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JEE • MATHEMATICS • PROPERTIES OF TRIANGLES

Properties of triangles practice · Mixed difficulty

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121 Medium
121 of 434
Application +4 | -1
In △ABC,(a−b)2cos⁡2C2+(a+b)2sin⁡2C2\triangle A B C,(a-b)^2 \cos ^2 \frac{C}{2}+(a+b)^2 \sin ^2 \frac{C}{2} equal to
View answer and explanation
Correct option: C
c2c^2
Option C (c2c^2) is correct because by expanding the given expression and using the cosine rule, it simplifies to c2c^2.
122 Medium
122 of 434
Application +4 | -1
In triangle ABCA B C if A:B:C=1:2:4A: B: C=1: 2: 4, then (a2−b2)(b2−c2)(c2−a2)=λa2b2c2\left(a^2-b^2\right)\left(b^2-c^2\right)\left(c^2-a^2\right)=\lambda a^2 b^2 c^2, where λ=\lambda= (where notations have their usual meaning)
View answer and explanation
Correct option: A
1
Option A (1) is correct because by substituting the angles and using trigonometric identities, the expression simplifies to 1.
123 Easy
123 of 434
Application +4 | -1
If c2=a2+b2c^2=a^2+b^2, then 4s(s−a)(s−b)(s−c)4 s(s-a)(s-b)(s-c) is equal to
View answer and explanation
Correct option: A
a2b2a^2 b^2
Option A (a2b2a^2 b^2) is correct because given c2=a2+b2c^2 = a^2 + b^2, the triangle is right-angled, and the area can be used to derive the expression.
124 Medium
124 of 434
Application +4 | -1
If sin⁡A+cos  B=a\sin A + cos\;B = a and sin⁡B+cos⁡A=b\sin B + \cos A = b, then sin⁡(A+B)\sin \left( {A + B} \right) is equal to
View answer and explanation
Correct option: C
a2+b2−22\frac{{{a^2} + {b^2} - 2}}{2}
Option C (a2+b2−22\frac{{{a^2} + {b^2} - 2}}{2}) is correct because squaring and adding the given equations allows us to express sin⁡(A+B)\sin(A+B) in terms of aa and bb.
125 Medium
125 of 434
Application +4 | -1
If in a ΔABC\Delta ABC, the sides ABAB and ACAC are perpendicular, then the true equation is
View answer and explanation
Correct option: D
tan⁡Btan⁡C=1\tan B\tan C = 1
Option D (tan⁡Btan⁡C=1\tan B\tan C = 1) is correct because with A = 90°, B + C = 90°, making B and C complementary angles, and the product of their tangents is 1.
126 Medium
126 of 434
Application +4 | -1
If in a △ABC,∠B=90∘\triangle A B C, \angle B=90^{\circ}, then tan⁡2(A2)\tan ^2\left(\frac{A}{2}\right) is equal to
View answer and explanation
Correct option: A
b−cb+c\frac{b-c}{b+c}
Option A (b−cb+c\frac{b-c}{b+c}) is correct because using trigonometric identities and the given right angle, the expression simplifies to this form.
127 Easy
127 of 434
Application +4 | -1
In a triangle ABCABC, if a=2,B−60∘a = 2,B - 60^\circ and C=75∘C = 75^\circ, then b=b =
View answer and explanation
Correct option: B
6\sqrt 6
Option B (6\sqrt{6}) is correct because using the sine rule, we can find the value of b as 6\sqrt{6}.
128 Easy
128 of 434
Application +4 | -1
In a Δ  ABC,2acsin⁡A−B+C2\Delta \;ABC,2ac\sin \frac{{A - B + C}}{2} is equal to
View answer and explanation
Correct option: B
c2+a2−b2{c^2} + {a^2} - {b^2}
Option B (c2+a2−b2{c^2} + {a^2} - {b^2}) is correct because it is derived from applying trigonometric identities and the cosine rule in the given triangle.
129 Easy
129 of 434
Application +4 | -1
In a triangle, if b=20,  c=21b = 20,\;c = 21 and sin⁡A=35\sin A = \frac{3}{5}, then aa is equal to
View answer and explanation
Correct option: B
13
Option B (13) is correct because it is the value of side 'a' calculated using the Law of Cosines with the given values of sides 'b', 'c', and angle A.
130 Easy
130 of 434
Application +4 | -1
If a triangle is right angled at BB, then the diameter of the incircle of the triangle is
View answer and explanation
Correct option: A
c+a−bc + a - b
Option A (c+a−bc + a - b) is correct because it represents the diameter of the incircle in a right-angled triangle, derived from the inradius formula.
131 Medium
131 of 434
Application +4 | -1
In a Δ ABC,   B=π8\Delta \,ABC,\,\;B = \frac{\pi }{8} and C=5  π8C = \frac{{5\;\pi }}{8}. The altitude from AA to the side BCBC, is

View answer and explanation
Correct option: A
a2\frac{a}{2}
Option A (a2\frac{a}{2}) is correct because using the sine rule and area of triangle formula, the altitude from A to BC is found to be a2\frac{a}{2}.
132 Easy
132 of 434
Application +4 | -1
In a △ABC\triangle A B C, the sides a,ba, b and cc are in AP.Then, (tan⁡A2+tan⁡C2):cot⁡B2\left(\tan \frac{A}{2}+\tan \frac{C}{2}\right): \cot \frac{B}{2} is equal to
View answer and explanation
Correct option: D
None of these
Option D (None of these) is correct because the calculated ratio is 2/3, which is not among the given options.
133 Easy
133 of 434
Application +4 | -1
In Δ  ABC,(b−c)sin⁡A+(c−a)sin⁡B+(a−b)sin⁡C\Delta \;ABC,\left( {b - c} \right)\sin A + \left( {c - a} \right)\sin B + \left( {a - b} \right)\sin C is equal to
View answer and explanation
Correct option: C
0
Option C (0) is correct because, using the sine rule, the expression simplifies to 0.
134 Easy
134 of 434
Application +4 | -1
In △ABC\triangle A B C, if c2+a2=2b2c^2+a^2=2 b^2, then value of cot⁡Bcot⁡C+cot⁡A=\frac{\cot B}{\cot C+\cot A}=
View answer and explanation
Correct option: B
12\frac{1}{2}
Option B (12\frac{1}{2}) is correct because it simplifies to 12\frac{1}{2} after applying trigonometric identities and the given condition c2+a2=2b2c^2 + a^2 = 2b^2.
135 Medium
135 of 434
Application +4 | -1
In a Δ ABC\Delta \,ABC, if the diameter of the incircle is a+c−ba + c - b, then ∠ B=\angle \,B =
View answer and explanation
Correct option: C
π2\frac{\pi }{2}
Option C (π2\frac{\pi}{2}) is correct because using the given condition 2r=a+c−b2r = a + c - b and relating it to the area of the triangle, we can deduce that angle B is 90 degrees.

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136 Easy
136 of 434
Application +4 | -1
If AA is the area and 2s2s the sum of three sides of a triangle, then
View answer and explanation
Correct option: A
A≤s233A \le \frac{{{s^2}}}{{3\sqrt 3 }}
Option A (A≤s233A \le \frac{{{s^2}}}{{3\sqrt 3 }}) is correct because it uses AM-GM inequality to find the maximum area of the triangle in terms of its semi-perimeter.
137 Easy
137 of 434
Application +4 | -1
In ΔABC,a≥b≥c\Delta A B C, a \geq b \geq c. If a3+b3+c3sin⁡3A+sin⁡3B+sin⁡3C=7\frac{a^3+b^3+c^3}{\sin ^3 A+\sin ^3 B+\sin ^3 C}=7, then the maximum possible value of aa is
View answer and explanation
Correct option: C
73\sqrt[3]{7}
Option C (73\sqrt[3]{7}) is correct because it represents the maximum possible value of side 'a' in the triangle, which cannot exceed the diameter of the circumscribed circle.
138 Easy
138 of 434
Application +4 | -1
In a Δ ABC\Delta \,ABC if a=13,  b=14a = 13,\,\,b = 14 and c=15c = 15, then reciprocals of r1,  r2{r_1},\,\,{r_2} and r3{r_3} are in the ratio
View answer and explanation
Correct option: C
8  :7  :68\;:7\;:6
Option B (nπ,   nπ+π4n\pi ,\;\,n\pi + \frac{\pi }{4}) is correct because it is the general solution of the trigonometric equation derived from the given condition.
139 Easy
139 of 434
Analyzing +4 | -1
In a triangle (1−r1r2)(1−r1r3)=2\left( {1 - \frac{{{r_1}}}{{{r_2}}}} \right)\left( {1 - \frac{{{r_1}}}{{{r_3}}}} \right) = 2, then the triangle is
View answer and explanation
Correct option: A
Right angled
Option A (Right angled) is correct because the given condition leads to the Pythagorean theorem, indicating a right-angled triangle.
140 Medium
140 of 434
Understanding +4 | -1
In a right angled triangle, RR is equal to
View answer and explanation
Correct option: B
s−r2\frac{s-r}{2}
Option C (s−rs-r) is correct because in a right-angled triangle, the circumradius RR is related to the semi-perimeter ss and inradius rr by the formula R=s−r2R = \frac{{s-r}}{{2}}, which simplifies to R=s−rR = s - r when A=90.
141 Easy
141 of 434
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.
142 Easy
142 of 434
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.
143 Easy
143 of 434
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.
144 Easy
144 of 434
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.
145 Medium
145 of 434
Application +4 | -1
In a triangle ABC\mathrm{ABC} if cot⁡A2cot⁡B2=c\cot ^{\frac{\mathrm{A}}{2}} \cot ^{\frac{\mathrm{B}}{2}}=\mathrm{c}, cot⁡B2cot⁡C2=a\cot ^{\frac{\mathrm{B}}{2}} \cot ^{\frac{\mathrm{C}}{2}}=\mathrm{a} and cot⁡C2cot⁡A2=b\cot \frac{C}{2} \cot ^{\frac{A}{2}}=b, then 1s−a+1s−b+1s−c=\frac{1}{s-a}+\frac{1}{s-b}+\frac{1}{s-c}=
View answer and explanation
Correct option: D
2
Option D (2) is correct because the given cotangent relationships lead to the simplification of the expression to 2.
146 Easy
146 of 434
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.
147 Easy
147 of 434
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.
148 Easy
148 of 434
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.
149 Medium
149 of 434
Analyzing +4 | -1
The sides of a △ABC\triangle A B C are in AP\mathrm{AP}, if the angles AA and CC are the greatest and the smallest angles respectively, then 4(1−cos⁡A)(1−cos⁡C)4(1-\cos A)(1-\cos C) is equal to
View answer and explanation
Correct option: B
4cos⁡2A+C24 \cos ^2 \frac{A+C}{2}
Option B (4cos⁡2A+C24 \cos ^2 \frac{{A+C}}{{2}}) is correct because by using the properties of arithmetic progressions and trigonometric identities, we can show that 4(1−cos⁡A)(1−cos⁡C)4(1-\cos A)(1-\cos C) simplifies to 4cos⁡2A+C24 \cos ^2 \frac{{A+C}}{{2}}.
150 Easy
150 of 434
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.
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