Question

in ∆ABC ✓3sinC=2secA-tanA and one of the sides has lengths 2 units then the maximum area of ∆ABC​

Answer 1

It has given that, in triangle ABC, √3 sinC = 2secA - tanA and one of the sides has lengths 2 units.

To find : The maximum area of ∆ABC

solution : given expression, √3 sinC = 2secA - tanA

⇒√3 sinC = 2/cosA - sinA/cosA

⇒√3 sinC = (2 - sinA)/cosA

⇒√3 sinC cosA = 2 - sinA

⇒√3 sinC cosA + sinA = 2

we know, -√(a² + b²) ≤ a sinx + b cosx ≤ √(a² + b²)

here, if we assume sinC = 1

then, maximum value of √3 cosA + sinA = √(√3² + 1²) = 2

it means, sinC = 1 is true. so C = π/2

now √3cosA + sinA = 2

⇒2(√3/2 cosA + 1/2 sinA ) = 2

⇒(sin60° cosA + cos60° sinA) = 1

⇒sin(A + π/3) = sinπ/2

⇒A = π/6

B = π/2 - π/6 = π/3

ABC is a right angled triangle.

using sine rule,

a/sinA = b/sinB = c/sinC

⇒a/sin30° = b/sin60° = c/sin90°

⇒2a = 2b/√3 = c

if we assume, smallest side is 2 units

i.e., a = 2

then, b = 2 × 2 × √3/2 = 2√3 units

and c = 2 × 2 = 4 units

now maximum area = 1/2 × a × b

= 1/2 × 2 × 2√3

= 2√3 sq units.

Therefore the maximum area of triangle ∆ABC is 2√3 sq units.

Answer 2

Step-by-step explanation:

the answer is 2√3 units