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 n, n+1, n+2 . i.e., A C = n , A B = n + 1 , B C = n + 2 AC = n, AB = n+1, BC = n+2 Smallest angle is B B and largest one is A A . Here ∠ A = 2 ∠ B \angle A = 2 \angle B Also, ∠ A + ∠ B + ∠ C = 180 ∘ \angle A + \angle B + \angle C = 180^\circ ⇒ 3 ∠ B + ∠ C = 180 ∘ ⇒ ∠ C = 180 ∘ − 3 ∠ B \Rightarrow 3 \angle B + \angle C = 180^\circ \Rightarrow \angle C = 180^\circ - 3 \angle B We have, sine law as, sin A n + 2 = sin B n = sin C n + 1 \frac{\sin A}{n+2} = \frac{\sin B}{n} = \frac{\sin C}{n+1} ⇒ sin 2 B n + 2 = sin B n = sin ( 180 ∘ − 3 B ) n + 1 \Rightarrow \frac{\sin 2B}{n+2} = \frac{\sin B}{n} = \frac{\sin (180^\circ - 3B)}{n+1} ⇒ sin 2 B n + 2 = sin B n = sin 3 B n + 1 \Rightarrow \frac{\sin 2B}{n+2} = \frac{\sin B}{n} = \frac{\sin 3B}{n+1} From (i)...