Question

If in triangle ABC, r₁ = 2r₂= 3r₃, D is the middle point of BC. Then cos Đ ADC is equal to-

• A. $\frac { 7 } { 25 }$
• B. –$\frac { 7 } { 25 }$
• C. $\frac { 24 } { 25 }$
• D. – $\frac { 24 } { 25 }$

Answer 1

Answer: (B)

–$\frac { 7 } { 25 }$

Answer 2

r₁ = 2r₂= 3r₃

̃ $\frac { \Delta } { s - a }$ = = $\frac { 3 \Delta } { s - c }$ = (say)

Then s – a = k, s – b = 2k, s – c = 3k

̃ 3s – (a + b + c) = 6k ̃ s = 6k

̃ = = = k

So that a² = b² + c²

̃ ABC is a right angle triangle with A = 90⁰, since D is the

middle point of BC,

AD = DC (radius of the circumcircle)

̃ Đ DAC = C

̃ Đ ADC = 180⁰ – 2C

̃ cos Đ ADC = cos (180⁰ – 2C) = – cos 2C

= – [2 cos²C –1] = 1– 2 cos²C

= 1– 2 × $\frac { 16 } { 25 }$= – $\frac { 7 } { 25 }$