Question

If in triangle ABC, r₁ = 2r₂ = 3r₃, D is the middle point of BC. Then cos (ĐADC) 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 = 2r₁ = 3r₃ ̃ = $\frac { 2 \Delta } { s - b }$ = $\frac { 3 \Delta } { s - c }$ =(say)

̃ s – a = k, s – b = 2k, s – c = 3k

̃ 3s – (a + b + c) = 6k ̃ s = 6k

̃ = = = k

So, that a² = b² + c²

DABC is right angled D, ĐA = 90⁰, since D is mid point of

BC

AD = DC (radius of circum circle)

̃ ĐDAC = C ̃ ĐADC = 180⁰ – 2C

̃ cos ĐADC = cos(180⁰ – 2C) = –cos2C

= – (2 cos²C – 1) = 1 – 2cos²C

= 1 – 2 × $\frac { 16 } { 25 }$ = $\frac { - 7 } { 25 }$[From D ABC cos

C = ].