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JEE • MATHEMATICS • PROPERTIES OF TRIANGLES

Properties of triangles practice · Easy

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Showing 61-90 of 287 questions

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61 Easy
61 of 287
Analyzing +4 | -1
If in a △ABC,atan⁡A+btan⁡B=(a+b)tan⁡(A+B)2\triangle A B C, a \tan A+b \tan B=(a+b) \tan \frac{(A+B)}{2}, then
View answer and explanation
Correct option: C
A=BA=B
Option C (A=BA=B) is correct because the given equation simplifies to tan⁡A=tan⁡B\tan A = \tan B, implying A=BA = B.
62 Easy
62 of 287
Application +4 | -1
In Δ  ABC,(a+b+c)(tan⁡A2+tan⁡B2)\Delta \;ABC,\left( {a + b + c} \right)\left( {\tan \frac{A}{2} + \tan \frac{B}{2}} \right) is equal to
View answer and explanation
Correct option: A
2c  cot⁡C22c\;\cot \frac{C}{2}
Option A (2c  cot⁡C22c\;\cot \frac{C}{2}) is correct because by using trigonometric identities and the area of a triangle formula, the expression simplifies to this form.
63 Easy
63 of 287
Application +4 | -1
Three poles whose feet lie on a circle subtend angle α,β,γ\alpha, \beta, \gamma respectively at the centre of the circle. If the height of the poles are in A.P. then cot⁡α,cot⁡β,cot⁡γ\cot \alpha, \cot \beta, \cot \gamma are in
View answer and explanation
Correct option: C
H.P.
Option C (H.P.) is correct because if the heights are in A.P., then their cotangents are in H.P. due to the relationship between height, radius, and angle.
64 Easy
64 of 287
Application +4 | -1
If A,  A1,A2,  A3A,\;{A_1},{A_2},\;{A_3} be the areas of the incircle and excircles, then 1A1+1A2+1A3\frac{1}{{\sqrt {{A_1}} }} + \frac{1}{{\sqrt {{A_2}} }} + \frac{1}{{\sqrt {{A_3}} }} is equal to
View answer and explanation
Correct option: A
1A\frac{1}{{\sqrt A }}
Option A 1A\frac{1}{{\sqrt A }} is correct because it is derived from the relationship between the inradius, exradii, and the area of the triangle.
65 Easy
65 of 287
Application +4 | -1
In a △ABC,2casin⁡(A−B+C2)\triangle A B C, 2 c a \sin \left(\frac{A-B+C}{2}\right) is equal to
View answer and explanation
Correct option: B
c2+a2−b2c^2+a^2-b^2
Option B (c2+a2−b2c^2+a^2-b^2) is correct because by simplifying the trigonometric expression and using the Law of Cosines, we can show that 2casin⁡(A−B+C2)=c2+a2−b22 c a \sin \left(\frac{A-B+C}{2}\right) = c^2+a^2-b^2.
66 Easy
66 of 287
Application +4 | -1
If a=22,  b=6,  A=45∘a = 2\sqrt 2 ,\;b = 6,\;A = 45^\circ, then
View answer and explanation
Correct option: A
No triangle is possible
Option A (No triangle is possible) is correct because the calculated value of sin B is greater than 1, which is not possible.
67 Easy
67 of 287
Application +4 | -1
In a ΔABC\Delta ABC, if 2s=a+b+c2s = a + b + c and (s−b)(s−c)=x  sin⁡2A2\left( {s - b} \right)\left( {s - c} \right) = x\;{\sin ^2}\frac{A}{2}, then the value of xx is
View answer and explanation
Correct option: A
bcbc
Option A (bcbc) is correct because by using the formula for sin⁡A2\sin \frac{A}{2} and comparing it with the given equation, we find that x=bcx = bc.
68 Easy
68 of 287
Application +4 | -1
The radii r1,r2,r3r_1, r_2, r_3 of escribed circles of the △ABC\triangle A B C are in HP. If its area is 24sqcm24 \mathrm{sq} \mathrm{cm} and its perimeter is 24 cm24 \mathrm{~cm}, then the lengths of its sides are
View answer and explanation
Correct option: C
6,8,106,8,10
Option C (6,8,10) is correct because these side lengths form an arithmetic progression and satisfy the area and perimeter conditions.
69 Easy
69 of 287
Application +4 | -1
In a triangle ABC,tan⁡A−B2cot⁡A+B2A B C, \tan \frac{A-B}{2} \cot \frac{A+B}{2} is equal to
View answer and explanation
Correct option: C
a−ba+b\frac{a-b}{a+b}
Option C (a−ba+b\frac{a-b}{a+b}) is correct because using Napier's analogy, tan⁡A−B2=a−ba+bcot⁡C2\tan \frac{A-B}{2} = \frac{a-b}{a+b} \cot \frac{C}{2}, and since A+B+C=πA+B+C = \pi, cot⁡A+B2=tan⁡C2\cot \frac{A+B}{2} = \tan \frac{C}{2}.
70 Easy
70 of 287
Understanding +4 | -1
If in Δ   ABC,   a  sin⁡  A=b  sin⁡  B\Delta \,\;ABC,\;\,a\,\,\sin \,\,A = b\,\,\sin \,\,B, then the triangle is

View answer and explanation
Correct option: A
Isosceles
Option A (Isosceles) is correct because if a sin A = b sin B, then by the sine rule, a²k = b²k, implying a = b, which defines an isosceles triangle.
71 Easy
71 of 287
Application +4 | -1
In a Δ ABC,  a=2b\mathrm{\Delta }\,ABC,\,\,a = 2b and A=3BA = 3B, the A=A =
View answer and explanation
Correct option: A
90∘90^\circ
Option A (90∘90^\circ) is correct because by applying the sine rule and trigonometric identities, we find that A = 90 degrees.
72 Easy
72 of 287
Analyzing +4 | -1
In a ΔABC,(a+b+c)(b+c−a)=kbc\Delta ABC,\left( {a + b + c} \right)\left( {b + c - a} \right) = kbc, if
View answer and explanation
Correct option: C
0<k<40 < k < 4
Option C (0<k<40 < k < 4) is correct because it is derived from the trigonometric identity relating the given expression to cos⁡2(A/2)\cos^2(A/2) and the range of cosine squared.
73 Easy
73 of 287
Application +4 | -1
In a △ABC,a=8 cm,b=10 cm,c=12 cm\triangle A B C, a=8 \mathrm{~cm}, b=10 \mathrm{~cm}, c=12 \mathrm{~cm}. Then, relation between angles of the triangle is
View answer and explanation
Correct option: C
C=2AC=2 A
Option C (C=2AC=2A) is correct because by using the cosine rule, we find that cos(C) = cos(2A), which implies C = 2A.
74 Easy
74 of 287
Application +4 | -1
The sides of a triangle are three consecutive natural numbers and its largest angle is twice the smallest one. Then, the sides of the triangle are
View answer and explanation
Correct option: D
4, 5, 6
Option D (4, 5, 6) is correct because by using the sine rule and the given condition C = 2A, we can derive a relationship between the sides and solve for the consecutive natural numbers.
75 Easy
75 of 287
Application +4 | -1
In a Δ ABC\Delta \,ABC if a=26,  b=30a = 26,\,\,b = 30 and cos⁡  C=6365\cos \,\,C = \frac{{63}}{{65}}, then r2={r_2} =
View answer and explanation
Correct option: C
48\mathrm{48}
Option C (48) is correct because by using the cosine rule to find c and then using Heron's formula to find the area, r2r_2 can be calculated as Δs−b\frac{\Delta}{s-b}.

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76 Easy
76 of 287
Application +4 | -1
In a Δ  ABC\Delta \;ABC, if a,  b,  ca,\;b,\;c are in AP, then the value of sin⁡A2sin⁡C2sin⁡B2\frac{{\sin \frac{A}{2}\sin \frac{C}{2}}}{{\sin \frac{B}{2}}} is
View answer and explanation
Correct option: B
12\frac{1}{2}
Option B 12\frac{1}{2} is correct because it is the simplified value of the given expression using the sine rule and the condition that a, b, c are in AP.
77 Easy
77 of 287
Application +4 | -1
In triangle ABC, the value of cot⁡A2cot⁡B2−1cot⁡A2cot⁡B2\frac{{\cot \frac{A}{2}\cot \frac{B}{2} - 1}}{{\cot \frac{A}{2}\cot \frac{B}{2}}} is
View answer and explanation
Correct option: D
2ca+b+c\frac{{2c}}{{a + b + c}}
Option D (2ca+b+c\frac{{2c}}{{a + b + c}}) is correct because simplifying the given expression using trigonometric identities and half-angle formulas leads to the result cs\frac{c}{s}, where ss is the semi-perimeter of the triangle, which is equal to 2ca+b+c\frac{{2c}}{{a + b + c}}.
78 Easy
78 of 287
Application +4 | -1
The angle of a right angled triangle are in A.P. The ratio of the in-radius and the perimeter is
View answer and explanation
Correct option: A
(2−3)  :23\left( {2 - \sqrt 3 } \right)\;:2\sqrt 3
Option A ((2−3) :23\left( {2 - \sqrt 3 } \right)\,:2\sqrt 3) is correct because the ratio of the in-radius and the perimeter of the right-angled triangle with angles in A.P. is (2−3) :23\left( {2 - \sqrt 3 } \right)\,:2\sqrt 3.
79 Easy
79 of 287
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.
80 Easy
80 of 287
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}.
81 Easy
81 of 287
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.
82 Easy
82 of 287
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.
83 Easy
83 of 287
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.
84 Easy
84 of 287
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.
85 Easy
85 of 287
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.
86 Easy
86 of 287
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.
87 Easy
87 of 287
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.
88 Easy
88 of 287
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.
89 Easy
89 of 287
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.
90 Easy
90 of 287
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.
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