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JEE • MATHEMATICS • PROPERTIES OF TRIANGLES

Properties of triangles practice · Easy

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These sets use single-correct-answer MCQs. They do not cover every JEE Main or JEE Advanced question format and are not full-length mock papers.

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Showing 211-240 of 287 questions

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211 Easy
211 of 287
Application +4 | -1
In a Δ ABC,  a=5,  b=4\Delta \,ABC,\,\,a = 5,\,\,b = 4 and cos⁡(A−B)=3132\cos \left( {A - B} \right) = \frac{{31}}{{32}}, then side cc is
View answer and explanation
Correct option: A
6\mathrm{6}
Option B (3\sqrt 3) is correct because simplifying the given trigonometric expression using identities results in 3\sqrt 3.
212 Easy
212 of 287
Application +4 | -1
In a Δ  ABC,∠C=60∘\Delta \;ABC,\angle \mathrm{ }C = 60^\circ then 1a+c+1b+c\frac{1}{{a + c}} + \frac{1}{{b + c}} is equal to
View answer and explanation
Correct option: C
3a+b+c\frac{3}{{a + b + c}}
Option C (3a+b+c\frac{3}{{a + b + c}}) is correct because by using the cosine rule and algebraic manipulation, the given expression simplifies to this form.
213 Easy
213 of 287
Application +4 | -1
In △ABC\triangle \mathrm{ABC}, if a:b:c=1:3:2\mathrm{a}: \mathrm{b}: \mathrm{c}=1: \sqrt{3}: 2, then the ratio A:B:C\mathrm{A}: \mathrm{B}: \mathrm{C} is equal to
View answer and explanation
Correct option: C
1:2:31: 2: 3
Option C (1:2:31: 2: 3) is correct because the ratio of sides corresponds to the ratio of sines of angles, allowing us to determine the angles and their ratio.
214 Easy
214 of 287
Analyzing +4 | -1
If the area of a Δ  ABC\Delta \;ABC is given by Δ=a2−(b−c)2\Delta = {a^2} - {\left( {b - c} \right)^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 simplifying the given area equation and using Heron's formula, we can find that tan⁡A2=14\tan \frac{A}{2} = \frac{1}{4}.
215 Easy
215 of 287
Analyzing +4 | -1
In any △ABC\triangle A B C, the simplified form of cos⁡2Aa2−cos⁡2Bb2\frac{\cos 2 A}{a^2}-\frac{\cos 2 B}{b^2} is
View answer and explanation
Correct option: C
1a2−1b2\frac{1}{a^2}-\frac{1}{b^2}
Option C (1a2−1b2\frac{1}{a^2}-\frac{1}{b^2}) is correct because the expression simplifies to this form using trigonometric identities and the sine rule.
216 Easy
216 of 287
Application +4 | -1
If in the Δ  ABC,  ∠B=45∘\Delta \;ABC,\;\angle B = 45^\circ, then a4+b4+c4{a^4} + {b^4} + {c^4} is equal to
View answer and explanation
Correct option: C
2b2(a2+c2)2{b^2}\left( {{a^2} + {c^2}} \right)
Option C (2b2(a2+c2)2{b^2}({{a^2} + {c^2}})) is correct because by applying the cosine rule with ∠B=45∘\angle B = 45^\circ, we can derive the relationship a4+b4+c4=2b2(a2+c2){a^4} + {b^4} + {c^4} = 2{b^2}({{a^2} + {c^2}}).
217 Easy
217 of 287
Application +4 | -1
The angle of depression of a point situated at a distance of 70 m70 \mathrm{~m} from the base of a tower is 45∘45^{\circ}. The height of the tower is
View answer and explanation
Correct option: A
70 m70 \mathrm{~m}
Option A (70 m70 \mathrm{~m}) is correct because the height of the tower is equal to the distance from the base when the angle of depression is 45 degrees.
218 Easy
218 of 287
Application +4 | -1
The perimeter of a Δ  ABC\Delta \;ABC is 6 times the arithmetic mean of the sine ratios of its angles. If a=1a = 1, then AA is equal to
View answer and explanation
Correct option: A
π6\frac{\pi }{6}
Option A (π6\frac{\pi }{6}) is correct because by using the given condition and the sine rule, we find that angle A is π6\frac{\pi }{6}.
219 Easy
219 of 287
Application +4 | -1
The value of 1r12+1r22+1r32+1r2\frac{1}{{r_1^2}} + \frac{1}{{r_2^2}} + \frac{1}{{r_3^2}} + \frac{1}{{{r^2}}} is
View answer and explanation
Correct option: B
a2+b2+c2Δ2\frac{{{a^2} + {b^2} + {c^2}}}{{{\Delta ^2}}}
Option B (a2+b2+c2Δ2\frac{{{a^2} + {b^2} + {c^2}}}{{{\Delta ^2}}}) is correct because by substituting the formulas for rr, r1r_1, r2r_2, and r3r_3 in terms of area and semi-perimeter, and simplifying, we arrive at the given expression.
220 Easy
220 of 287
Application +4 | -1
In a Δ  ABC,  a=5,  b=7\Delta\; ABC,\;a = 5,\;b = 7 and sin⁡A=34\sin A = \frac{3}{4}, then the number of possible triangles are
View answer and explanation
Correct option: B
0
Option B (0) is correct because the value of sin B is greater than 1, which is not possible, indicating no such triangle exists.
221 Easy
221 of 287
Understanding +4 | -1
In a Δ  ABC,  cosec  A(sin⁡Bcos⁡C+cos⁡Bsin⁡C)\Delta \;ABC,\;\mathrm{cosec}\;A\left( {\sin B\cos C + \cos B\sin C} \right) is equal to
View answer and explanation
Correct option: C
1
Option C (1) is correct because the expression simplifies to sin(B+C)/sin(A), and since A+B+C = 180, sin(B+C) = sin(A).
222 Easy
222 of 287
Application +4 | -1
If AD,  BEAD,\,\,BE and CFCF are the medians of a Δ ABC\Delta \,ABC, then (AD2+BE2+CF2)  :  (BC2+CA2+AB2)\left( {A{D^2} + B{E^2} + C{F^2}} \right)\;:\,\,\left( {B{C^2} + C{A^2} + A{B^2}} \right) is equal to
View answer and explanation
Correct option: C
3  :43\;:4
Option C (3:43:4) is correct because the ratio of the sum of squares of medians to the sum of squares of sides of a triangle is always 3:43:4.
223 Easy
223 of 287
Application +4 | -1
In an obtuse angle triangle ABC,a=3+1,B=π3A B C, a=\sqrt{3}+1, B=\frac{\pi}{3} and R=2R=\sqrt{2}, then angle CC is
View answer and explanation
Correct option: D
None of these
None of these is correct because the calculated angle C is π12\frac{\pi}{12}, which is not among the given options.
224 Easy
224 of 287
Application +4 | -1
In Δ  ABC,∠A=π2,b=4,c=3\Delta \;ABC,\angle A = \frac{\pi }{2},b = 4,c = 3, then the value of Rr\frac{R}{r} is equal to
View answer and explanation
Correct option: A
52\frac{5}{2}
Option A (5/2) is correct because it is the result of calculating R/r using the given triangle properties and trigonometric relationships.
225 Easy
225 of 287
Application +4 | -1
In a Δ  ABC,a(cos⁡2B+cos⁡2C)+cos⁡A(ccos⁡C+bcos⁡B)\Delta \;ABC,a\left( {{{\cos }^2}B + {{\cos }^2}C} \right) + \cos A\left( {c\cos C + b\cos B} \right) is equal to
View answer and explanation
Correct option: A
aa
Option A (aa) is correct because the given expression simplifies to aa using trigonometric identities and projection formulas.
226 Easy
226 of 287
Application +4 | -1
If r1,r2,r3r_1, r_2, r_3 are ex-radii of the encircles of triangle ABCA B C then (s−a)r1+(s−b)r2+(s−c)r3(s-a) r_1+(s-b) r_2+(s-c) r_3 is
View answer and explanation
Correct option: B
3rs3 r s
Option B (3rs) is correct because the sum of the given expression simplifies to 3 times the area of the triangle, which is 3rs.
227 Easy
227 of 287
Analyzing +4 | -1
In any △ABC,1r12+1r22+1r32+1r2\triangle A B C, \frac{1}{r_1^2}+\frac{1}{r_2^2}+\frac{1}{r_3^2}+\frac{1}{r^2} is equal to
View answer and explanation
Correct option: B
a2+b2+c2Δ2\frac{a^2+b^2+c^2}{\Delta^2}
Option B (a2+b2+c2Δ2\frac{a^2+b^2+c^2}{\Delta^2}) is correct because the given expression simplifies to this form using the properties of exradii and inradius.
228 Easy
228 of 287
Application +4 | -1
If in Δ  ABC,sin⁡A2sin⁡C2=sin⁡B2\Delta \;ABC,\sin \frac{A}{2}\sin \frac{C}{2} = \sin \frac{B}{2} and 2s2s is the perimeter of the triangle, then s is
View answer and explanation
Correct option: A
2b2b
Option A (2b2b) is correct because, given sin⁡A2sin⁡C2=sin⁡B2\sin \frac{A}{2}\sin \frac{C}{2} = \sin \frac{B}{2}, it can be shown that s=2bs = 2b using trigonometric identities and the definition of ss as the semi-perimeter.
229 Easy
229 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: D
a2b2a^2 b^2
Option D (a2b2a^2 b^2) is correct because the area of a right-angled triangle with sides a and b is 12ab\frac{1}{2}ab, and 4(12ab)2=a2b24(\frac{1}{2}ab)^2 = a^2b^2.
230 Easy
230 of 287
Application +4 | -1
If in a Δ  ABC  a=5,  b=4,  A=π2+B\Delta \;ABC\;a = 5,\;b = 4,\;A = \frac{\pi }{2} + B, then CC
View answer and explanation
Correct option: B
is  tan⁡−1(940)\mathrm{is\;}{\tan ^{ - 1}}\left( {\frac{9}{{40}}} \right)
Option B is  tan⁡−1(940)\mathrm{is\;}{\tan ^{ - 1}}\left( {\frac{9}{{40}}} \right) is correct because it is found by using the sine rule and trigonometric identities to find angle C.
231 Easy
231 of 287
Application +4 | -1
If in a Δ ABC,  3  sin⁡A=6  sin⁡B=2 3  sin⁡C\Delta \,ABC,\,\,3\,\,\sin A = 6\,\,\sin B = 2\,\sqrt 3 \,\,\sin C then angle AA is
View answer and explanation
Correct option: D
90∘90^\circ
Option D (90∘90^\circ) is correct because the given ratios of sines imply that the sides of the triangle are in the ratio 2:1:3\sqrt{3}, which corresponds to a 30-60-90 triangle, making angle A the right angle.
232 Easy
232 of 287
Application +4 | -1
If the area of a triangle ABCABC is Δ\Delta, then a2sin⁡2B+b2sin⁡2A{a^2}\sin 2B + {b^2}\sin 2A is equal to
View answer and explanation
Correct option: C
4Δ4\Delta
Option C (4Δ4\Delta) is correct because it is derived from the area of a triangle and trigonometric identities.
233 Easy
233 of 287
Analyzing +4 | -1
Let p,qp, q and rr be the sides opposite to the angles P,QP, Q and RR, respectively in a △PQR\triangle P Q R. If r2sin⁡Psin⁡Q=pqr^2 \sin P \sin Q=p q, then the triangle is
View answer and explanation
Correct option: D
right angled
Option D (right angled) is correct because the given condition simplifies to sin⁡2R=1\sin^2 R = 1, implying R=90∘R = 90^{\circ}.
234 Easy
234 of 287
Analyzing +4 | -1
a3cos⁡(B−C)+b3cos⁡(C−A)+c3cos⁡(A−B)a^3 \cos (B-C)+b^3 \cos (C-A)+c^3 \cos (A-B) is equal to
View answer and explanation
Correct option: A
3abc3 a b c
Option A (3abc3abc) is correct because the expression simplifies to 3abc3abc using trigonometric identities and the sine rule.
235 Easy
235 of 287
Application +4 | -1
In a ΔABC,let∠C=π2\Delta ABC,\mathrm{let}\angle C = \frac{\pi }{2}, if rr is the inradius and RR is the circumradius of the Δ  ABC\Delta \;ABC, then 2(r+R)2\left( {r + R} \right) equals
View answer and explanation
Correct option: C
a+ba + b
Option C a+ba + b is correct because it is derived by using the properties of a right-angled triangle and the formulas for inradius and circumradius.
236 Easy
236 of 287
Application +4 | -1
In a triangle, the lengths of the two larger sides are 10 cm and 9 cm respectively. If the angles of the triangle are in AP, then the length of the third side in cm can be
View answer and explanation
Correct option: C
5−65 - \sqrt 6 or 5+65 + \sqrt 6
Option C (5−65 - \sqrt 6 or 5+65 + \sqrt 6) is correct because both values satisfy the conditions derived from the cosine rule and the fact that the angles are in A.P.
237 Easy
237 of 287
Application +4 | -1
The angle of elevation of the top of a vertical tower from a point A\mathrm{A}, due east of it is 45∘45^{\circ}. The angle of elevation of the top of the same tower from a point BB, due south of AA is 30∘30^{\circ}. If the distance between AA and BB is 542 m54 \sqrt{2} \mathrm{~m}, then the height of the tower (in metres), is :
View answer and explanation
Correct option: D
54
Option D (54) is correct because using trigonometric relationships and the given distances, the height of the tower is found to be 54 meters.
238 Easy
238 of 287
Application +4 | -1
If the sides of a triangle are in ratio 3:7:8, then R:rR:r is equal to
View answer and explanation
Correct option: B
7:27:2
Option B (7:27:2) is correct because the ratio of circumradius (R) to inradius (r) is calculated using the formulas R=abc4ΔR = \frac{{abc}}{{4\Delta}} and r=Δsr = \frac{\Delta}{s}, where Δ\Delta is the area and s is the semi-perimeter.
239 Easy
239 of 287
Analyzing +4 | -1
In the ambiguous case, given a,  ba,\,\,b and AA. The difference between the two values of CC is
View answer and explanation
Correct option: C
2 a2−b2  sin⁡2  A2\,\sqrt {{a^2} - {b^2}\,\,{{\sin }^2}\,\,A}
Option C (2,a2−b2 sin⁡2 A2\\,\sqrt {{{{a^2} - {b^2}\,{{\sin }^2}\,A}}}) is correct because it represents the difference between the two possible values of side c in the ambiguous case of triangle solving, derived using the cosine rule and properties of quadratic equations.
240 Easy
240 of 287
Application +4 | -1
In a △ABC,a=2b\triangle A B C, a=2 b and ∣A−B∣=π3|A-B|=\frac{\pi}{3}. The measure of ∠C\angle C is
View answer and explanation
Correct option: B
π3\frac{\pi}{3}
Option B (π3\frac{\pi}{3}) is correct because using the tangent half-angle formula and the given conditions, the angle C is found to be π3\frac{\pi}{3}.
Revision notes and key takeaways

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