Q24

q24 - The sides of a Triangle are three consecutive natural numbers and its largest angle is twice the smallest one. Determine perimeter of the triangle.
Answer: 15

Solution:

Let the sides be $n, n+1, n+2$.
i.e., $AC = n, AB = n+1, BC = n+2$
Smallest angle is $B$ and largest one is $A$.

Here $\angle A = 2 \angle B$

Also, $\angle A + \angle B + \angle C = 180^\circ$

$$\Rightarrow 3 \angle B + \angle C = 180^\circ \Rightarrow \angle C = 180^\circ - 3 \angle B$$

We have, sine law as,

$$\frac{\sin A}{n+2} = \frac{\sin B}{n} = \frac{\sin C}{n+1}$$ $$\Rightarrow \frac{\sin 2B}{n+2} = \frac{\sin B}{n} = \frac{\sin (180^\circ - 3B)}{n+1}$$ $$\Rightarrow \frac{\sin 2B}{n+2} = \frac{\sin B}{n} = \frac{\sin 3B}{n+1}$$

From (i) and (ii):

$$\frac{2 \sin B \cos B}{n+2} = \frac{\sin B}{n}$$ $$\Rightarrow \cos B = \frac{n+2}{2n} \quad \text{----- (iv)}$$

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and from (ii) and (iii)

        sin B        3 sin B - 4sin³B
       —— = —————————————
         n               n + 1

⇒   sin B     =  (3 sin B (3 − 4sin²B)) / (n + 1)
     —— 
      n

⇒   (n + 1)/n = 3 − 4 (1 − cos²B)   ------ (v)


from (iv) and (v), we get

      (n + 1)/n = −1 + 4 ((n + 2)/2n)²

⇒   (n + 1)/n + 1 = (n² + 4n + 4)/n²

⇒   (2n + 1)/n = (n² + 4n + 4)/n²

⇒   2n² + n = n² + 4n + 4

⇒   n² − 3n − 4 = 0

⇒   (n − 4)(n + 1) = 0

n = 4 or −1
n = 4
Hence the sides are 4, 5, 6
Perimeter = 4 + 5 + 6 = 15