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JEE • MATHEMATICS • PROPERTIES OF TRIANGLES

Properties of triangles practice · Mixed difficulty

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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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301 Easy
301 of 434
Application +4 | -1
In Δ  ABC,1+cos⁡(A−B)cos⁡C1+cos⁡(A−C)cos⁡B\Delta \;ABC,\frac{{1 + \cos \left( {A - B} \right)\cos C}}{{1 + \cos \left( {A - C} \right)\cos B}} is equal to
View answer and explanation
Correct option: D
a2+b2a2+c2\frac{{{a^2} + {b^2}}}{{{a^2} + {c^2}}}
Option D (a2+b2a2+c2\frac{{{a^2} + {b^2}}}{{{a^2} + {c^2}}}) is correct because by using trigonometric identities and the sine rule, the given expression can be simplified to a2+b2a2+c2\frac{{{a^2} + {b^2}}}{{{a^2} + {c^2}}}.
302 Medium
302 of 434
Application +4 | -1
In any △ABC\triangle A B C, if C=90∘C=90^{\circ}, then tan⁡B2\tan \frac{B}{2} is equal to
View answer and explanation
Correct option: A
c−ac+a\sqrt{\frac{c-a}{c+a}}
Option A (c−ac+a\sqrt{\frac{c-a}{c+a}}) is correct because it is derived from Napier's analogy and the given condition C=90∘C=90^{\circ}.
303 Easy
303 of 434
Application +4 | -1
In a Δ ABC\Delta \,ABC, if a=8,  b=10a = 8,\,\,b = 10 and c=12c = 12, then CC is equal to
View answer and explanation
Correct option: B
2A2A
Option B (2A2A) is correct because by using the Law of Cosines, we find that cos⁡C=18\cos C = \frac{1}{8} and cos⁡2A=18\cos 2A = \frac{1}{8}, thus C=2AC = 2A.
304 Easy
304 of 434
Application +4 | -1
Let a,ba, b and cc be real numbers such that 4a+2b+c4 a+2 b+c =0=0 and ab>0a b>0. Then the quadratic equation ax2+bxa x^2+b x +c=0+c=0 has
View answer and explanation
Correct option: C
Real roots
Option C (Real roots) is correct because given the condition 4a+2b+c=04a + 2b + c = 0, x=2x=2 is a real root of the quadratic equation ax2+bx+c=0ax^2 + bx + c = 0. Since one root is real, the other must also be real.
305 Easy
305 of 434
Application +4 | -1
If the sides of a triangle are the roots of the equation x3−2x2−x−16=0{x^3} - 2{x^2} - x - 16 = 0, then the product of the in-radius and circum-radius of the triangle is
View answer and explanation
Correct option: C
4\mathrm{4}
Option C (4) is correct because the product of the in-radius and circum-radius is calculated using the relationship between the sides of the triangle and the roots of the given equation.
306 Easy
306 of 434
Application +4 | -1
If A=30∘,a=100,c=1002A=30^{\circ}, a=100, c=100 \sqrt{2}, find the number of triangles that can be formed.
View answer and explanation
Correct option: A
2
Option A (2) is correct because two triangles are possible when a<ca < c and a>csin⁡Aa > c \sin A.
307 Easy
307 of 434
Application +4 | -1
The smallest angle of the triangle whose sides are 6+12, 48,246 + \sqrt {12} ,\,\sqrt {48} ,\sqrt {24} is
View answer and explanation
Correct option: C
π/6\pi /6
Option C (π/6\pi /6) is correct because it represents the smallest angle of the triangle, calculated using the law of cosines with the given side lengths.
308 Easy
308 of 434
Application +4 | -1
If a ΔABC,  2casin⁡A−B+C2\Delta ABC,\;2ca\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 the expression simplifies to 2accos⁡B2ac \cos B, which equals c2+a2−b2{c^2} + {a^2} - {b^2} by the cosine rule.
309 Easy
309 of 434
Application +4 | -1
In a triangle with one angle if 120∘120^\circ, the length of the sides forms an AP. If the length of the greatest sides is 7 cm, then area of triangle is
View answer and explanation
Correct option: B
1534  cm2\frac{{15\sqrt 3 }}{4}\mathrm{\;c}{\mathrm{m}^2}
Option C (6 cm) is correct because the radius of the incircle is calculated as the area of the triangle divided by its semi-perimeter, which gives 6 cm.
310 Easy
310 of 434
Application +4 | -1
If in a Δ  ABC,cot⁡A2=b+ca\Delta \;ABC,\cot \frac{A}{2} = \frac{{b + c}}{a}, then the Δ  ABC\Delta \;ABC is
View answer and explanation
Correct option: C
Right angled
Option C (Right angled) is correct because given cot⁡A2=b+ca\cot \frac{A}{2} = \frac{{b + c}}{a}, it can be shown that angle B is equal to π2\frac{\pi}{2}.
311 Easy
311 of 434
Application +4 | -1
In a Δ ABC,  cos⁡2  A2  +  cos⁡2  B2 + cos⁡2  C2=\Delta \,ABC,\,\,{\cos ^2}\,\,\frac{A}{2}\,\, + \,\,{\cos ^2}\,\,\frac{B}{2}\, + \,{\cos ^2}\,\,\frac{C}{2} =
View answer and explanation
Correct option: C
2+r2R2 + \frac{r}{{2R}}
Option C (2+r2R2 + \frac{r}{{2R}}) is correct because the expression cos⁡2 A2 + cos⁡2 B2 + cos⁡2 C2{\cos ^2}\,\frac{A}{2}\, + \,{\cos ^2}\,\frac{B}{2}\, + \,{\cos ^2}\,\frac{C}{2} simplifies to 2+r2R2 + \frac{r}{{2R}} using trigonometric identities and the relationship between inradius and circumradius.
312 Easy
312 of 434
Application +4 | -1
In a Δ  ABC\Delta \;ABC, if cos⁡Aa=cos⁡Bb=cos⁡Cc\frac{{\cos A}}{a} = \frac{{\cos B}}{b} = \frac{{\cos C}}{c} and the side a=2a = 2, then area of the triangle is
View answer and explanation
Correct option: D
3\sqrt 3 sq unit
Option D (3\sqrt{3}) is correct because the given condition implies that the triangle is equilateral, and the area can be calculated using the side length.
313 Easy
313 of 434
Understanding +4 | -1
If RR is the radius of the circumcircle of the △ABC\triangle A B C and its area is Δ\Delta, then
View answer and explanation
Correct option: C
R=abc4ΔR=\frac{a b c}{4 \Delta}
Option C (R=abc4ΔR=\frac{{a b c}}{{4 \Delta}}) is correct because it is the standard formula relating the circumradius (R) to the sides (a, b, c) and area (Δ\Delta) of a triangle.
314 Easy
314 of 434
Application +4 | -1
The area of the circle and the area of a regular polygon of nn sides and of perimeter equal to that of the circle are in the ratio of
View answer and explanation
Correct option: A
tan⁡  (πn)  :  πn\tan \,\,\left( {\frac{\pi }{n}} \right)\;:\,\,\frac{\pi }{n}
Option A (tan⁡,(πn);:,πn\tan \\,\left( {\frac{\pi }{n}} \right)\\;:\\,\frac{\pi }{n}) is correct because the ratio of the area of the circle to the area of the regular polygon can be expressed in terms of tan⁡(πn)\tan \left( {\frac{\pi }{n}} \right) and πn\frac{\pi }{n}.
315 Easy
315 of 434
Application +4 | -1
If in a △ABC,r1=2,r2=3\triangle A B C, r_1=2, r_2=3 and r3=6r_3=6, then aa equals
View answer and explanation
Correct option: D
3
Option D (3) is correct because it is the value of side 'a' calculated using the given exradii and the properties of triangles.
316 Easy
316 of 434
Application +4 | -1
If sides of a triangle are 18,24,30 cm18,24,30 \mathrm{~cm}, then radius of circumcircle is
View answer and explanation
Correct option: D
None of these
Option D (None of these) is correct because the circumradius is calculated as 15 cm, which is not among the given options.
317 Easy
317 of 434
Understanding +4 | -1
In a ∆ABC, the correct formulae among the following are

III.     r=4Rsin⁡A2sin⁡B2sin⁡C2\;\;r = 4R\sin \frac{A}{2}\sin \frac{B}{2}\sin \frac{C}{2}

IV.     r1=(s−a)tan⁡A2\;\;{r_1} = \left( {s - a} \right)\tan \frac{A}{2}

V.     r3=Δs−c\;\;{r_3} = \frac{\Delta }{{s - c}}
View answer and explanation
Correct option: C
Only I, III
Option C (Only I, III) is correct because these are the standard formulae for inradius and exradius of a triangle.
318 Easy
318 of 434
Application +4 | -1
If, in a Δ ABC,  (a+b+c)  (b+c−a)=λ   bc\Delta \,ABC,\,\,\left( {a + b + c} \right)\,\,\left( {b + c - a} \right) = \lambda \;\,bc, then
View answer and explanation
Correct option: D
0<λ<40 < \lambda < 4
The correct answer is D (0<λ<40 < \lambda < 4) because the given equation can be rewritten in terms of cos⁡2(A2)\cos^2(\frac{A}{2}), which must be between 0 and 1.
319 Medium
319 of 434
Application +4 | -1
In a triangle ABC,2a2+4b2+c2=4ab+2ac\mathrm{ABC}, 2 a^2+4 b^2+c^2=4 a b+2 a c, then cos⁡B\cos \mathrm{B} is equal to
View answer and explanation
Correct option: A
0.875
Option A (0.875) is correct because by manipulating the given equation, we can find the relationship between a, b, and c, and then calculate cos B using the Law of Cosines.
320 Easy
320 of 434
Application +4 | -1
Angles A,BA,B and CC of a triangle are in AP with common difference 15 degree, then angle AA is equal to
View answer and explanation
Correct option: B
60∘60^\circ
Option A (60∘60^\circ) is incorrect. The correct answer is B (60∘60^\circ) because the angles are in AP and their sum is 180 degrees.
321 Easy
321 of 434
Application +4 | -1
In a ΔABC\Delta ABC, if tan⁡A2=56,tan⁡C2=25\tan \frac{A}{2} = \frac{5}{6},\tan \frac{C}{2} = \frac{2}{5}, then
View answer and explanation
Correct option: D
a,b,ca,b,c are in AP
Option A (a,b,ca,b,c are in AP) is correct because given the values of tan⁡A2\tan \frac{A}{2} and tan⁡C2\tan \frac{C}{2}, we can derive that 2b=a+c2b = a + c, which means a,b,ca, b, c are in arithmetic progression.
322 Medium
322 of 434
Analyzing +4 | -1
In Δ  ABC,a2(cos⁡2B−cos⁡2C)+b2(cos⁡2C−cos⁡2A)+c2(cos⁡2A−cos⁡2B)\Delta \;ABC,{a^2}\left( {{{\cos }^2}B - {{\cos }^2}C} \right) + {b^2}\left( {{{\cos }^2}C - {{\cos }^2}A} \right) + {c^2}\left( {{{\cos }^2}A - {{\cos }^2}B} \right) is equal to
View answer and explanation
Correct option: A
0
Option A (0) is correct because by using the sine rule and trigonometric identities, the expression simplifies to zero.
323 Easy
323 of 434
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.
324 Medium
324 of 434
Analyzing +4 | -1
In a Δ ABC\Delta \,ABC, angles A,  B,  CA,\,\,B,\,\,C are in A.P., then
lim⁡A→C  3−4  sin⁡  A  sin⁡C∣ A−C ∣\lim\limits_{A \to C} \,\,\frac{{\sqrt {3 - 4\,\,\sin \,\,A\,\,\sin C} }}{{\left| {\,A - C\,} \right|}} is equal to
View answer and explanation
Correct option: A
1\mathrm{1}
Option A (1) is correct because after applying the A.P. condition and trigonometric identities, the limit simplifies to 1.
325 Easy
325 of 434
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.
326 Easy
326 of 434
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.
327 Easy
327 of 434
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}.
328 Easy
328 of 434
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.
329 Medium
329 of 434
Application +4 | -1
The angles of elevation of a vertical tower standing inside a triangular field at the vertices of the field are each equal to θ\theta. If the length of the sides of the field are 30 m,50 m30 \mathrm{~m}, 50 \mathrm{~m} and 70 m70 \mathrm{~m}, the height of the tower is
View answer and explanation
Correct option: B
(70/3)tan⁡θm(70 / \sqrt{3}) \tan \theta \mathrm{m}
Option B ((70/3)tan⁡θm(70 / \sqrt{3}) \tan \theta \mathrm{m}) is correct because it is derived by finding the circumradius of the triangle and relating it to the height of the tower and the angle of elevation.
330 Easy
330 of 434
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}}).
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