Let the sides of the cyclid quadrilateral A B C D ABCD A BC D having sides A B = 3 , B C = 3 , C D = 4 , D A = 4. AB = 3, BC = 3, CD = 4, DA = 4. A B = 3 , BC = 3 , C D = 4 , D A = 4.
Join B B B with D . D. D . Clearly, B D BD B D is diameter of circum-circle so it will subtend a right-angle at A A A
and C . C. C .
Clearly △ A B D ≅ △ B C D \triangle ABD\cong \triangle BCD △ A B D ≅ △ BC D
∴ B D = 3 2 + 4 2 = 5 \therefore BD = \sqrt{3^2 + 4^2} = 5 ∴ B D = 3 2 + 4 2 = 5
Thus, radius of the circumcircle = 2.5 = 2.5 = 2.5 cm
Also, A D + B C = A B + C D AD + BC = AB + CD A D + BC = A B + C D
∴ \therefore ∴ Area of the quadrilateral = a b c d = 12 = \sqrt{abcd} = 12 = ab c d = 12 sq. cm.
We know that 12 = r s ⇒ r = 12 / 7 12 = rs \Rightarrow r = 12/7 12 = rs ⇒ r = 12/7 cm where r r r is radius of in-circle.
Let the quadrilateral be A B C D ABCD A BC D and let points be E , F , G , H , I , J , K , L E, F, G, H, I, J, K, L E , F , G , H , I , J , K , L on sides in order.
∴ E F = F G = G H = H I = I J = J K = K L = L E = x \therefore EF=FG=GH=HI=IJ=JK=KL=LE=x ∴ EF = FG = G H = H I = I J = J K = K L = L E = x
By symmetry, A E = A L = B G = B F = C H = C I = D J = D K = a AE=AL=BG=BF=CH=CI=DJ=DK=a A E = A L = BG = BF = C H = C I = D J = DK = a
Using Pythagoras theorem in △ A L E , \triangle ALE, △ A L E ,
L E 2 = A L 2 + A E 2 ⇒ x 2 = 2 a LE^2 = AL^2 + AE^2 \Rightarrow x^2 = \sqrt{2}a L E 2 = A L 2 + A E 2 ⇒ x 2 = 2 a
∴ A B = a + x + a = 2 a + x = 2 \therefore AB = a + x + a = 2a + x = 2 ∴ A B = a + x + a = 2 a + x = 2
⇒ a = 2 2 + 1 \Rightarrow a = \frac{\sqrt{2}}{\sqrt{2} + 1} ⇒ a = 2 + 1 2
⇒ x = 2 2 + 1 \Rightarrow x = \frac{2}{\sqrt{2} + 1} ⇒ x = 2 + 1 2
Thus, length of each side of octagon is 2 2 + 1 . \frac{2}{\sqrt{2} + 1}. 2 + 1 2 .
Let the perimeter be 6 a 6a 6 a the sides of hexagon will be a a a and sides of triangle will be 2 a . 2a. 2 a .
We know that for a regular polygon having n n n sides with each side’s length a a a the area is
n a 2 4 cot π n \frac{na^2}{4}\cot\frac{\pi}{n} 4 n a 2 cot n π
Area of regular hexagon = 6 a 2 4 cot 3 0 ∘ = 3 3 a 2 2 = \frac{6a^2}{4}\cot30^\circ = \frac{3\sqrt{3}a^2}{2} = 4 6 a 2 cot 3 0 ∘ = 2 3 3 a 2
Area of equilateral triangle = 3.4 a 2 4 cot 6 0 ∘ = 3 a 2 = \frac{3.4a^2}{4}\cot60^\circ = \sqrt{3}a^2 = 4 3.4 a 2 cot 6 0 ∘ = 3 a 2
∴ \therefore ∴ Ratio of areas = 2 / 3 = 2/3 = 2/3
Following the above problem n = 3 n = 3 n = 3
Ratio of areas = 3 4 = \frac{3}{4} = 4 3
∴ \therefore ∴ Ratio of sides = 3 2 = \frac{\sqrt{3}}{2} = 2 3
∴ \therefore ∴ Ratio of altitudes = 3 / 2 = \sqrt{3}/2 = 3 /2
⇒ sin θ = 3 / 2 \Rightarrow \sin\theta = \sqrt{3}/2 ⇒ sin θ = 3 /2
θ = 6 0 ∘ , 12 0 ∘ \theta = 60^\circ, 120^\circ θ = 6 0 ∘ , 12 0 ∘ so it is an equilateral triangle and a hexogon.
We know that angle of a polygon having n n n sides is ( n − 2 ) π / n (n - 2)\pi/n ( n − 2 ) π / n
Let there be n n n sides in one polygon and 2 n 2n 2 n in another.
Ratio = 2 n − 2 n − 2 . n 2 n = 9 8 = \frac{2n - 2}{n - 2}.\frac{n}{2n} = \frac{9}{8} = n − 2 2 n − 2 . 2 n n = 8 9
( n − 1 ) / n − 2 = 9 / 8 ⇒ n = 10 ⇒ 2 n = 20 (n - 1)/n - 2 = 9/8 \Rightarrow n = 10 \Rightarrow 2n = 20 ( n − 1 ) / n − 2 = 9/8 ⇒ n = 10 ⇒ 2 n = 20
The diagram is given below:
The six touching circle will form a hexgon. The sector internal to hexgon is of angle 12 0 ∘ . 120^\circ. 12 0 ∘ .
Side of hexagon will be double the radius of circles i.e. 2 a 2a 2 a as shown in figure.
Area inside circles = Area of hexgon - 6*area of sector
= 3 3 2 4 a 2 − 6. π a 2 3 = 6 3 a 2 − 2 π a 2 = \frac{3\sqrt{3}}{2}4a^2 - 6.\frac{\pi a^2}{3} = 6\sqrt{3}a^2 - 2\pi a^2 = 2 3 3 4 a 2 − 6. 3 π a 2 = 6 3 a 2 − 2 π a 2
= 2 a 2 ( 3 3 − π ) = 2a^2(3\sqrt{3} - \pi) = 2 a 2 ( 3 3 − π )
Let O O O is the radius of circumcircle of the square. Given,
A B = 1 , B D = 3 AB = 1, BD = \sqrt{3} A B = 1 , B D = 3 Also, O A = O B = O D = 1 OA = OB = OD = 1 O A = OB = O D = 1
Thus, for △ A B D , R = 1 , a sin A = 2 R = 2 \triangle ABD, R = 1, \frac{a}{\sin A = 2R} = 2 △ A B D , R = 1 , s i n A = 2 R a = 2
3 sin A = 2 ∴ A = 6 0 ∘ \frac{\sqrt{3}}{\sin A} = 2 \therefore A = 60^\circ s i n A 3 = 2 ∴ A = 6 0 ∘
⇒ C = 12 0 c i r c \Rightarrow C = 120^circ ⇒ C = 12 0 c i rc [opposite angle in cyclic quadrilateral]
By cosine law in △ A B D \triangle ABD △ A B D
3 = 1 + x 2 − 2 x cos 6 0 ∘ 3 = 1 + x^2 - 2x\cos60^\circ 3 = 1 + x 2 − 2 x cos 6 0 ∘
x 2 − x − 2 = 0 ⇒ x = 2 x^2 - x - 2 = 0 \Rightarrow x = 2 x 2 − x − 2 = 0 ⇒ x = 2
Thus, Δ = 3 3 4 = 1 2 1.2. sin 6 0 ∘ + 1 2 c . d sin 6 0 ∘ \Delta = \frac{3\sqrt{3}}{4} = \frac{1}{2}1.2.\sin60^\circ +
\frac{1}{2}c.d\sin60^\circ Δ = 4 3 3 = 2 1 1.2. sin 6 0 ∘ + 2 1 c . d sin 6 0 ∘
∴ c d = 1 \therefore cd = 1 ∴ c d = 1
By cosine law in △ B C D , \triangle BCD, △ BC D ,
3 = c 2 + d 2 − 2 c d cos 12 0 ∘ ⇒ c 2 + d 2 = 2 3 = c^2 + d^2 - 2cd\cos 120^\circ \Rightarrow c^2 + d^2 = 2 3 = c 2 + d 2 − 2 c d cos 12 0 ∘ ⇒ c 2 + d 2 = 2
⇒ c = d = 1 \Rightarrow c = d = 1 ⇒ c = d = 1
B C = 1 , C D = 1 , A D = x = 2 BC = 1, CD = 1, AD = x = 2 BC = 1 , C D = 1 , A D = x = 2
Let A B = a , B C = b , C D = c , D A = d AB = a, BC = b, CD = c, DA = d A B = a , BC = b , C D = c , D A = d
In △ A B C , \triangle ABC, △ A BC ,
A C 2 = a 2 + b 2 − 2 a b cos B AC^2 = a^2 + b^2 - 2ab\cos B A C 2 = a 2 + b 2 − 2 ab cos B
In △ A D C , \triangle ADC, △ A D C ,
A C 2 = c 2 + d 2 − 2 c d cos D AC^2 = c^2 + d^2 - 2cd\cos D A C 2 = c 2 + d 2 − 2 c d cos D
B + D = π B + D = \pi B + D = π [Angles opposite in a cyclic quadrilateral]
⇒ A C 2 ( a b + c d ) = ( a 2 + b 2 ) c d + ( c 2 + d 2 ) a b \Rightarrow AC^2(ab + cd) = (a^2 + b^2)cd + (c^2 + d^2)ab ⇒ A C 2 ( ab + c d ) = ( a 2 + b 2 ) c d + ( c 2 + d 2 ) ab
Simirlarly, we can find B D BD B D and then we find that
( A C . B D ) 2 = ( a c + b d ) 62 (AC.BD)^2 = (ac + bd)62 ( A C . B D ) 2 = ( a c + b d ) 62
⇒ A C . B D = A B . C D + B C . A D \Rightarrow AC.BD = AB.CD + BC.AD ⇒ A C . B D = A B . C D + BC . A D
Let p p p be the perimeter of both the polygons. So the length of the
sides will be p / n p/n p / n and p / 2 n . p/2n. p /2 n . Let A 1 A_1 A 1 and
A 2 A_2 A 2 denote the areas for them.
A 1 = 1 4 . n . p 2 n 2 cot π n A_1 = \frac{1}{4}.n.\frac{p^2}{n^2}\cot\frac{\pi}{n} A 1 = 4 1 . n . n 2 p 2 cot n π
A 2 = 1 4 . 2 n . p 2 4 n 2 cot π 2 n A_2 = \frac{1}{4}.2n.\frac{p^2}{4n^2}\cot\frac{\pi}{2n} A 2 = 4 1 .2 n . 4 n 2 p 2 cot 2 n π
A 1 A 2 = 2 cot π n cot π 2 n \frac{A_1}{A_2} = \frac{2\cot\frac{\pi}{n}}{\cot\frac{\pi}{2n}} A 2 A 1 = c o t 2 n π 2 c o t n π
= 2 cos π n 1 + cos π 2 n = \frac{2\cos\frac{\pi}{n}}{1 + \cos\frac{\pi}{2n}} = 1 + c o s 2 n π 2 c o s n π
Let P = sin A 2 sin B 2 sin C 2 P = \sin\frac{A}{2}\sin\frac{B}{2}\sin\frac{C}{2} P = sin 2 A sin 2 B sin 2 C and C C C is fixed.
P = 1 2 [ cos A − B 2 − cos A + B 2 ] sin C 2 P = \frac{1}{2}\left[\cos\frac{A - B}{2} - \cos\frac{A + B}{2}\right]\sin\frac{C}{2} P = 2 1 [ cos 2 A − B − cos 2 A + B ] sin 2 C
= 1 2 [ cos A − B 2 − sin C 2 ] sin C 2 = \frac{1}{2}\left[\cos \frac{A - B}{2} - \sin\frac{C}{2}\right]\sin\frac{C}{2} = 2 1 [ cos 2 A − B − sin 2 C ] sin 2 C
As C C C is fixed, the value P P P will depend on the value of cos A − B 2 \cos\frac{A - B}{2} cos 2 A − B and P P P will be
maxmimum when A = B A = B A = B
Similalry, when B B B is fixed P P P will be maxed when A = C , A = C, A = C , and when A A A is fixed P P P will be
maxed when B = C B = C B = C
Thus, P P P will be maximum when A = B = C = π / 3 A = B = C = \pi/3 A = B = C = π /3
⇒ P m a x = 1 / 8 \Rightarrow P_{max} = 1/8 ⇒ P ma x = 1/8
Hence proved.
Let A A A be the arithmetic mean. Then A = cos ( α + π 2 ) + cos ( β + π 2 ) + cos ( γ + π 2 ) 3 A = \frac{\cos\left(\alpha + \frac{\pi}{2}\right) + \cos\left(\beta +
\frac{\pi}{2}\right) + \cos\left(\gamma + \frac{\pi}{2}\right)}{3} A = 3 c o s ( α + 2 π ) + c o s ( β + 2 π ) + c o s ( γ + 2 π )
= − sin α + sin β + sin γ 3 = -\frac{\sin\alpha + \sin\beta + \sin\gamma}{3} = − 3 s i n α + s i n β + s i n γ
Clearly, sin α + sin β + sin γ \sin\alpha + \sin\beta + \sin\gamma sin α + sin β + sin γ will be maximum when α = β = γ \alpha = \beta = \gamma α = β = γ as proved in last
problem making A A A minimum.
α + β + γ = 2 π \alpha + \beta + \gamma = 2\pi α + β + γ = 2 π
A m i n = 3 sin 2 π 3 . 1 3 = 3 2 A_{min} = 3\sin\frac{2\pi}{3}.\frac{1}{3} = \frac{\sqrt{3}}{2} A min = 3 sin 3 2 π . 3 1 = 2 3
Let tan A 2 = x , tan B 2 = y , tan C 2 = z \tan\frac{A}{2} = x, \tan\frac{B}{2} = y, \tan\frac{C}{2} = z tan 2 A = x , tan 2 B = y , tan 2 C = z
We know that x 2 + y 2 + z 2 − x y − y z − x z ≥ 0 x^2 + y^2 + z^2 - xy - yz - xz \geq 0 x 2 + y 2 + z 2 − x y − yz − x z ≥ 0
x 2 + y 2 + z 2 ≥ x y + y z + x z x^2 + y^2 + z^2 \geq xy + yz + xz x 2 + y 2 + z 2 ≥ x y + yz + x z
A + B + C = π A + B + C = \pi A + B + C = π
⇒ tan ( A 2 + B 2 ) = cot C 2 \Rightarrow \tan\left(\frac{A}{2} + \frac{B}{2}\right) = \cot\frac{C}{2} ⇒ tan ( 2 A + 2 B ) = cot 2 C
⇒ tan A 2 tan B 2 + tan B 2 tan C 2 + tan C 2 tan A 2 = 1 \Rightarrow \tan\frac{A}{2}\tan\frac{B}{2} + \tan\frac{B}{2}\tan\frac{C}{2} + \tan\frac{C}{2}\tan\frac{A}{2} = 1 ⇒ tan 2 A tan 2 B + tan 2 B tan 2 C + tan 2 C tan 2 A = 1
⇒ x y + y z + x z = 1 \Rightarrow xy + yz + xz = 1 ⇒ x y + yz + x z = 1
⇒ x 2 + y 2 + z 2 ≥ 1 \Rightarrow x^2 + y^2 + z^2 \geq 1 ⇒ x 2 + y 2 + z 2 ≥ 1
Given, 2 b = ( m + 1 ) a ⇒ m = 2 b a − 1 2b = (m + 1)a \Rightarrow m = \frac{2b}{a} - 1 2 b = ( m + 1 ) a ⇒ m = a 2 b − 1
⇒ cos A = 1 2 ( m − 1 ) ( m + 3 ) m \Rightarrow \cos A = \frac{1}{2}\sqrt{\frac{(m - 1)(m + 3)}{m}} ⇒ cos A = 2 1 m ( m − 1 ) ( m + 3 )
⇒ b 2 + c 2 − a 2 2 b c = 1 2 ( 2 b a − 2 ) ( 2 b a + 2 ) m \Rightarrow \frac{b^2 + c^2 - a^2}{2bc} = \frac{1}{2}\sqrt{\frac{\left(\frac{2b}{a} - 2\right)\left(\frac{2b}{a} + 2\right)}{m}} ⇒ 2 b c b 2 + c 2 − a 2 = 2 1 m ( a 2 b − 2 ) ( a 2 b + 2 )
⇒ b 2 + c 2 − a 2 ( m + 1 ) a c = 1 a ( b − a ) ( b + a ) m \Rightarrow \frac{b^2 + c^2 - a^2}{(m + 1)ac} = \frac{1}{a}\sqrt{\frac{(b - a)(b + a)}{m}} ⇒ ( m + 1 ) a c b 2 + c 2 − a 2 = a 1 m ( b − a ) ( b + a )
⇒ c 2 m − ( m + 1 ) p . c + p 2 m = 0 \Rightarrow c^2\sqrt{m} - (m + 1)p.c + p^2\sqrt{m} = 0 ⇒ c 2 m − ( m + 1 ) p . c + p 2 m = 0
This is a quadratic equation in c c c and thus it will have two values.
⇒ c 1 , c 2 = p / m , m p \Rightarrow c_1, c_2 = p/m, \sqrt{m}p ⇒ c 1 , c 2 = p / m , m p
⇒ c 2 / c 1 = m \Rightarrow c_2/c_1 = m ⇒ c 2 / c 1 = m
Let a , b , c a, b, c a , b , c be the sides of the triangle.
⇒ s = ( a + b + c ) / 2 \Rightarrow s = (a + b + c)/2 ⇒ s = ( a + b + c ) /2 and s − a , s − b , s − c s - a, s- b, s - c s − a , s − b , s − c will be all greater than zero.
For positive quantities A.M. > > > G.M.
∴ s + s − a + s − b + s − c 4 > [ s ( s − a ) ( s − b ) ( s − c ) ] 1 / 4 \therefore \frac{s + s - a + s - b + s - c}{4} > [s(s - a)(s - b)(s - c)]^{1/4} ∴ 4 s + s − a + s − b + s − c > [ s ( s − a ) ( s − b ) ( s − c ) ] 1/4
⇒ 2 s 4 > Δ 1 / 2 \Rightarrow \frac{2s}{4}>\Delta^{1/2} ⇒ 4 2 s > Δ 1/2
⇒ Δ < s 2 4 \Rightarrow \Delta < \frac{s^2}{4} ⇒ Δ < 4 s 2
A + B + C = π A + B + C = \pi A + B + C = π
B + C = π − π 4 ⇒ C = 3 π 4 − B B + C = \pi - \frac{\pi}{4} \Rightarrow C = \frac{3\pi}{4} - B B + C = π − 4 π ⇒ C = 4 3 π − B
Let p = tan A tan B tan C = tan B tan ( 3 π / 4 − B ) p = \tan A\tan B\tan C = \tan B\tan\left(3\pi/4 - B\right) p = tan A tan B tan C = tan B tan ( 3 π /4 − B )
p = tan tan 3 π / 4 − tan B 1 + tan 3 π / 4 tan B p = \tan\frac{\tan3\pi/4 - \tan B}{1 + \tan3\pi/4\tan B} p = tan 1 + t a n 3 π /4 t a n B t a n 3 π /4 − t a n B
= tan B ( − 1 − tan B 1 − tan B ) = \tan B\left(\frac{-1 - \tan B}{1 - \tan B}\right) = tan B ( 1 − t a n B − 1 − t a n B )
p − p tan B = − tan B − tan 2 B p - p\tan B = -\tan B - \tan^2B p − p tan B = − tan B − tan 2 B
tan 2 B + ( 1 − p ) tan B + p = 0 \tan^2B + (1 - p)\tan B + p = 0 tan 2 B + ( 1 − p ) tan B + p = 0
For tan B \tan B tan B to be real, D ≥ 0 D\geq 0 D ≥ 0
⇒ ( 1 − p ) 2 − 4 p ≥ 0 \Rightarrow (1 - p)^2 - 4p \geq 0 ⇒ ( 1 − p ) 2 − 4 p ≥ 0
p = 3 ± 2 2 p = 3\pm 2\sqrt{2} p = 3 ± 2 2
Clearly, both B B B and C C C both cannot be obtuse.
If either of B B B or C C C is obtuse angle, then
tan B tan C < 0 \tan B\tan C < 0 tan B tan C < 0 ⇒ p < 0 \Rightarrow p < 0 ⇒ p < 0
If both are acute then
π / 4 < B < π / 2 , π / 4 < C < π / 2 \pi/4 < B < \pi/2, \pi/4 < C < \pi/2 π /4 < B < π /2 , π /4 < C < π /2
⇒ tan B > 1 , tan C > 1 \Rightarrow \tan B>1, \tan C> 1 ⇒ tan B > 1 , tan C > 1
⇒ tan B tan C > 1 ⇒ p > 1 \Rightarrow \tan B\tan C > 1 \Rightarrow p > 1 ⇒ tan B tan C > 1 ⇒ p > 1
⇒ p < 0 , p ≥ 3 + 2 2 \Rightarrow p < 0, p\geq 3 + 2\sqrt{2} ⇒ p < 0 , p ≥ 3 + 2 2
Let A B C ABC A BC be a triangle and A D , B E , C F AD, BE, CF A D , BE , CF be lines drawn from vertices to opposite sides such that ∠ A D C = ∠ B E S = ∠ C F B = α \angle
ADC = \angle BES = \angle CFB = \alpha ∠ A D C = ∠ BES = ∠ CFB = α
Let the triangle formed by A D , B E , C F AD, BE, CF A D , BE , CF be A ′ B ′ C ′ A'B'C' A ′ B ′ C ′
Clearly, ∠ B ′ A ′ C ′ = ∠ F A ′ B = π − ( ∠ B F A ′ + ∠ F B A ′ ) = π − [ α + π − ( α + A ) ] = A \angle B'A'C' = \angle FA'B = \pi - (\angle BFA' + \angle FBA') = \pi - [\alpha + \pi - (\alpha + A)] = A ∠ B ′ A ′ C ′ = ∠ F A ′ B = π − ( ∠ BF A ′ + ∠ FB A ′ ) = π − [ α + π − ( α + A )] = A
Similarly, ∠ A ′ B ′ C ′ = B \angle A'B'C'= B ∠ A ′ B ′ C ′ = B and A ′ C ′ B ′ = C A'C'B' = C A ′ C ′ B ′ = C
Thus, △ A B C △ A ′ B ′ C ′ \triangle ABC ~ \triangle A'B'C' △ A BC △ A ′ B ′ C ′
Area of △ A ′ B ′ C ′ Area of △ A B C = B ′ C ′ 2 a 2 \frac{\text{Area of }\triangle A'B'C'}{\text{Area of }\triangle ABC} = \frac{B'C'^2}{a^2} Area of △ A BC Area of △ A ′ B ′ C ′ = a 2 B ′ C ′2
Applying sine rule in A C ′ B , AC'B, A C ′ B ,
A C ′ sin [ π − ( A + α ) ] = A B sin ( π − C ) \frac{AC'}{\sin[\pi - (A + \alpha)]} = \frac{AB}{\sin(\pi - C)} s i n [ π − ( A + α )] A C ′ = s i n ( π − C ) A B
⇒ A C ′ = 2 R sin ( A + α ) \Rightarrow AC' = 2R\sin(A + \alpha) ⇒ A C ′ = 2 R sin ( A + α )
Similarly, B C ′ = 2 R sin ( α − A ) BC' = 2R\sin(\alpha - A) B C ′ = 2 R sin ( α − A )
⇒ B ′ C ′ = A C ′ − A B ′ = 2 a cos α \Rightarrow B'C' = AC' - AB' = 2a\cos\alpha ⇒ B ′ C ′ = A C ′ − A B ′ = 2 a cos α
Thus, ratio of areas = 4 cos 2 α : 1 = 4\cos^2\alpha:1 = 4 cos 2 α : 1
Given, a , b , c a,b,c a , b , c and Δ \Delta Δ are rational.
s = ( a + b + c ) / 2 s = (a + b + c)/2 s = ( a + b + c ) /2 will be rational.
tan B 2 = Δ s ( s − a ) \tan\frac{B}{2} = \frac{\Delta}{s(s - a)} tan 2 B = s ( s − a ) Δ will be rational as all terms involved are rational.
Similarly, tan C 2 \tan\frac{C}{2} tan 2 C will be rational.
sin B = 2 tan B 2 1 + tan 2 B 2 \sin B = \frac{2\tan\frac{B}{2}}{1 + \tan^2\frac{B}{2}} sin B = 1 + t a n 2 2 B 2 t a n 2 B will be rational as tan B 2 \tan\frac{B}{2} tan 2 B is rational.
Likewise sin C \sin C sin C will be rational.
A 2 = 9 0 ∘ − ( C / 2 + B / 2 ) \frac{A}{2} = 90^\circ - (C/2 + B/2) 2 A = 9 0 ∘ − ( C /2 + B /2 )
tan A 2 = cot ( B 2 + C 2 ) \tan\frac{A}{2} = \cot\left(\frac{B}{2} + \frac{C}{2}\right) tan 2 A = cot ( 2 B + 2 C )
= 1 − tan B 2 tan C 2 tan B 2 + tan C 2 = \frac{1 - \tan\frac{B}{2}\tan\frac{C}{2}}{\tan\frac{B}{2} + \tan\frac{C}{2}} = t a n 2 B + t a n 2 C 1 − t a n 2 B t a n 2 C will be rational as all the terms
involved are rational.
Thus, sin A \sin A sin A will also be rational.
a sin A = b sin B = c sin C = 2 R \frac{a}{\sin A} = \frac{b}{\sin B} = \frac{c}{\sin C} = 2R s i n A a = s i n B b = s i n C c = 2 R are rational as all the terms involved are rational.
R = a b c 4 Δ ⇒ Δ = a b c 4 R R = \frac{abc}{4\Delta} \Rightarrow \Delta = \frac{abc}{4R} R = 4Δ ab c ⇒ Δ = 4 R ab c will be rational as well.
Applying sine rule, b sin B = c sin C \frac{b}{\sin B} = \frac{c}{\sin C} s i n B b = s i n C c
⇒ 6 sin 3 0 ∘ = 4 sin C \Rightarrow \frac{\sqrt{6}}{\sin 30^\circ} = \frac{4}{\sin C} ⇒ s i n 3 0 ∘ 6 = s i n C 4
sin C = 2 6 < 1 \sin C = \frac{2}{\sqrt{6}} < 1 sin C = 6 2 < 1
So C C C may be acute or obtuse.
We observed that b < c ⇒ B < C , b < c \Rightarrow B < C, b < c ⇒ B < C , so B B B may be acute or obtuse.
If C C C is obtuse B B B may be acute. Hence two triangles are possible.
Applying cosine rule, cos B = c 2 + a 2 − b 2 2 a c = 16 + a 2 − 6 2.4. a \cos B = \frac{c^2 + a^2 - b^2}{2ac} = \frac{16 + a^2 - 6}{2.4.a} cos B = 2 a c c 2 + a 2 − b 2 = 2.4. a 16 + a 2 − 6
3 2 = 10 + a 2 8 a ⇒ a 2 − 4 3 a + 10 = 0 \frac{\sqrt{3}}{2} = \frac{10 + a^2}{8a} \Rightarrow a^2 - 4\sqrt{3}a + 10 = 0 2 3 = 8 a 10 + a 2 ⇒ a 2 − 4 3 a + 10 = 0
a = 2 3 ± 2 a = 2\sqrt{3}\pm 2 a = 2 3 ± 2
∴ Δ 1 , Δ 2 = 1 2 a . c . sin 3 0 ∘ = 2 3 ± 2 \therefore \Delta_1, \Delta_2 = \frac{1}{2}a.c.\sin30^\circ = 2\sqrt{3}\pm\sqrt{2} ∴ Δ 1 , Δ 2 = 2 1 a . c . sin 3 0 ∘ = 2 3 ± 2
Let △ A B C \triangle ABC △ A BC be the equilateral triangle such that its sides have length a . a. a .
s = 3 a 2 , Δ = 3 2 a s = \frac{3a}{2}, \Delta = \frac{\sqrt{3}}{2}a s = 2 3 a , Δ = 2 3 a
r = Δ s = a 2 3 r = \frac{\Delta}{s} = \frac{a}{2\sqrt{3}} r = s Δ = 2 3 a
Diameter of incircle will be diagonal of inscribed square i.e. 2 r = a 3 2r = \frac{a}{\sqrt{3}} 2 r = 3 a
Thus, side of square = a 6 = \frac{a}{\sqrt{6}} = 6 a
∴ \therefore ∴ Area of square = a 2 6 = \frac{a^2}{6} = 6 a 2
Given, A D = a b c b 2 − c 2 AD = \frac{abc}{b^2 - c^2} A D = b 2 − c 2 ab c
Also, A D = b sin 2 3 ∘ ⇒ a b c b 2 − c 2 = b sin 2 3 ∘ AD = b\sin 23^\circ \Rightarrow \frac{abc}{b^2 - c^2} = b\sin23^\circ A D = b sin 2 3 ∘ ⇒ b 2 − c 2 ab c = b sin 2 3 ∘
⇒ a c b 2 − c 2 = sin 2 3 ∘ \Rightarrow \frac{ac}{b^2 - c^2} = \sin23^\circ ⇒ b 2 − c 2 a c = sin 2 3 ∘
⇒ sin A sin C sin 2 B − sin 2 C = sin 2 3 ∘ \Rightarrow \frac{\sin A\sin C}{\sin^2B - \sin^2C} = \sin23^\circ ⇒ s i n 2 B − s i n 2 C s i n A s i n C = sin 2 3 ∘
⇒ sin C sin 2 3 ∘ sin ( B + C ) sin ( B − C ) = sin 2 3 ∘ \Rightarrow \frac{\sin C\sin 23^\circ}{\sin(B + C)\sin(B - C)} = \sin23^\circ ⇒ s i n ( B + C ) s i n ( B − C ) s i n C s i n 2 3 ∘ = sin 2 3 ∘
⇒ sin ( B − 2 3 ∘ ) = 1 = sin 9 0 ∘ \Rightarrow \sin(B - 23^\circ) = 1 = \sin90^\circ ⇒ sin ( B − 2 3 ∘ ) = 1 = sin 9 0 ∘
⇒ B = 11 3 ∘ \Rightarrow B = 113^\circ ⇒ B = 11 3 ∘
Given a : b : c = 4 : 5 : 6 ⇒ a = 4 k , b = 5 k , c = 6 k a:b:c = 4:5:6 \Rightarrow a = 4k, b = 5k, c = 6k a : b : c = 4 : 5 : 6 ⇒ a = 4 k , b = 5 k , c = 6 k (let)
R r = a b c 4 Δ . s Δ = a b c . a + b + c 2 4. s ( s − a ) ( s − b ) ( s − c ) \frac{R}{r} = \frac{abc}{4\Delta}.\frac{s}{\Delta} = \frac{abc.\frac{a + b + c}{2}}{4.s(s - a)(s - b)(s - c)} r R = 4Δ ab c . Δ s = 4. s ( s − a ) ( s − b ) ( s − c ) ab c . 2 a + b + c
= 16 7 = \frac{16}{7} = 7 16
Let ∠ B A D = α , ∠ C A D = β \angle BAD = \alpha, \angle CAD = \beta ∠ B A D = α , ∠ C A D = β
Applying sine law in △ A D B , B D sin α = A D sin B \triangle ADB, \frac{BD}{\sin\alpha} = \frac{AD}{\sin B} △ A D B , s i n α B D = s i n B A D
⇒ A D = B D sin α . 3 2 \Rightarrow AD = \frac{BD}{\sin\alpha}.\frac{\sqrt{3}}{2} ⇒ A D = s i n α B D . 2 3
Applying sine law in △ A D C , C D sin β = A D sin C \triangle ADC, \frac{CD}{\sin\beta} = \frac{AD}{\sin C} △ A D C , s i n β C D = s i n C A D
⇒ A D = C D sin β . 1 2 \Rightarrow AD = \frac{CD}{\sin\beta}.\frac{1}{\sqrt{2}} ⇒ A D = s i n β C D . 2 1
⇒ B D C D . 3 2 = sin α sin β \Rightarrow \frac{BD}{CD}.\frac{\sqrt{3}}{\sqrt{2}} = \frac{\sin\alpha}{\sin\beta} ⇒ C D B D . 2 3 = s i n β s i n α
⇒ sin α sin β = 1 6 \Rightarrow \frac{\sin\alpha}{\sin\beta} = \frac{1}{\sqrt{6}} ⇒ s i n β s i n α = 6 1
Given, 3 sin x − 4 sin 2 x − k = 0 3\sin x - 4\sin^2x - k = 0 3 sin x − 4 sin 2 x − k = 0
sin 3 x = k \sin3x = k sin 3 x = k
Since A A A and B B B satisfy the equations ∴ sin 3 A = sin 3 B = k \therefore \sin3A = \sin3B = k ∴ sin 3 A = sin 3 B = k
sin 3 A − sin 3 B = 0 \sin3A - \sin3B = 0 sin 3 A − sin 3 B = 0
∴ − 2 sin 3 C 2 sin 3 ( A − B ) 2 = 0 \therefore -2\sin\frac{3C}{2}\sin\frac{3(A - B)}{2} = 0 ∴ − 2 sin 2 3 C sin 2 3 ( A − B ) = 0
Since A ≠ B A\neq B A = B and also both A A A and B B B are less that π / 3 [ ∵ 0 < k < 1 ] \pi/3[\because 0<k<1] π /3 [ ∵ 0 < k < 1 ]
⇒ sin 3 C 2 = 0 ⇒ C = 2 π 3 \Rightarrow \sin\frac{3C}{2} = 0 \Rightarrow C = \frac{2\pi}{3} ⇒ sin 2 3 C = 0 ⇒ C = 3 2 π
Since A , B , C A,B,C A , B , C are in A.P. ∴ 2 B = A + C \therefore 2B = A + C ∴ 2 B = A + C
A + B + C = π ⇒ B = π / 3 A + B + C = \pi \Rightarrow B = \pi/3 A + B + C = π ⇒ B = π /3
sin ( 2 A + B ) = 1 2 = sin π 6 \sin(2A + B) = \frac{1}{2} = \sin\frac{\pi}{6} sin ( 2 A + B ) = 2 1 = sin 6 π
⇒ 2 A + B = n π + ( − 1 ) n π 6 \Rightarrow 2A + B = n\pi + (-1)^n\frac{\pi}{6} ⇒ 2 A + B = nπ + ( − 1 ) n 6 π
A = π 4 , 11 π 12 A = \frac{\pi}{4},\frac{11\pi}{12} A = 4 π , 12 11 π these are the values between 0 0 0 and π . \pi. π .
But 11 π 12 \frac{11\pi}{12} 12 11 π is not possible as B = π / 3 B = \pi/3 B = π /3
∴ A = π / 4 \therefore A = \pi/4 ∴ A = π /4
Let A B C ABC A BC be the triangle having right angle at B B B . From question, A C = 2 2 B D AC = 2\sqrt{2}BD A C = 2 2 B D
Let B D = x ∴ A C = 2 2 x BD = x \therefore AC = 2\sqrt{2}x B D = x ∴ A C = 2 2 x and ∠ C = θ \angle C = \theta ∠ C = θ
tan C = B D C D = x C D ⇒ C D = x cot θ \tan C = \frac{BD}{CD} = \frac{x}{CD}\Rightarrow CD = x\cot\theta tan C = C D B D = C D x ⇒ C D = x cot θ
tan ( 9 0 ∘ − C ) = B D A D ∴ A D = x tan θ \tan(90^\circ - C) = \frac{BD}{AD} \therefore AD = x\tan\theta tan ( 9 0 ∘ − C ) = A D B D ∴ A D = x tan θ
A D + C D = A C ⇒ tan θ + cot θ = 2 2 AD + CD = AC \Rightarrow \tan\theta + \cot\theta = 2\sqrt{2} A D + C D = A C ⇒ tan θ + cot θ = 2 2
⇒ 2 sin θ cos θ = 1 2 \Rightarrow 2\sin\theta\cos\theta = \frac{1}{\sqrt{2}} ⇒ 2 sin θ cos θ = 2 1
⇒ sin 2 θ = sin π 4 \Rightarrow \sin2\theta = \sin\frac{\pi}{4} ⇒ sin 2 θ = sin 4 π
⇒ θ = π 8 , 3 π 8 \Rightarrow \theta = \frac{\pi}{8}, \frac{3\pi}{8} ⇒ θ = 8 π , 8 3 π
⇒ A = 3 π 8 , π 8 \Rightarrow A = \frac{3\pi}{8}, \frac{\pi}{8} ⇒ A = 8 3 π , 8 π
P + Q + R = π P + Q + R = \pi P + Q + R = π
∴ P + Q = π / 2 [ ∵ R = π / 2 ] \therefore P + Q = \pi/2 [\because R = \pi/2] ∴ P + Q = π /2 [ ∵ R = π /2 ]
⇒ tan ( P + Q 2 ) = 1 \Rightarrow \tan\left(\frac{P + Q}{2}\right) = 1 ⇒ tan ( 2 P + Q ) = 1
⇒ tan P 2 + tan Q 2 = 1 − tan P 2 tan Q 2 \Rightarrow \tan\frac{P}{2} + \tan\frac{Q}{2} = 1 - \tan\frac{P}{2}\tan\frac{Q}{2} ⇒ tan 2 P + tan 2 Q = 1 − tan 2 P tan 2 Q
Since tan P 2 \tan\frac{P}{2} tan 2 P and tan Q 2 \tan\frac{Q}{2} tan 2 Q are roots of the equation a x 2 + b x + c = 0 ax^2 + bx + c = 0 a x 2 + b x + c = 0
⇒ tan P 2 + tan Q 2 = − b a , tan P 2 tan Q 2 = c 2 \Rightarrow \tan\frac{P}{2} + \tan\frac{Q}{2} = -\frac{b}{a}, \tan\frac{P}{2}\tan\frac{Q}{2} = \frac{c}{2} ⇒ tan 2 P + tan 2 Q = − a b , tan 2 P tan 2 Q = 2 c
⇒ − b a = 1 − c a \Rightarrow -\frac{b}{a} = 1 - \frac{c}{a} ⇒ − a b = 1 − a c
⇒ a + b = c \Rightarrow a + b = c ⇒ a + b = c
Let A D AD A D be the median such that ∠ B A D = 3 0 ∘ , ∠ D A C = 4 5 ∘ \angle BAD = 30^\circ, \angle DAC = 45^\circ ∠ B A D = 3 0 ∘ , ∠ D A C = 4 5 ∘ and ∠ A D C = θ \angle ADC = \theta ∠ A D C = θ
Applying m n mn mn theorem, we get
2 cot θ = cot 3 0 ∘ − cot 4 5 ∘ 2\cot\theta = \cot30^\circ - \cot45^\circ 2 cot θ = cot 3 0 ∘ − cot 4 5 ∘
⇒ tan θ = 3 + 1 \Rightarrow \tan\theta = \sqrt{3} + 1 ⇒ tan θ = 3 + 1
sin C = 3 + 2 2 5 + 2 3 \sin C = \frac{\sqrt{3} + 2}{\sqrt{2}\sqrt{5 + 2\sqrt{3}}} sin C = 2 5 + 2 3 3 + 2
Applying sine law in △ A D C , \triangle ADC, △ A D C , we get
A D sin C = D C sin 4 5 ∘ \frac{AD}{\sin C} = \frac{DC}{\sin45^\circ} s i n C A D = s i n 4 5 ∘ D C
⇒ D C = 1 \Rightarrow DC = 1 ⇒ D C = 1
⇒ D C = B D = 1 ⇒ B C = 2 \Rightarrow DC = BD = 1 \Rightarrow BC = 2 ⇒ D C = B D = 1 ⇒ BC = 2
We know that in a triangle tan A + tan B + tan C = tan A + tan B + tan C \tan A + \tan B + \tan C = \tan A + \tan B + \tan C tan A + tan B + tan C = tan A + tan B + tan C
Also, since A.M ≥ \geq ≥ G.M.
⇒ tan A + tan B + tan C 3 ≥ tan A tan B tan C 3 \Rightarrow \frac{\tan A + \tan B + \tan C}{3}\geq \sqrt[3]{\tan A\tan B\tan C} ⇒ 3 t a n A + t a n B + t a n C ≥ 3 tan A tan B tan C
⇒ tan A tan B tan C ≥ 3 tan A tan B tan C 3 \Rightarrow \tan A\tan B\tan C\geq 3\sqrt[3]{\tan A\tan B\tan C} ⇒ tan A tan B tan C ≥ 3 3 tan A tan B tan C
⇒ tan 2 A tan 2 B tan 2 C ≥ 27 \Rightarrow \tan^2A\tan^2B\tan^2C\geq 27 ⇒ tan 2 A tan 2 B tan 2 C ≥ 27
⇒ tan A + tan B + tan C ≥ 3 3 \Rightarrow \tan A + \tan B + \tan C\geq 3\sqrt{3} ⇒ tan A + tan B + tan C ≥ 3 3
Given, cos A 2 = 1 2 b c + c b \cos\frac{A}{2} = \frac{1}{2}\sqrt{\frac{b}{c} + \frac{c}{b}} cos 2 A = 2 1 c b + b c
s ( s − a ) b c = 1 2 b c + c b \sqrt{\frac{s(s - a)}{bc}} = \frac{1}{2}\sqrt{\frac{b}{c} + \frac{c}{b}} b c s ( s − a ) = 2 1 c b + b c
( a + b + c ) ( b + c − a ) 4 b c = b 2 + c 2 4 b c \frac{(a + b + c)(b + c - a)}{4bc} = \frac{b^2 + c^2}{4bc} 4 b c ( a + b + c ) ( b + c − a ) = 4 b c b 2 + c 2
⇒ a 2 = 2 b c \Rightarrow a^2 = 2bc ⇒ a 2 = 2 b c
Thus, square described on side a a a is twice the rectangle contained by two other sides.
Given, cos θ = a − b c ⇒ sin θ = 1 c c 2 − ( a − b ) 2 \cos\theta = \frac{a - b}{c} \Rightarrow \sin\theta = \frac{1}{c}\sqrt{c^2 - (a - b)^2} cos θ = c a − b ⇒ sin θ = c 1 c 2 − ( a − b ) 2
( a + b ) sin θ 2 a b = ( a + b ) c 2 − ( a − b ) 2 2 c a b \frac{(a + b)\sin\theta}{2\sqrt{ab}} = \frac{(a + b)\sqrt{c^2 - (a - b)^2}}{2c\sqrt{ab}} 2 ab ( a + b ) s i n θ = 2 c ab ( a + b ) c 2 − ( a − b ) 2
( a + b ) 2 a b ( 1 − cos C ) 2 c a b = a + b 2 c . 2 sin C 2 \frac{(a + b)\sqrt{2ab(1 - \cos C)}}{2c\sqrt{ab}} = \frac{a + b}{\sqrt{2}c}.\sqrt{2}\sin\frac{C}{2} 2 c ab ( a + b ) 2 ab ( 1 − c o s C ) = 2 c a + b . 2 sin 2 C
= sin A + sin B sin C . sin C 2 = \frac{\sin A + \sin B}{\sin C}.\sin\frac{C}{2} = s i n C s i n A + s i n B . sin 2 C
= cos A − B 2 = \cos\frac{A - B}{2} = cos 2 A − B
c sin θ 2 a b = c c 2 − ( a − b ) 2 2 a b \frac{c\sin\theta}{2\sqrt{ab}} = \frac{c\sqrt{c^2 - (a - b)^2}}{2\sqrt{ab}} 2 ab c s i n θ = 2 ab c c 2 − ( a − b ) 2
= c 2 a b ( 1 − cos C ) 2 a b = sin C 2 = cos A + B 2 = \frac{c\sqrt{2ab(1 - \cos C)}}{2\sqrt{ab}} = \sin \frac{C}{2} = \cos \frac{A + B}{2} = 2 ab c 2 ab ( 1 − c o s C ) = sin 2 C = cos 2 A + B
We have proven earlier that distance between circumcenter and incenter is R 2 − 2 r R \sqrt{R^2 - 2rR} R 2 − 2 r R
Clearly, R 2 − 2 r R ≥ 0 \sqrt{R^2 - 2rR}\geq 0 R 2 − 2 r R ≥ 0
⇒ R ≥ 2 r \Rightarrow R\geq 2r ⇒ R ≥ 2 r
Given, tan B = cos A \tan B = \cos A tan B = cos A
⇒ 1 − cos 2 B cos B = cos A \Rightarrow \frac{\sqrt{1 - \cos^2B}}{\cos B} = \cos A ⇒ c o s B 1 − c o s 2 B = cos A
⇒ cos 2 B = 1 2 − sin 2 A \Rightarrow \cos^2B = \frac{1}{2 - \sin^2A} ⇒ cos 2 B = 2 − s i n 2 A 1
Also given that cos C = tan A \cos C = \tan A cos C = tan A
⇒ tan 2 C = 1 − 2 sin 2 A sin 2 A \Rightarrow \tan^2 C = \frac{1 - 2\sin^2A}{\sin^2A} ⇒ tan 2 C = s i n 2 A 1 − 2 s i n 2 A
Now given that cos B = tan C \cos B = \tan C cos B = tan C
⇒ 1 2 − sin 2 A = 1 − 2 sin 2 A sin 2 A \Rightarrow \frac{1}{2 - \sin^2A} = \frac{1 - 2\sin^2A}{\sin^2A} ⇒ 2 − s i n 2 A 1 = s i n 2 A 1 − 2 s i n 2 A
⇒ sin A = ± 3 ± 5 2 \Rightarrow \sin A = \pm\sqrt{\frac{3\pm\sqrt{5}}{2}} ⇒ sin A = ± 2 3 ± 5
Same value will be obtained for sin B \sin B sin B and sin C \sin C sin C and that is equal to 2 sin 1 8 ∘ 2\sin 18^\circ 2 sin 1 8 ∘
A + B + C = 18 0 ∘ A + B + C = 180^\circ A + B + C = 18 0 ∘
⇒ cos ( A + B ) = − cot C \Rightarrow \cos(A + B) = -\cot C ⇒ cos ( A + B ) = − cot C
⇒ cot A cot B + cot B cot C + cot C cot A = 1 \Rightarrow \cot A\cot B + \cot B\cot C + \cot C\cot A = 1 ⇒ cot A cot B + cot B cot C + cot C cot A = 1
Now cot 2 A + cot 2 B + cot 2 C − 1 = cot 2 A + cot 2 B + cot 2 C − ( cot A cot B + cot B cot C + cot C cot A ) \cot^2A + \cot^2B + \cot^2C - 1 = \cot^2A + \cot^2B + \cot^2C - (\cot A\cot B + \cot B\cot C + \cot C\cot A) cot 2 A + cot 2 B + cot 2 C − 1 = cot 2 A + cot 2 B + cot 2 C − ( cot A cot B + cot B cot C + cot C cot A )
= 1 2 [ ( cot A − cot B ) 2 + ( cot B − cot C ) 2 + ( cot C − cot A ) 2 ] ≥ 0 = \frac{1}{2}[(\cot A - \cot B)^2 + (\cot B - \cot C)^2 + (\cot C - \cot A)^2] \geq 0 = 2 1 [( cot A − cot B ) 2 + ( cot B − cot C ) 2 + ( cot C − cot A ) 2 ] ≥ 0
⇒ cot 2 A + cot 2 B + cot 2 C ≥ 1 \Rightarrow \cot^2A + \cot^2B + \cot^2C \geq 1 ⇒ cot 2 A + cot 2 B + cot 2 C ≥ 1
We have proven in 229 that tan A + tan B + tan C ≥ 3 3 \tan A + \tan B + \tan C \geq 3\sqrt{3} tan A + tan B + tan C ≥ 3 3
We apply A.M. ≥ \geq ≥ G.M. again on tan 2 A , tan 2 B , tan 2 C \tan^2A, \tan^2B, \tan^2C tan 2 A , tan 2 B , tan 2 C
tan 2 A + tan 2 B + tan 2 C 3 ≥ ( tan 2 A tan B tan 2 C ) 1 / 3 \frac{\tan^2A + \tan^2B + \tan^2C}{3}\geq (\tan^2A\tan^B\tan^2C)^{1/3} 3 t a n 2 A + t a n 2 B + t a n 2 C ≥ ( tan 2 A tan B tan 2 C ) 1/3
⇒ tan 2 A + tan 2 B + tan C ≥ 9 \Rightarrow \tan^2A + \tan^2B + \tan^C \geq 9 ⇒ tan 2 A + tan 2 B + tan C ≥ 9
We know that in a △ A B C , sin A 2 + sin B 2 + sin C 2 ≤ 3 2 \triangle ABC, \sin\frac{A}{2} + \sin\frac{B}{2} + \sin\frac{C}{2} \leq \frac{3}{2} △ A BC , sin 2 A + sin 2 B + sin 2 C ≤ 2 3
Now we know that A.M. ≥ \geq ≥ H.M.
cosec A 2 + cosec B 2 + cosec C 2 3 ≥ 3 sin A 2 + sin B 2 + sin C 2 \frac{\cosec\frac{A}{2} + \cosec\frac{B}{2} + \cosec\frac{C}{2}}{3}\geq \frac{3}{\sin\frac{A}{2} + \sin\frac{B}{2} + \sin\frac{C}{2}} 3 c o s e c 2 A + c o s e c 2 B + c o s e c 2 C ≥ s i n 2 A + s i n 2 B + s i n 2 C 3
⇒ cosec A 2 + cosec B 2 + cosec C 2 ≥ 6 \Rightarrow \cosec\frac{A}{2} + \cosec\frac{B}{2} + \cosec\frac{C}{2} \geq 6 ⇒ cosec 2 A + cosec 2 B + cosec 2 C ≥ 6
cos A + cos B + cos C − 3 2 = 2 cos A + B 2 cos A − B 2 + 1 − 2 sin 2 C 2 − 3 2 \cos A + \cos B + \cos C -\frac{3}{2} = 2\cos\frac{A + B}{2}\cos\frac{A - B}{2} + 1 - 2\sin^2\frac{C}{2} - \frac{3}{2} cos A + cos B + cos C − 2 3 = 2 cos 2 A + B cos 2 A − B + 1 − 2 sin 2 2 C − 2 3
⇒ − 2 sin 2 C 2 + 2 cos A − B 2 sin C 2 − 1 2 = 0 \Rightarrow -2\sin^2\frac{C}{2} + 2\cos\frac{A - B}{2}\sin\frac{C}{2} - \frac{1}{2} = 0 ⇒ − 2 sin 2 2 C + 2 cos 2 A − B sin 2 C − 2 1 = 0
Clearly D = 4 cos 2 A − B 2 − 4 < 0 D = 4\cos^2\frac{A - B}{2} - 4 < 0 D = 4 cos 2 2 A − B − 4 < 0 so sign of quadratic equation will be same as that of first term.
⇒ cos A + cos B + cos C ≤ 3 2 \Rightarrow \cos A + \cos B + \cos C \leq \frac{3}{2} ⇒ cos A + cos B + cos C ≤ 2 3
Now cos A + cos B + cos C − 1 = 2 sin C 2 cos A − B 2 − 2 sin C 2 \cos A + \cos B + \cos C - 1 = 2\sin\frac{C}{2}\cos\frac{A - B}{2} - 2\sin^\frac{C}{2} cos A + cos B + cos C − 1 = 2 sin 2 C cos 2 A − B − 2 sin 2 C
= 2 sin C 2 [ cos A − B 2 − cos A + B 2 ] = 2\sin\frac{C}{2}\left[\cos\frac{A - B}{2} - \cos\frac{A + B}{2}\right] = 2 sin 2 C [ cos 2 A − B − cos 2 A + B ]
= 4 sin A 2 sin B 2 sin C 2 > 0 = 4\sin\frac{A}{2}\sin\frac{B}{2}\sin\frac{C}{2} > 0 = 4 sin 2 A sin 2 B sin 2 C > 0
∴ cos A + cos B + cos C > 1 \therefore \cos A + \cos B + \cos C > 1 ∴ cos A + cos B + cos C > 1
y = 2 cos A cos B cos C = [ cos ( A − B ) + cos ( A + B ) ] cos C = [ cos ( A − B ) − cos C ] cos C y=2\cos A\cos B\cos C=[\cos(A-B)+\cos(A+B)]\cos C=[\cos(A-B)-\cos C]\cos C y = 2 cos A cos B cos C = [ cos ( A − B ) + cos ( A + B )] cos C = [ cos ( A − B ) − cos C ] cos C
⟹ cos 2 C − cos ( A − B ) cos C + y = 0 \implies\cos^2C-\cos(A-B)\cos C+y=0 ⟹ cos 2 C − cos ( A − B ) cos C + y = 0
which is quadratic equation in cos C \cos C cos C which would be real.
⇒ D ≥ 0 ⟹ cos 2 ( A − B ) − 4 y ≥ 0 ⟺ y ≤ cos 2 ( A − B ) 4 ≤ 1 4 \Rightarrow D \geq 0 \implies \cos^2(A-B)-4y\ge0\iff y\le\dfrac{\cos^2(A-B)}4\le\dfrac14 ⇒ D ≥ 0 ⟹ cos 2 ( A − B ) − 4 y ≥ 0 ⟺ y ≤ 4 cos 2 ( A − B ) ≤ 4 1
Hence proved.
Let A A A be the center of first circle having radius a a a and B B B be the center of second circle having radius b . b. b .
Let the two circles intersect at C C C and D D D so ∠ A C B = ∠ A D B = θ \angle ACB = \angle ADB = \theta ∠ A CB = ∠ A D B = θ
We know that perpendicular from the center of a circle divides the chord in two equal parts. So A B AB A B will divide C D CD C D in two equal parts.
Let C D = 2 x . CD = 2x. C D = 2 x . Let A B AB A B cut C D CD C D at O O O then O C = O D = x OC = OD = x OC = O D = x
A B = O A + O B = a 2 − x 2 + b 2 − x 2 AB = OA + OB = \sqrt{a^2 - x^2} + \sqrt{b^2 - x^2} A B = O A + OB = a 2 − x 2 + b 2 − x 2
Clearly, ∠ A C B = θ \angle ACB = \theta ∠ A CB = θ
Applying cosine law in △ A B C \triangle ABC △ A BC
A B 2 = A C 2 + B C 2 − 2. A C . B C cos ( π − θ ) AB^2 = AC^2 + BC^2 - 2.AC.BC\cos(\pi - \theta) A B 2 = A C 2 + B C 2 − 2. A C . BC cos ( π − θ )
⇒ a 2 − x 2 + b 2 − x 2 + 2 a 2 − x 2 b 2 − x 2 = a 2 + b 2 + 2 a b cos θ \Rightarrow a^2 - x^2 + b^2 - x^2 + 2\sqrt{a^2 - x^2}\sqrt{b^2 - x^2} = a^2 + b^2 + 2ab\cos\theta ⇒ a 2 − x 2 + b 2 − x 2 + 2 a 2 − x 2 b 2 − x 2 = a 2 + b 2 + 2 ab cos θ
⇒ C D = 2 x = 2 a b sin θ a 2 + b 2 + 2 a b cos θ \Rightarrow CD = 2x = \frac{2ab\sin\theta}{\sqrt{a^2 + b^2 + 2ab\cos\theta}} ⇒ C D = 2 x = a 2 + b 2 + 2 ab c o s θ 2 ab s i n θ
The diagram is given below: