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JEE • MATHEMATICS • PROPERTIES OF TRIANGLES

Properties of triangles practice · Easy

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Showing 151-180 of 287 questions

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151 Easy
151 of 287
Application +4 | -1
In an ambiguous case of solving a triangle when a=5,  b=2,  ∠A=π6a = \sqrt 5 ,\;b = 2,\;\angle A = \frac{\pi }{6}, the two possible values of third side are c1{c_1} and c2{c_2}, then
View answer and explanation
Correct option: C
∣c1−c2∣=4\left| {{c_1} - {c_2}} \right| = 4
Option C (|c1 - c2| = 4) is correct because it is the difference between the two possible values of the third side, calculated using the cosine rule and solving the quadratic equation.
152 Easy
152 of 287
Application +4 | -1
If in a Δ  ABC,  2b2=a2+c2\Delta \;ABC,\;2{b^2} = {a^2} + {c^2}, then sin⁡3Bsin⁡B\frac{{\sin 3B}}{{\sin B}} is equal to
View answer and explanation
Correct option: D
(c2−a22ca)2{\left( {\frac{{{c^2} - {a^2}}}{{2ca}}} \right)^2}
Option D, (c2−a22ca)2{\left( {\frac{{{c^2} - {a^2}}}{{2ca}}} \right)^2}, is the correct answer because it is derived by using the trigonometric identity for sin⁡3B\sin 3B and the given condition 2b2=a2+c22b^2 = a^2 + c^2 to simplify the expression.
153 Easy
153 of 287
Application +4 | -1
In a right-angled △ABC,sin⁡2A+sin⁡2B+sin⁡2C\triangle A B C, \sin ^2 A+\sin ^2 B+\sin ^2 C is equal to
View answer and explanation
Correct option: D
None of these
The correct answer is None of these because in a right-angled triangle, sin⁡2A+sin⁡2B+sin⁡2C=2\sin^2 A + \sin^2 B + \sin^2 C = 2 when C=90∘C = 90^{\circ}.
154 Easy
154 of 287
Application +4 | -1
In a ΔABC,2a2+4b2+c2=4ab+2ac\Delta ABC,2{a^2} + 4{b^2} + {c^2} = 4ab + 2ac, then cos⁡B\cos B is equal to
View answer and explanation
Correct option: D
78\frac{7}{8}
Option D (78\frac{7}{8}) is correct because by rearranging the given equation and using the cosine rule, we can find the value of cos B.
155 Easy
155 of 287
Application +4 | -1
If in a Δ ABC,   3a=b+c\Delta \,ABC,\,\;3a = b + c, then the value of cot⁡  B2  cot⁡  C2\cot \,\,\frac{B}{2}\,\,\cot \,\,\frac{C}{2} is
View answer and explanation
Correct option: C
2
Option C (2) is correct because the value of cot⁡B2cot⁡C2\cot \frac{B}{2} \cot \frac{C}{2} simplifies to 2 when 3a=b+c3a = b + c is used in the formula.
156 Easy
156 of 287
Application +4 | -1
In a triangle ABCa=6,b=3&cos⁡(A−B)=4/5A B C a=6, b=3 \& \cos (A-B)=4 / 5. Find area of Δ\Delta.
View answer and explanation
Correct option: A
9
Option A (9) is correct because using the given information and trigonometric identities, we can find that the triangle is right-angled, and then calculate the area.
157 Easy
157 of 287
Analyzing +4 | -1
In a Δ ABC,  cot⁡  A2+cot⁡  B2+cot⁡  C2cot⁡  A+cot⁡  B+cot⁡  C=\Delta \,ABC,\,\,\frac{{\cot \,\,\frac{A}{2} + \cot \,\,\frac{B}{2} + \cot \,\,\frac{C}{2}}}{{\cot \,\,A + \cot \,\,B + \cot \,\,C}} =
View answer and explanation
Correct option: A
(a+b+c)2a2+b2+c2\frac{{{{\left( {a + b + c} \right)}^2}}}{{{a^2} + {b^2} + {c^2}}}
The correct answer is A ((a+b+c)2a2+b2+c2\frac{{(a + b + c)^2}}{{a^2 + b^2 + c^2}}) because using trigonometric identities and the properties of triangles, the given expression simplifies to this ratio.
158 Easy
158 of 287
Application +4 | -1
In triangle ABC,  A=30∘,  b=8,  a=6ABC,\,\,A = 30^\circ ,\,\,b = 8,\,\,a = 6, then B=sin⁡−1xB = {\sin ^{ - 1}}x, where x=x =
View answer and explanation
Correct option: C
2/32/3
Option C (2/32/3) is correct because by using the sine rule, the value of sin⁡B\sin B is calculated to be 2/32/3.
159 Easy
159 of 287
Analyzing +4 | -1
In a Δ  ABC,  if  r1=2r2=3r3\Delta \;ABC,\;\mathrm{if}\;{r_1} = 2{r_2} = 3{r_3}, then
View answer and explanation
Correct option: B
ab=54\frac{a}{b} = \frac{5}{4}
Option B (ab=54\frac{a}{b} = \frac{5}{4}) is correct because by using the given relationship between the exradii and the formula for exradii, we can derive the relationship between sides a and b.
160 Easy
160 of 287
Application +4 | -1
In any triangle ABC,  c2sin⁡2B+b2sin⁡2CABC,\;{c^2}\sin 2B + {b^2}\sin 2C is equal to
View answer and explanation
Correct option: D
4Δ4\Delta
Option D (4Δ4\Delta) is correct because by using the sine rule and area of triangle formula, the expression c2sin⁡2B+b2sin⁡2C{c^2}\sin 2B + {b^2}\sin 2C simplifies to 4Δ4\Delta.
161 Easy
161 of 287
Application +4 | -1
In a Δ ABC,  b=2,  C=60∘,  c=6\Delta \,ABC,\,\,b = 2,\,\,C = 60^\circ ,\,\,c = \sqrt 6, then a=a =
View answer and explanation
Correct option: C
3+1\sqrt 3 + 1
Option C (3+1\sqrt 3 + 1) is correct because it is obtained by applying the sine rule and solving for side 'a' using the given values.
162 Easy
162 of 287
Application +4 | -1
In a triangle vertex angles are A,  B,  CA,\;B,\;C and side BCBC are given. The area of ΔABC\Delta ABC is
View answer and explanation
Correct option: D
12.a2sin⁡Bsin⁡Csin⁡A\frac{1}{2}.\frac{{{a^2}\sin B\sin C}}{{\sin A}}
Option D (12.a2sin⁡Bsin⁡Csin⁡A\frac{1}{2}.\frac{{{a^2}\sin B\sin C}}{{\sin A}}) is correct because it represents the area of the triangle in terms of one side and all three angles, derived using the sine rule.
163 Easy
163 of 287
Application +4 | -1
If the lengths of the sides of a triangle are in AP and the greatest angle is double the smallest, then a ratio of lengths of the sides of this triangle is
View answer and explanation
Correct option: B
04:05:06
Option B (04:05:06) is correct because when the sides are in AP and the greatest angle is double the smallest, the ratio of the sides is 4:5:6.
164 Easy
164 of 287
Application +4 | -1
If the altitudes of a triangle are in AP, then the sides of the triangle are in
View answer and explanation
Correct option: C
H.P.
Option C (H.P.) is correct because the sides are inversely proportional to the altitudes, and if altitudes are in A.P., the sides are in H.P.
165 Easy
165 of 287
Application +4 | -1
If the radius of the incircle of a triangle with its sides 5 k,  6 k5\,k,\,\,6\,k, and 5k5k is 6\mathrm{6}, then kk is equal to
View answer and explanation
Correct option: B
4\mathrm{4}
Option B, 4\mathrm{4}, is correct because it is the value of k calculated by equating the area of the triangle using Heron's formula and the formula involving inradius.

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166 Easy
166 of 287
Application +4 | -1
The area of a triangle ABCA B C, where a=2(3+1),B=45∘,C=60∘a=2(\sqrt{3}+1), B=45^{\circ}, C=60^{\circ} is
View answer and explanation
Correct option: C
23(3+1)2 \sqrt{3}(\sqrt{3}+1) square unit
Option C (23(3+1)2 \sqrt{3}(\sqrt{3}+1) square unit) is correct because it is the area calculated using the sine rule and the given values of aa, BB, and CC.
167 Easy
167 of 287
Application +4 | -1
In a ΔABC,a:b:c=4:5:6\Delta ABC,a:b:c = 4:5:6. The ratio of the radius of the circumcircle to that of the incircle is
View answer and explanation
Correct option: B
167\frac{{16}}{7}
Option B 167\frac{{16}}{7} is correct because it is derived by using the formulas for the circumradius and inradius of a triangle and the given ratio of sides.
168 Easy
168 of 287
Analyzing +4 | -1
If a2,b2,c2{a^2},{b^2},{c^2} are in AP, then which of the following are also in AP?
View answer and explanation
Correct option: C
cot⁡A,cot⁡B,cot⁡C\cot A,\cot B,\cot C
Option C (cot⁡A,cot⁡B,cot⁡C\cot A,\cot B,\cot C) is correct because if a2,b2,c2{a^2},{b^2},{c^2} are in AP, then cot⁡A,cot⁡B,cot⁡C\cot A,\cot B,\cot C are also in AP.
169 Easy
169 of 287
Application +4 | -1
In a triangle ABCA B C, if (1+ab+cb)(1+bc−ac)=3\left(1+\frac{a}{b}+\frac{c}{b}\right)\left(1+\frac{b}{c}-\frac{a}{c}\right)=3, then the angle AA is equal to
View answer and explanation
Correct option: A
π3\frac{\pi}{3}
Option A (π3\frac{\pi}{3}) is correct because solving the given equation using trigonometric identities and the cosine rule leads to angle A being 60 degrees.
170 Easy
170 of 287
Analyzing +4 | -1
If cos⁡2A+cos⁡2C=sin⁡2B{\cos ^2}A + {\cos ^2}C = {\sin ^2}B, then ΔABC\Delta ABC is
View answer and explanation
Correct option: B
Right angled
Option B (Right angled) is correct because the given trigonometric identity holds true for a right-angled triangle, specifically when one of the angles is 90 degrees.
171 Easy
171 of 287
Application +4 | -1
The radius of the incircle of triangle when sides are 18, 24 and 30 cm is
View answer and explanation
Correct option: C
6 cm
Option C (6 cm) is correct because the inradius is calculated by dividing the area of the triangle by its semi-perimeter.
172 Easy
172 of 287
Application +4 | -1
A triangle ABCA B C is inscribed in a circle of radius 4 cm.∠A=60∘AC−AB=4 cm4 \mathrm{~cm} . \angle A=60^{\circ} A C-A B=4 \mathrm{~cm}, then area of △ABC\triangle \mathrm{ABC} will be?
View answer and explanation
Correct option: B
838 \sqrt{3}
Option B (838 \sqrt{3}) is correct because it is the area of the triangle calculated using the given conditions and trigonometric relationships.
173 Easy
173 of 287
Analyzing +4 | -1
In any △ABC,rr1+r2r3\triangle A B C, r r_1+r_2 r_3 is equal to
View answer and explanation
Correct option: C
bcb c
Option C (bcbc) is correct because the expression rr1+r2r3rr_1 + r_2r_3 simplifies to bcbc using the properties of inradius and exradii.
174 Easy
174 of 287
Analyzing +4 | -1
In any Δ ABC\Delta \,ABC, the distance of the orthocentre from the vertices A,B,CA,B,C are in the ratio
View answer and explanation
Correct option: B
cos⁡  A  :cos⁡  B  :cos⁡  C\cos \,\,A\;:\cos \,\,B\;:\cos \,\,C
Option B (cos⁡A:cos⁡B:cos⁡C\cos A:\cos B:\cos C) is correct because the distances of the orthocenter from the vertices of a triangle are proportional to the cosines of the respective angles.
175 Easy
175 of 287
Analyzing +4 | -1
In a ΔABC,  ∑(b+c)tan⁡A2tan⁡(B−C2)\Delta ABC,\;\sum \left( {b + c} \right)\tan \frac{A}{2}\tan \left( {\frac{{B - C}}{2}} \right) is equal to
View answer and explanation
Correct option: D
0
Option D (0) is correct because the expression simplifies to 0 using trigonometric identities and the sine rule.
176 Easy
176 of 287
Application +4 | -1
In a triangle ABC,if  ∠A=60∘,a=5,b=4ABC,\mathrm{if}\;\angle A = 60^\circ ,a = 5,b = 4, then cc is a root of the equation
View answer and explanation
Correct option: B
c2−4c−9=0{c^2} - 4c - 9 = 0
Option B (c2−4c−9=0{c^2} - 4c - 9 = 0) is correct because it is the quadratic equation derived from applying the cosine rule with the given triangle properties.
177 Easy
177 of 287
Application +4 | -1
If a,b,ca,b,c the sides of a Δ  ABC\Delta \;ABC are in AP and aa is the smallest side, then cos⁡A\cos A equals
View answer and explanation
Correct option: C
4c−3b2c\frac{{4c - 3b}}{{2c}}
Option C 4c−3b2c\frac{{4c - 3b}}{{2c}} is correct because it is derived by using the cosine rule and the given condition that a, b, c are in AP.
178 Easy
178 of 287
Application +4 | -1
In a Δ  ABC\Delta \;ABC, if the sides are a=3,b=5a = 3,b = 5 and c=4c = 4, then sin⁡B2+cos⁡B2\sin \frac{B}{2} + \cos \frac{B}{2} is equal to
View answer and explanation
Correct option: A
2\sqrt 2
Option A (2\sqrt{2}) is correct because, given the sides a=3,b=5,c=4a = 3, b = 5, c = 4, angle B is 90 degrees, and sin⁡B2+cos⁡B2=2\sin \frac{B}{2} + \cos \frac{B}{2} = \sqrt{2}.
179 Easy
179 of 287
Understanding +4 | -1
Two triangles are possible if
View answer and explanation
Correct option: B
A<π2,a>csin⁡A,a<cA < \frac{\pi}{2}, a>c \sin A, a < c
Option B (A<π2,a>csin⁡A,a<cA < \frac{\pi}{2}, a>c \sin A, a < c) is correct because it represents the condition where two triangles can be formed given angle A, side a, and side c.
180 Easy
180 of 287
Analyzing +4 | -1
If angles A,  BA,\;B and CC are in AP, then a+bc\frac{{a + b}}{c} is equal to
View answer and explanation
Correct option: B
2cos⁡A−C22\cos \frac{{A - C}}{2}
Option B (2cos⁡A−C22\cos \frac{{A - C}}{2}) is correct because using the sine rule and the given condition that A, B, and C are in AP, the expression simplifies to 2cos⁡A−C22\cos \frac{{A - C}}{2}.
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