The sides of a triangle are three consecutive natural numbers and its largest angle is twice the smallest one, then the sides of the triangle (in units) are
‌ Let ‌AC=n,AB=n+1,BC=n+2 ∴‌‌ Largest angle is A and smallest angle is B. ∴‌‌A=2B Since A+B+C=180∘ ‌∴‌‌3B+C=180∘ ‌⇒‌‌C=180∘−3B ‌⇒‌‌sin‌C=sin‌(180∘−3B)=sin‌3B By sine rule, ‌‌
sin‌A
n+2
=‌
sin‌B
n
=‌
sin‌C
n+1
‌‌
sin‌2B
n+2
=‌
sin‌B
n
=‌
sin‌3B
n+1
‌⇒‌
2‌sin‌B‌cos‌B
n+2
=‌
sin‌B
n
=‌
3‌sin‌B−4‌sin‌3B
n+1
‌‌
2‌cos‌B
n+2
=‌
1
n
=‌
3−4‌sin‌2B
n+1
∴‌‌cos‌B=‌
n+2
2n
,3−4‌sin‌2B=‌
n+1
n
∴‌‌3−4(1−cos2B)=‌
n+1
n
∴‌‌3−4+4(‌
n+2
2n
)2=‌
n+1
n
⇒−1+‌
n2+4n+4
n2
=‌
n+1
n
‌⇒−n2+n2+4n+4=n2+n ‌⇒n2−3n−4=0 ‌⇒(n+1)(n−4)=0 ‌⇒n=−1‌ or ‌n=4 But n cannot be negative. ∴n=4 ∴ The sides of the ∆ are 4,5,6.