There exists a unique triangle whose side lengths are consecutive natural number and one of the whose angle is twice the measure of one of the other find the perimeter of this triangle
Answers
Solution :-
Let us assume that, three sides of given ∆ are :-
- a(smallest side) = (x - 1) unit .
- b = x unit .
- c(Largest side) = (x + 1) unit .
since angle opposite to larger side is greater in measure .
- ∠C = 2∠A .
applying sine rule we get,
→ a/sin A = c / sin C
→ (x - 1)/sin A = (x + 1) / sin 2A
→ sin 2A / sin A = (x + 1)/(x - 1)
→ (2•sin A•cos A)/sin A = (x + 1)/(x - 1)
→ 2•cos A = (x + 1)/(x - 1) --------- Eqn.(1)
now, using cosine rule for cos A ,
→ cos A = (b² + c² - a²)/2bc
→ cos A = [(x)² + (x + 1)² - (x - 1)²] [2 * x * (x + 1)]
→ 2•cos A = [x² + x² + 2x + 1 - x² + 2x - 1] / [x(x + 1)]
→ 2•cos A = (x² + 4x)/x(x + 1)
→ 2•cos A = x(x + 4)/x(x + 1)
→ 2•cos A = (x + 4)/(x + 1) ---------- Eqn.(2)
comparing Eqn.(1) and Eqn.(2) we get,
→ (x + 1)/(x - 1) = (x + 4)/(x + 1)
→ (x + 1)(x + 1) = (x - 1)(x + 4)
→ x² + 2x + 1 = x² + 3x - 4
→ 3x - 2x = 1 + 4
→ x = 5 unit .
therefore,
→ Perimeter of given ∆ = (x - 1) + x + (x + 1) = 3x = 3 * 5 = 15 units (Ans.)
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