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In `Delta ABC` if `r_1=2r_2=3r_3` and `D` is the mid point of `BC` then `cos/_ADC=`A. `(7)/(25)`B. `-(7)/(25)`C. `(24)/(25)`D. `-(24)/(25)` |
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Answer» Correct Answer - D `r_(1)=2r_(1)=3r_(3)` `rArr (Delta)/(s-a)=(2Delta)/(s-b)=(3Delta)/(s-c)=(Delta)/(k)` (say) `rArr s-a=k, s-b=2k, s-c=3k` `rArr 3s-(a+b+c)=6k rArr s=6k` `rArr (a)/(5)=(b)/(4)=(c )/(3)=k` So, `a^(2)=b^(2)+c^(2)` `Delta ABC` is right angled `Delta, angle A =90^(@)` and D is midpoint of BC, AD = DC (radius of circumicrcle) `rArr angle DAC = C rArr angle ADC = 180^(@)-2C` `rArr cos angle ADC` `=cos(180^(@)-2C)` `=-cos 2C` `-(2cos^(2)C-1)` `=1-2cos^(2)C` `=1-2xx(16)/(25)=(-7)/(25)` `["from " Delta ABC, cos C=(b)/(a)=(4)/(5)]` |
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