In a triangle A B C , the median to the side B C is of length 1 11 - 6 3 and it divides the angle A into angles of 30 ° and 45 ° . Find the length of the side B C .

Let AD be the median to side BC , which divides ∠ A in angle of 30 o & 45 o Also, Let BC = x ∴ BD = CD = x 2 Let ∠ ADC be θ . ∴ x 2 + x 2 cot θ = x 2 cot 30 o - x 2 cot 45 o ⇒ cot θ = 3 - 1 2 ∴ cos θ = 3 - 1 8 - 2 3 , sin θ = 2 8 - 2 3 Applying sine-rule in ∆ ADC , we get AD sin π - θ - 45 o = DC sin 45 o ⇒ AD sin θ + 45 o = x 2 1 2 ⇒ AD = x 2 sin 45 o · cos θ + cos 45 o sin θ ⇒ AD = x 2 cos θ + sin θ ⇒ 1 11 - 6 3 = x 2 3 - 1 8 - 2 3 + 2 8 - 2 3 ⇒ x = 2 8 - 2 3 11 - 6 3 3 + 1 ⇒ x = 2 8 - 2 3 3 + 1 2 11 - 6 3 ⇒ x = 2 8 - 2 3 44 - 24 3 + 22 3 - 36 ⇒ x = 2 8 - 2 3 8 - 2 3 = 2

Prove that : r 1 b c + r 2 c a + r 3 a b = 1 r - 1 2 R

L.H.S = r 1 bc + r 2 ca + r 3 ab = 1 abc ar 1 + br 2 + cr 3 = 1 abc a s tan A 2 + b s tan B 2 + c s tan C 2 = s abc 2 R sinA tan A 2 + 2 R sinB tan B 2 + 2 R sinC tan C 2 = s 2 R abc 2 sin A 2 · cos A 2 · tan A 2 + 2 sin B 2 · cos B 2 · tan B 2 + 2 sin C 2 · cos C 2 · tan C 2 = s 4 R abc sin 2 A 2 + sin 2 B 2 + sin 2 C 2 = s ∆ 1 - cosA 2 + 1 - cosB 2 + 1 - cosC 2 = 1 2 r 3 - cosA + cosB + cosC = 1 2 r 3 - 1 + 4 sin A 2 · sin B 2 · sin C 2 = 1 2 r 2 - 4 Rsin A 2 · sin B 2 · sin C 2 R = 1 2 r 2 - r R = 1 r - 1 2 R = R . H . S .

Prove that : tan A 2 ( a - b ) ( a - c ) + tan B 2 ( b - a ) ( b - c ) + tan C 2 ( c - a ) ( c - b ) = 1 Δ

Consider a triangle ABC having right angle at A and consider sides A B = c , A C = b & B C = a Let a = 2 , b = 1 & c = 3 As ∠ A = 90 ° Thus, area of △ A B C = 1 2 × 1 × 3 = 3 2 ⋯ 1 Now tan B = b c = 1 3 = tan 30 ° ⇒ ∠ B = 30 ° Since ∠ B = 30 ° & ∠ A = 90 ° t h e n ∠ C = 60 ° Now tan 15 ° = tan 45 ° - 30 ° = tan 45 ° - tan 30 ° 1 + tan 45 ° tan 30 ° = 1 - 1 3 1 + 1 3 = 3 - 1 3 + 1 = 3 + 1 - 2 3 2 = 4 - 2 3 2 = 2 - 3 ⋯ 2 Now we simplify L.H.S i.e., tan A 2 ( a - b ) ( a - c ) + tan B 2 ( b - a ) ( b - c ) + tan C 2 ( c - a ) ( c - b ) = tan 45 ° ( 2 - 1 ) ( 2 - 3 ) + tan 15 ° ( 1 - 2 ) ( 1 - 3 ) + tan 30 ° ( 3 - 2 ) ( 3 - 1 ) = 1 ( 2 - 1 ) ( 2 - 3 ) + 2 - 3 ( 1 - 2 ) ( 1 - 3 ) + 1 3 ( 3 - 2 ) ( 3 - 1 ) U s e 2 = 1 ( 2 - 3 ) + 2 - 3 3 - 1 + 1 3 ( 3 - 2 3 - 3 + 2 ) = 1 2 - 3 × 2 + 3 2 + 3 + 2 - 3 3 - 1 × 3 + 1 3 + 1 + 1 3 ( 5 - 3 3 ) × 5 + 3 3 5 + 3 3 = 2 + 3 1 + 2 3 - 3 + 2 - 3 3 - 1 + 5 + 3 3 3 25 - 27 = 2 + 3 + 3 - 1 2 + 5 3 + 9 - 6 = 2 + 3 + 3 2 - 1 2 - 5 3 6 - 3 2 = 3 3 2 - 5 3 6 = 3 3 2 - 5 6 = 3 × 2 3 = 2 3 = 1 △ (Using 1 ) Hence, proved.

Prove that in a triangle b c r 1 + c a r 2 + a b r 3 = 2 R a b + b a + b c + c b + c a + a c - 3 .

As we know that circum radius R = a b c 4 △ ⋯ 1 and inradius r 1 = △ s - a , r 2 = △ s - b & r 3 = △ s - c ⋯ 2 , where △ is area of triangle and s be semi perimeter and a , b & c are the sides of a triangle. Now we simplify L.H.S i.e., b c r 1 + c a r 2 + a b r 3 = s - a △ b c + s - b △ c a + s - c △ a b (Using 2 ) = 1 △ s b c + c a + a b - 3 a b c Multiply and divide by a b c = a b c △ s b c a b c + c a a b c + a b a b c - 3 = 4 R s 1 a + 1 b + 1 c - 3 (Using 1 ) = 4 R a + b + c 2 1 a + 1 b + 1 c - 3 (Using s = a + b + c 2 ) = 4 R 1 2 1 + 1 + 1 + a b + a c + b c + b a + c a + c b - 3 = 4 R 3 2 - 3 + 1 2 a b + b a + b c + c b + c a + a c = 4 R - 3 2 + 1 2 a b + b a + b c + c b + c a + a c = 2 R a b + b a + b c + c b + c a + a c - 3 = R.H.S Hence, proved.

In a triangle the angles A , B , C are in A.P. Show that 2 cos A - C 2 = a + c a 2 - a c + c 2 .

Given that angles A , B , C of ∆ ABC are in A . P . ∴ 2 B = A + C Sum of all angles of ∆ ABC is 180 ° . ⇒ A + B + C = 180 ° ⇒ 3 B = 180 ° ⇒ B = 60 ° ⇒ cos B = cos 60 ° We know that, by cosine rule cos B = a 2 + c 2 - b 2 2 a c ⇒ a 2 + c 2 - b 2 2 a c = 1 2 ⇒ a 2 + c 2 - a c = b 2 (i) We know that, by sine rule a sin A = b sin B = c sin C = k ( say ) ⇒ k sinA = a , k sinB = b , k sinC = c ∴ R . H . S = a + c a 2 - a c + c 2 = a + c b 2 = a + c b = k sinA + k sinC k sinB = sinA + sinC sinB = 2 sin A + C 2 cos A - C 2 sin B = 2 sin B cos A - C 2 sin B [ ∵ A + C = 2 B ] = 2 cos A - C 2 = L · H . S Hence, proved.

Let a , b , c be the sides of a trianlge & Δ its area. Prove that a 2 + b 2 + c 2 ≥ 4 3 Δ , and find when does the equality hold?

Consider a , b , c be the sides of a triangle then by using cosine rule cos C = a 2 + b 2 - c 2 2 a b ⇒ c 2 = a 2 + b 2 - 2 a b cos C ⋯ 1 As we know that area of triangle ABC = △ = 1 2 a b sin C ⋯ 2 Now, a 2 + b 2 + c 2 - 4 3 △ = 2 a 2 + b 2 - a b cos C - 4 3 × 1 2 a b sin C (Using 1 & 2 ) = 2 a 2 + b 2 - a b cos C + 3 sin C = 2 a 2 + b 2 - 2 a b 1 2 × cos C + 3 2 × sin C = 2 a 2 + b 2 - 2 a b × sin π 6 + C ∵ sin A cos B + cos A sin B = sin A + B ≥ 2 a 2 + b 2 - 2 a b ∵ M a x i m u m v a l u e o f sin π 6 + C = 1 ⋯ 3 ≥ 2 a - b 2 ≥ 0 ∵ M i n i m u m v a l u e o f a - b 2 = 0 ⋯ 4 Thus, a 2 + b 2 + c 2 ≥ 4 3 △ And this equality holds only when 3 & 4 hold i.e., when sin π 6 + C = sin π 2 ⇒ C = π 2 - π 6 ⇒ C = π 3 and a = b .