Answer:
x = 3.08 (3 s.f.)
Step-by-step explanation:
Area of a Triangle (using sine)
[tex]\sf Area=\dfrac{1}{2}ac \sin B[/tex]
where:
- a and c are adjacent sides
- B is the included angle
Given:
- Area = 4√2 m²
- a = (2x - 3) m
- c = (x + 2) m
- B = 45°
Substitute the given values into the formula:
[tex]\implies 4\sqrt{2}=\dfrac{1}{2}(2x-3)(x+2) \sin 45^{\circ}[/tex]
[tex]\implies 8\sqrt{2}=(2x-3)(x+2) \sin 45^{\circ}[/tex]
[tex]\implies 8\sqrt{2}=(2x-3)(x+2) \left(\dfrac{\sqrt{2}}{2}\right)[/tex]
[tex]\implies 8\sqrt{2}\left(\dfrac{2}{\sqrt{2}}\right)=(2x-3)(x+2)[/tex]
[tex]\implies (2x-3)(x+2) =16[/tex]
[tex]\implies 2x^2+4x-3x-6=16[/tex]
[tex]\implies 2x^2+x-22=0[/tex]
Use the Quadratic Formula to solve for x:
[tex]x=\dfrac{-b \pm \sqrt{b^2-4ac} }{2a}\quad\textsf{when }\:ax^2+bx+c=0[/tex]
[tex]\implies a=2, \quad b=1, \quad c=-22[/tex]
[tex]\implies x=\dfrac{-1 \pm \sqrt{1^2-4(2)(-22)} }{2(2)}[/tex]
[tex]\implies x=\dfrac{-1 \pm \sqrt{177}}{4}[/tex]
[tex]\implies x=3.08, -3.58\:\: \sf (3 \:s.f.)[/tex]
As distance is positive:
[tex]\implies x=3.08 \:\:(3 \sf \:s.f.) \:\textsf{only}[/tex]
