Question

In ∆ABC BC=4, and angle A is 120° the maximum of area of the ∆le is​

Answer 1

Answer:

 Area of ΔCAB ≅ 6.93( to 2 decimal places)

Step-by-step explanation:

This question really tasking, I need a dig. for it but I'll try my best, it has to be split such that ∠CAB=60° and there will be a line drawn from A to touch BC(let the point be called D) thereby forming a right angled triangle.

∴  ∠CAB=60°,  ∠BTA=90°( cos it is a right angle triangle), TC=2 (cos it is split into two equal halves)

   using PYTHAGORAS theorem; Hypotenuse= AB(dats I am looking for)(using  ∠TAB=60° )

                                                 Opposite= 2  (using  ∠TAB=60° )

                                                  Adjacent= AT(using  ∠TAB=60° )

                                                ∠TAB=60°

so, I have decided to use       Adjacent and Hypotenuse(cos area of Δ is half base × height)=SOH= sine

recall; tanФ= Adjacent ÷ Hypotenuse

         tan 60°= AT÷ 2

        1.732050808 =  AT ÷ 2  ..  (tan 60°= 1.732050808  )

              cross- multiply

      AT = 2 * 1.732050808

      AT = 3.464101615

∴ Area of ΔABT =     1       ×base(2) ×height(3.464101615)

                                   2

           

    Area of ΔABT=     1     × 2 × 3.464101615

                                   2

   

   Area of ΔABT=  3.464101615

∴ Area of ΔCAB = 2× Area of ΔABT(it was divided by 2 previously)

   Area of ΔCAB = 2× 3.464101615

  Area of ΔCAB = 6.92820323

 Area of ΔCAB ≅ 6.93( to 2 decimal places)