Properties of Triangle | JEE Advanced - Free MCQ Practice Test with solutions,

QS : SR = PQ : PR (as bisector of an angle, in a triangle, divides the opposite side in the same ratio as the sides containing the angle.)

Given that A > B and 3 sin x – 4 sin 3 x – k = 0, 0 <  k < 1
⇒ sin 3 x = k
As A and B satisfy above eq. (given)
∴ sin 3A = k,  sin 3B = k
⇒ sin 3A – sin 3B = 0

⇒ A + B = 60°  or  A = B
But given that A  >  B,   ∴  A ≠ B
Thus, A + B = 60°
But A + B + C = 180°  ⇒ C = 180° –  60° =  120°
∴ C = 2π3

Let  a = 3, b = 5, c = 7 then the largest ∠ is  opp. to the longest side, i.e., ∠C

In ΔABD, applying Sine law we get

⇒  

In ΔACD, applying Sine law, we get

⇒      ....(2)

We know that A + B + C = 180°
⇒ A + C – B = 180° – 2B

We know by Sine rule

⇒ r = s - c (∴ ∠C = 90º)
⇒ a + b - c= 2r
or   2r + c = a+b
or   2r +2R = a+b (Using C = 2R)

Let OP = Pole,    
∠PAO = ∠PBO = ∠PCQ = α

Similarly OA = OP  cot a .... (2)
Similarly OC = OP  cot a .... (3)
From  (1),  (2)  and (3), OA = OB = OC ⇒ O  is the point of circumcentre of the triangle ABC.

We know by Sine law in ΔABC as

(a) If we know a  , sin A  , sin B we can find b , c ; values of ∠’s  A , B and C all.
(b) Using a , b , c we can find ∠A  , ÐB, ∠C using cosine law.
(c) a , sin B , R are given then sin A , b and hence sin (A + B) and then C can be found.
(d) If we know a , sin A , R then we know  only the ratio  we can  not determine the values of b , c , sin B , sin C seperately.
∴  D can not be determined in  this case.

Given that  A : B : C = 4 : 1 : 1
Let A = 4x,  B =  x , C = x
But A + B + C = 180°
∴ 4x + x + x = 180°  ⇒ x = 30°
∴ A = 120° ,  B = 30° , C  = 30°

∴ Ratio of longest side to the perimeter

Sides are in the Ratio 1 : √3 :2

Let a = k , b = √3 k and c = 2k

The situation is as shown in the figure. For circle with centre C₂, BP and BP' are two tangents from B to circle, therefore BC₂ must be the ∠bisector of ∠B. But ∠B = 60° (∴ ΔABC is an equilateral Δ)

Given: In ΔABC, sides a,b,ca,b,ca,b,c opposite angles A,B,C.

Use the Sine rule: a = 2Rsin⁡A, b = 2Rsin⁡B, c = 2Rsin⁡C (where R is circumradius).

Compute the LHS of option (b):

So option (b) is correct.
Options (c) and (d) reduce to b + c = a and b − c = 2a respectively (false in general), and (a) does not hold identically.

By Sine law in ΔABC

Given AB || CD,  CD = 2AB Let AB = a then CD = 2a Let radius of circle be r.
Let circle touches AB at P, BC at Q, AD at R and CD at S.
Then AR = AP = r, BP = BQ = a – r
DR = DS = r  and CQ = CS = 2a – r
In ΔBEC BC² = BE² + EC² ⇒ (a – r + 2a – r)² = (2r)² + (a)²
⇒  9a² + 4r² – 12ar = 4r² + a²

⇒      ....(1)

∵ A, B, C are in A.P..

= 2(sin A cos C + cos A sin C ) =2 sin( A + C )=2 sin B

We have,

Let two sides of Δ be a and b.
Then a + b = x and ab = y
Also given x² – c² = y, where c is the third side of Δ.