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JEE • MATHEMATICS • PROPERTIES OF TRIANGLES

Properties of triangles practice · Medium

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Showing 121-140 of 140 questions

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121 Medium
121 of 140
Analyzing +4 | -1
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
AP
Option A (AP) is correct because the given condition leads to the relationship a2+c2=2b2{a^2} + {c^2} = 2{b^2}, which is the condition for a2,b2,c2{a^2}, {b^2}, {c^2} to be in arithmetic progression.
122 Medium
122 of 140
Application +4 | -1
In a Δ ABC\Delta \,ABC, if c=2,  A=120∘,  a=6c = 2,\,\,A = 120^\circ ,\,\,a = \sqrt 6, then C=C =
View answer and explanation
Correct option: C
45∘45^\circ
Option C is correct because using the sine rule, we can find the angle C to be 45 degrees.
123 Medium
123 of 140
Analyzing +4 | -1
If PP is a point on the altitude ADA D of the △ABC\triangle A B C such that ∠CDP=B3\angle C D P=\frac{B}{3}, then APA P is equal to
View answer and explanation
Correct option: C
2csin⁡B32 c \sin \frac{B}{3}
Option C (2csin⁡B32 c \sin \frac{B}{3}) is correct because by applying the sine rule in triangle ABP, AP can be expressed as 2csin⁡B32 c \sin \frac{B}{3}.
124 Medium
124 of 140
Application +4 | -1
The area of the segment of a circle of radius aa subtending an angle of 2α2\alpha at the centre is
View answer and explanation
Correct option: D
a2α{a^2}\mathrm{\alpha }
The correct answer is a2α{a^2}\mathrm{\alpha } because the area of the segment is calculated as the area of the sector minus the area of the triangle, which simplifies to a2α{a^2}\mathrm{\alpha }.
125 Medium
125 of 140
Application +4 | -1
In any triangle ABC\mathrm{ABC}, acot⁡A+bcot⁡B+ccot⁡C=\mathrm{a} \cot \mathrm{A}+\mathrm{b} \cot \mathrm{B}+\mathrm{c} \cot \mathrm{C}=
View answer and explanation
Correct option: C
2(r+R)2(\mathrm{r}+\mathrm{R})
Option C (2(r+R)2(r+R)) is correct because, for an equilateral triangle, the expression simplifies to a value that matches the given relationship between inradius and circumradius.
126 Medium
126 of 140
Application +4 | -1
In a△ABC,a=1a \triangle A B C, a=1 and the perimeter is six times the AMA M of the sines of the angles. The measure of ∠A\angle A is
View answer and explanation
Correct option: C
π6\frac{\pi}{6}
Option C (π6\frac{\pi}{6}) is correct because the given condition leads to sin⁡A=12\sin A = \frac{1}{2}, which implies A=π6A = \frac{\pi}{6}.
127 Medium
127 of 140
Application +4 | -1
If the area of a triangle ABCA B C is given by Δ=a2−(b−c)2\Delta=a^2-(b-c)^2, then tan⁡(A2)\tan \left(\frac{A}{2}\right) is equal to
View answer and explanation
Correct option: C
14\frac{1}{4}
Option C (14\frac{1}{4}) is correct because by using the given area formula and Heron's formula, we can derive the value of tan⁡(A2)\tan(\frac{A}{2}).
128 Medium
128 of 140
Application +4 | -1
If the area of △ABC\triangle A B C is a2−(b−c)2a^2-(b-c)^2, then find the value of tan⁡A\tan A.
View answer and explanation
Correct option: A
815\frac{8}{15}
Option A (815\frac{8}{15}) is correct because it is the value of tan⁡A\tan A calculated using the given area of the triangle and trigonometric identities.
129 Medium
129 of 140
Analyzing +4 | -1
For a triangle ABCA B C, the value of cos⁡2A+cos⁡2B+\cos 2 A+\cos 2 B+ cos⁡2C\cos 2 C is least. If its inradius is 3 and incentre is MM, then which of the following is NOT correct?
View answer and explanation
Correct option: C
Area of ΔABC\Delta A B C is 2732\frac{27 \sqrt{3}}{2}
Option C (Area of ΔABC\Delta A B C is 2732\frac{{27 \sqrt{{3}}}}{{2}}) is correct because the area of an equilateral triangle with inradius 3 is 27327\sqrt{3}, not 2732\frac{{27 \sqrt{{3}}}}{{2}}.
130 Medium
130 of 140
Application +4 | -1
In a triangle ABC,2a2+4b2+c2=4ab+2acA B C, 2 a^2+4 b^2+c^2=4 a b+2 a c, then the numerical value of cos⁡B\cos B is equal to
View answer and explanation
Correct option: D
78\frac{7}{8}
Option D (7/8) is correct because it is the result of using the cosine rule and the given equation to find the value of cos B.
131 Medium
131 of 140
Application +4 | -1
In a Δ  ABC\Delta \;ABC, if a=2x,  b=2ya = 2x,\;b = 2y and ∠C=120∘\angle C = 120^\circ, then the area of the triangle is
View answer and explanation
Correct option: B
xy  3xy\;\sqrt 3 sq unit
Option B (xy  3xy\;\sqrt 3 sq unit) is correct because the area of the triangle is calculated using the formula 12absin⁡C\frac{1}{2}ab\sin C with the given values.
132 Medium
132 of 140
Application +4 | -1
If a2+b2+c2=1a^2+b^2+c^2=1, then ab+bc+caa b+b c+c a lies in the interval :
View answer and explanation
Correct option: A
[−12,1]\left[-\frac{1}{2}, 1\right]
Option A ([−12,1]\left[-\frac{1}{2}, 1\right]) is correct because it defines the interval in which the expression ab+bc+caa b+b c+c a lies, given the constraint a2+b2+c2=1a^2+b^2+c^2=1.
133 Medium
133 of 140
Application +4 | -1
The sides of two angles of a triangle are equal to 513\frac{5}{13} and 99101\frac{99}{101}. The cosine of the third angle is
View answer and explanation
Correct option: C
2551313\frac{255}{1313}
Option C (2551313\frac{255}{1313}) is correct because it is the result of applying the cosine subtraction formula after finding the sines and cosines of the given angles.
134 Medium
134 of 140
Application +4 | -1
If the angles of a triangle be in the ratio 1:2:7, then the ratio of its greatest side to the least side is
View answer and explanation
Correct option: C
(5+1):(5−1)\left( {\sqrt 5 + 1} \right):\left( {\sqrt 5 - 1} \right)
Option C ((5+1):(5−1)\left( {\sqrt 5 + 1} \right):\left( {\sqrt 5 - 1} \right)) is correct because it represents the ratio of the sines of the greatest and least angles, which corresponds to the ratio of the greatest and least sides.
135 Medium
135 of 140
Application +4 | -1
If in a △ABC,CD\triangle A B C, C D is the angle bisector of the ∠ABC\angle A B C, then CDC D is equal to
View answer and explanation
Correct option: C
2aba+bcos⁡C2\frac{2 a b}{a+b} \cos \frac{C}{2}
Option C (2aba+bcos⁡C2\frac{2ab}{a+b} \cos \frac{C}{2}) is correct because the length of the angle bisector can be expressed in terms of the sides and the angle.

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SEQUENCE OF SERIES 1764 MCQs DEFINITE INTEGRATION 1443 MCQs MATRICES 1363 MCQs INDEFINITE INTEGRATION 1179 MCQs PROBABILITY 1146 MCQs COMPLEX NUMBERS 980 MCQs
136 Medium
136 of 140
Application +4 | -1
In a △ABC\triangle \mathrm{ABC} if ∠A=60∘,bc=3+12\angle \mathrm{A}=60^{\circ}, \frac{\mathrm{b}}{\mathrm{c}}=\frac{\sqrt{3}+1}{2} then ∠B−∠C\angle \mathrm{B}-\angle \mathrm{C} has value equal to
View answer and explanation
Correct option: B
30∘30^{\circ}
Option B (30∘30^{\circ}) is correct because it is the result of applying the tangent half-angle formula and simplifying.
137 Medium
137 of 140
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: B
1\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).
138 Medium
138 of 140
Application +4 | -1
In a triangle, if r1+r3=kcos2B2{r_1} + {r_3} = kco{s^2}\frac{B}{2}, then kk is equal to
View answer and explanation
Correct option: D
4R4R
Option D (4R4R) is correct because, after simplification using trigonometric identities and the area of a triangle formula, k is found to be equal to 4R.
139 Medium
139 of 140
Application +4 | -1
In a △ABC,A=2π3,b−c=33 cm\triangle A B C, A=\frac{2 \pi}{3}, b-c=3 \sqrt{3} \mathrm{~cm} and ar⁡(△ABC)=932 cm2\operatorname{ar}(\triangle A B C)=\frac{9 \sqrt{3}}{2} \mathrm{~cm}^2. Then, aa is
View answer and explanation
Correct option: B
9 cm9 \mathrm{~cm}
Option B (9 cm9 \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.
140 Medium
140 of 140
Application +4 | -1
The sides of a triangle are 8 cm8 \mathrm{~cm}, 10 cm10 \mathrm{~cm} and 12 cm12 \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.
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