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JEE • MATHEMATICS • PROPERTIES OF TRIANGLES

Properties of triangles practice · Easy

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Showing 271-287 of 287 questions

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271 Easy
271 of 287
Analyzing +4 | -1
In a ΔABC\Delta ABC, if (3−1)a=2b,  A=3B\left( {\sqrt 3 - 1} \right)a = 2b,\;A = 3B, then ∠C\angle C is
View answer and explanation
Correct option: B
120∘120^\circ
Option B (120∘120^\circ) is correct because by using the sine rule and the given relationship between the sides and angles, we can find the angles of the triangle and determine that angle C is 120 degrees.
272 Easy
272 of 287
Application +4 | -1
In Δ  ABC,  2R2sin⁡Asin⁡Bsin⁡C\Delta \;ABC,\;2{R^2}\sin A\sin B\sin C is equal to
View answer and explanation
Correct option: C
Δ\Delta
Option C (Δ\Delta) is correct because using the sine rule, the expression simplifies to the area of the triangle.
273 Easy
273 of 287
Analyzing +4 | -1
In a △PQR\triangle P Q R, If 3sin⁡P+4cos⁡Q=63 \sin P+4 \cos Q=6 and 4sin⁡Q+3cos⁡4 \sin Q+3 \cos P=1P=1, then the angle RR is equal to:
View answer and explanation
Correct option: B
π6\frac{\pi}{6}
Option B (π6\frac{{\pi}}{{6}}) is correct because it is the only value of angle R that satisfies the given trigonometric equations and the triangle property.
274 Easy
274 of 287
Application +4 | -1
If in a ΔABC,(s−a)(s−b)=s(s−c)\Delta ABC,\left( {s - a} \right)\left( {s - b} \right) = s\left( {s - c} \right), then angle CC is equal to
View answer and explanation
Correct option: A
90∘90^\circ
Option A (90∘90^\circ) is correct because it satisfies the given condition when using the tangent half-angle formula.
275 Easy
275 of 287
Application +4 | -1
In a Δ  ABC\Delta \;ABC, if sin⁡A+sin⁡B+sin⁡C)(sin⁡A+sin⁡B−sin⁡C)=3sin⁡Asin⁡B\sin A + \sin B + \sin C)\left( {\sin A + \sin B - \sin C} \right) = 3\sin A\sin B, then the angle CC is equal to
View answer and explanation
Correct option: B
π3\frac{\pi }{3}
Option B (π3\frac{\pi }{3}) is correct because using sine rule and cosine rule, the given equation simplifies to cos⁡C=12\cos C = \frac{1}{2}, which implies C=π3C = \frac{\pi }{3}.
276 Easy
276 of 287
Application +4 | -1
In a ΔABC,a,c,A\Delta ABC,a,c,A are given and b1,b2{b_1},{b_2} are two values, if the third side bb such that b2=2b1{b_2} = 2{b_1}, then sin⁡A\sin A is equal to
View answer and explanation
Correct option: B
9a2−c28c2\sqrt {\frac{{9{a^2} - {c^2}}}{{8{c^2}}}}
Option B 9a2−c28c2\sqrt {\frac{{9{a^2} - {c^2}}}{{8{c^2}}}} is correct because it is derived by using the cosine rule and the given condition b2=2b1{b_2} = 2{b_1}.
277 Easy
277 of 287
Application +4 | -1
If in a triangle ABC,  b+c11=c+a12=a+b13ABC,\,\,\frac{{b + c}}{{11}} = \frac{{c + a}}{{12}} = \frac{{a + b}}{{13}} then cos⁡  A\cos \,\,A is equal to
View answer and explanation
Correct option: A
1/51/5
Option A (1/5) is correct because by setting up ratios and solving for the sides of the triangle, we can use the cosine rule to find cos A.
278 Easy
278 of 287
Application +4 | -1
The angles A,BA,B and CC of a Δ  ABC\Delta \;ABC are in A.P. If AB=6,BC=7,AB = 6,BC = 7, then AC=AC =
View answer and explanation
Correct option: D
None of these
Option D (None of these) is correct because the calculated value of AC is 43\sqrt{43}, which is not among the given options.
279 Easy
279 of 287
Application +4 | -1
In a Δ ABC\Delta \,ABC, the HM of the ex-radii is equal to
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Correct option: A
3 r3\,r
Option A (3r3r) is correct because the harmonic mean of the ex-radii is equal to 3r3r, where rr is the inradius of the triangle.
280 Easy
280 of 287
Analyzing +4 | -1
If in a △ABC,acos⁡A=bcos⁡B\triangle A B C, \frac{a}{\cos A}=\frac{b}{\cos B}, then
View answer and explanation
Correct option: B
2sin⁡Acos⁡B=sin⁡C2 \sin A \cos B=\sin C
Option D (None of the above) is incorrect. Option B (2sin⁡Acos⁡B=sin⁡C2 \sin A \cos B = \sin C) is the correct answer because it is derived from the given condition using trigonometric identities and the sine rule.
281 Easy
281 of 287
Application +4 | -1
In Δ  ABC\Delta \;ABC, if 1b+c+1c+a=3a+b+c\frac{1}{{b + c}} + \frac{1}{{c + a}} = \frac{3}{{a + b + c}}, then CC is equal to
View answer and explanation
Correct option: B
60∘60^\circ
Option B (60∘60^\circ) is correct because the given condition leads to the equation a2+b2−c2=aba^2 + b^2 - c^2 = ab, which implies cos⁡C=12\cos C = \frac{1}{2}.
282 Easy
282 of 287
Application +4 | -1
In Δ  ABC,(b+c)cos⁡A+(c+a)cos⁡B+(a+b)cos⁡C\Delta \;ABC,\left( {b + c} \right)\cos A + \left( {c + a} \right)\cos B + \left( {a + b} \right)\cos C is equal to
View answer and explanation
Correct option: C
a+b+ca + b + c
Option C a+b+ca + b + c is correct because it is derived by expanding the expression and using the projection formula.
283 Easy
283 of 287
Application +4 | -1
In a Δ  ABC,(b+c−a)tan⁡A2\Delta \;ABC,\left( {b + c - a} \right)\tan \frac{A}{2} is equal to
View answer and explanation
Correct option: A
2Δs\frac{{2\Delta }}{s}
Option A (2Δs\frac{{2\Delta }}{s}) is correct because (b+c−a)tan⁡A2\left( {b + c - a} \right)\tan \frac{A}{2} can be simplified using trigonometric identities and the area formula to equal 2Δs\frac{{2\Delta }}{s}.
284 Easy
284 of 287
Application +4 | -1
If angles A,BA, B and CC are in AP, then a+cb\frac{a+c}{b} is equal to
View answer and explanation
Correct option: B
cos⁡(A−C2)\cos \left(\frac{A-C}{2}\right)
Option B (cos⁡(A−C2)\cos \left(\frac{A-C}{2}\right)) is correct because using the properties of AP and trigonometric identities, the expression simplifies to this result.
285 Easy
285 of 287
Analyzing +4 | -1
If p1,  p2,  p3{p_1},\,\,{p_2},\,\,{p_3} are altitude of a triangle ABCABC from the vertices A,  B,  CA,\,\,B,\,\,C and Δ\Delta, the area of the triangle, then 1p12+1p22+1p32=\frac{1}{{p_1^2}} + \frac{1}{{p_2^2}} + \frac{1}{{p_3^2}} =
View answer and explanation
Correct option: A
cot⁡  A+cos⁡  B+cot⁡  CΔ\frac{{\cot \,\,A + \cos \,\,B + \cot \,\,C}}{\Delta }
Option A is correct because it correctly relates the altitudes of a triangle to its area and cotangents of its angles.

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286 Easy
286 of 287
Application +4 | -1
In a triangle ABC,2acsin⁡12(A−B+C)=A B C, 2 a c \sin \frac{1}{2}(A-B+C)=
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 using the given condition and trigonometric identities, the expression simplifies to this form.
287 Easy
287 of 287
Application +4 | -1
If the sides of the triangle are p,q,p2+q2+pqp,q,\sqrt {{p^2} + {q^2} + pq}, then the greatest angle is
View answer and explanation
Correct option: C
2π3\frac{{2\pi }}{3}
Option C (2π3\frac{{2\pi }}{3}) is correct because the greatest angle is found using the cosine rule.
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