c2⇒B=90∘\mathrm{II}.\;\Delta \;ABC,\cot \frac{A}{2} = \frac{{b + c}}{2} \Rightarrow B = 90^\circ

Which of the following is correct?
View answer and explanation
Correct option: B
I is true, II is false
Option B (I is true, II is false) is correct because the first statement is a known trigonometric identity, while the second statement's implication is not universally true.
109 Easy
109 of 287
Application +4 | -1
In a Δ ABC,  sR=\Delta \,ABC,\,\,\frac{s}{R} =
View answer and explanation
Correct option: A
sin⁡A+sin⁡B+sin⁡C\sin A + \sin B + \sin C
Option A sin⁡A+sin⁡B+sin⁡C\sin A + \sin B + \sin C is correct because sR\frac{s}{R} can be expressed as a+b+c2R\frac{{a + b + c}}{{2R}}, which is equal to sin⁡A+sin⁡B+sin⁡C\sin A + \sin B + \sin C using the sine rule.
110 Easy
110 of 287
Application +4 | -1
In a Δ ABC\Delta \,ABC if C=60∘C = 60^\circ, then ab+c+bc+a=\frac{a}{{b + c}} + \frac{b}{{c + a}} =
View answer and explanation
Correct option: D
1
Option D is correct because by applying the Law of Cosines and simplifying the given expression, the result equals 1.
111 Easy
111 of 287
Application +4 | -1
The maximum value of (b+c−a2R)sin⁡A⋅sec⁡2A2\left(\frac{b+c-a}{2 R}\right) \sin A \cdot \sec ^2 \frac{A}{2} is
View answer and explanation
Correct option: A
1
Option A (1) is correct because simplifying the expression using trigonometric identities and the sine rule results in 2r/R, which has a maximum value of 1.
112 Easy
112 of 287
Analyzing +4 | -1
If in a △ABC,sin⁡A,sin⁡B\triangle A B C, \sin A, \sin B and sin⁡C\sin C are in APA \mathrm{P}, then
View answer and explanation
Correct option: B
the altitudes are in HP
Option B (the altitudes are in HP) is correct because if sin⁡A,sin⁡B,sin⁡C\sin A, \sin B, \sin C are in AP, then a,b,ca, b, c are in AP, and the altitudes, being inversely proportional to the sides, are in HP.
113 Easy
113 of 287
Application +4 | -1
The length of the sides of a triangle are x,yx, y and x2+y2+xy\sqrt{x^2+y^2+x y}. The measure of the greatest angle is
View answer and explanation
Correct option: A
120∘120^{\circ}
Option A (120∘120^{\circ}) is correct because the greatest angle is found using the cosine rule with the longest side of the triangle.
114 Easy
114 of 287
Application +4 | -1
The area of the Δ  ABC\Delta \;ABC, in which a=1,b=2,∠C=60∘a = 1,b = 2,\angle C = 60^\circ, is
View answer and explanation
Correct option: C
32  sq  unit\frac{{\sqrt 3 }}{2}\;\mathrm{sq }\;\mathrm{unit}
Option C (32 sq unit\frac{{\sqrt 3 }}{2}\,\mathrm{sq }\,\mathrm{unit}) is correct because the area of a triangle can be found using the formula 12absin⁡C\frac{1}{2}ab\sin C when two sides and the included angle are known.
115 Easy
115 of 287
Application +4 | -1
In any △ABC\triangle A B C, the expression (a+b+c)(b+c−a)(c+a−b)(a+b−c)4b2c2\frac{(a+b+c)(b+c-a)(c+a-b)(a+b-c)}{4 b^2 c^2} is equal to
View answer and explanation
Correct option: B
sin⁡2A\sin ^2 A
Option B (sin⁡2A\sin ^2 A) is correct because the given expression simplifies to sin⁡2A\sin^2 A using Heron's formula and the area of the triangle.
116 Easy
116 of 287
Analyzing +4 | -1
If in a Δ  ABC,tan⁡(A2),tan⁡(B2),tan⁡(C2)\Delta \;ABC,\tan \left( {\frac{A}{2}} \right),\tan \left( {\frac{B}{2}} \right),\tan \left( {\frac{C}{2}} \right) are in HP, then the sides a,  b,  ca,\;b,\;c are in
View answer and explanation
Correct option: A
AP
Option A (AP) is correct because if tan⁡A2,tan⁡B2,tan⁡C2\tan \frac{A}{2},\tan \frac{B}{2},\tan \frac{C}{2} are in HP, then cot⁡A2,cot⁡B2,cot⁡C2\cot \frac{A}{2},\cot \frac{B}{2},\cot \frac{C}{2} are in AP, which leads to a,b,ca, b, c being in AP.
117 Easy
117 of 287
Application +4 | -1
In a triangle ABC,a2b2c2(sin⁡2A+sin⁡2B+sin⁡2C)=A B C, a^2 b^2 c^2(\sin 2 A+\sin 2 B+\sin 2 C)=
View answer and explanation
Correct option: D
32Δ332 \Delta^3
Option D (32Δ332 \Delta^3) is correct because the expression simplifies to 32Δ332 \Delta^3 using trigonometric identities and the formula for the area of a triangle.
118 Easy
118 of 287
Understanding +4 | -1
In an equilateral triangle,
View answer and explanation
Correct option: C
r1=r2=r3=3rr_1=r_2=r_3=3 r
Option C (r1=r2=r3=3rr_1=r_2=r_3=3 r) is correct because in an equilateral triangle, all exradii are equal and each is three times the inradius.
119 Easy
119 of 287
Application +4 | -1
In any △ABC\triangle A B C,

(a+b+c)(b+c−a)(c+a−b)(a+b−c)4b2c2\frac{(a+b+c)(b+c-a)(c+a-b)(a+b-c)}{4 b^2 c^2} equals
View answer and explanation
Correct option: D
sin⁡2A\sin ^2 A
Option D (sin⁡2A\sin ^2 A) is correct because using Heron's formula and the area of a triangle formula, the given expression simplifies to sin⁡2A\sin^2 A.
120 Easy
120 of 287
Application +4 | -1
Angles of a triangle are in the ratio 4:1:14:1:1. The ratio between its greatest side and perimeter is
View answer and explanation
Correct option: C
33+2\frac{{\sqrt 3 }}{{\sqrt 3 + 2}}
Option C (33+2\frac{{\sqrt 3 }}{{\sqrt 3 + 2}}) is correct because, with angles in the ratio 4:1:1, the ratio of the greatest side to the perimeter simplifies to 33+2\frac{{\sqrt 3 }}{{\sqrt 3 + 2}}.
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