Question

In ∆ABC, given above, AB = 8 cm, BC = 6 cm and AC = 3 cm. Calculate the length of OC.

Answer 1

Applying cosine rule in triangle ACB:
AB=8cm. AC= 3cm. BC = 6cm.
COS ACB = (AC^2+BC^2-AB^2)/(2×AC×BC)
= (9+36-64)/(2×3×6) = -19/36.

TRiangle ACO:

Cos ACO = cos(180°-ACB) = - Cos ACB
= 19/36
Cos ACO = CO/ AC

CO = 3×19/36= 19/12 cm
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If trigonometry is not known then we find the answer by finding area of triangle ABC in two ways.

s = semiperimeter
= (3+6+8)/2= 17/2 cm

Ar(ABC) = sqrt [s (s-a)(s-b)(s-c) ]
= SQRT[17/2 × 11/2 × 5/2 × 1/2) = sqrt (55×17/16).

Ar(ACB) = 1/2× BC × AO. = 3 × AO. cm^2.

=> AO^2 = 55×17/(9×16) cm^2.
=> Using Pythagoras theorem in triangle ACO:

CO^2 = 3^2 - 55×17/144)
= 361/144 cm^2.
=> CO = 19/12 cm...

Answer 2

Heya User,

Consider AO = x || OC = y
--> We have -> AB = 8cm || BC = 6cm || AC = 3cm

We apply Pythagoras :->
--> OA² + OB² = AB²
=> x² + ( y + 6 )² = 8²  ---> (i)

Again , 
--> OA² + OC² = AC²
=> x² + y² = 3²             ---> (ii)

--> Subtracting (ii) from (i) :->
     ( y + 6 )² - y² = 64 - 9
=> ( 12y + 36 ) = 55
=> 12y = 55 - 36 = 19
=> OC = y = 19/12 cm ... Ans.