Question

ABC is a line segment of length 8 cm locate at a point c on a b such that AC is equal to 1/3 CB please solve this question in notebook with diagram please please please please need your help please help​

Answer 1

Let the lenth of AD,BD and DC be m, x and y respectively.

Now in ∆ ABC ,applying cosine rule to find out the values of cos B and cos C.

c2=a2+b2-2ab cos C⇒cos C= c2-(a2+b2)-2ab= 9-(36+25)-2×6×5=5260 and sin C = 1-cos2C=0.498Similarly b2=a2+c2-2ac cos B ⇒cos B = b2-(a2+c2)-2×a×c=25-(36+9)-2×6×3=2036 and sin B =1-cos2B=0.831Now DA bisects ∠BAC So ∠BAD = ∠DACSo applying sine rule in ∆ABD and ∆ADC xsin(A2)=msin B.........(1) and ysin(A2)=msin C............(2)Dividing (1) by (2) we get xy=sin Csin B ⇒xy = 0.4980.831 =0.599.....................(3)And from the figure x+y = a= 6..............(4)Using (3) and (4)0.599y+y =6 or y = 61.599 =3.75Putting the value of y in (4) , we get x = 6-3.75 = 2.25 cmHence length of BD = 2.25 cmSo option (b) is correct