16N.2.AHL.TZ0.H_7

* This question is from an exam for a previous syllabus, and may contain minor differences in marking or structure.

METHOD 1

let $\text{AC}=x$

$3^2=x^2+4^2-8x\cos \frac{\pi }{9}$    M1A1

attempting to solve for $x$     (M1)

$x=1.09,6.43$    A1A1

METHOD 2

let $\text{AC}=x$

using the sine rule to find a value of $C$     M1

$4^2=x^2+3^2-6x\cos (152.869{\dots }^{\circ })\Rightarrow x=1.09$    (M1)A1

$4^2=x^2+3^2-6x\cos (27.131{\dots }^{\circ })\Rightarrow x=6.43$    (M1)A1

METHOD 3

let $\text{AC}=x$

using the sine rule to find a value of $B$ and a value of $C$     M1

obtaining $B=132.869{\dots }^{\circ },7.131{\dots }^{\circ }$ and $C=27.131{\dots }^{\circ },152.869{\dots }^{\circ }$     A1

$(B=2.319\dots ,0.124\dots$ and $C=0.473\dots ,2.668\dots )$

attempting to find a value of $x$ using the cosine rule     (M1)

$x=1.09,6.43$    A1A1

 

Note: Award M1A0(M1)A1A0 for one correct value of $x$

 

[5 marks]

$\frac{1}{2}\times 4\times 6.428\dots \times \sin \frac{\pi }{9}$ and $\frac{1}{2}\times 4\times 1.088\dots \times \sin \frac{\pi }{9}$     (A1)

($4.39747\dots$ and $0.744833\dots$)

let $D$ be the difference between the two areas

$D=\frac{1}{2}\times 4\times 6.428\dots \times \sin \frac{\pi }{9}-\frac{1}{2}\times 4\times 1.088\dots \times \sin \frac{\pi }{9}$    (M1)

$(D=4.39747\dots -0.744833\dots )$

$=3.65\text{ (c}{\text{m}}^2)$    A1

[3 marks]

[N/A]

[N/A]