Find AM.

Find ${\text{A}}\mathop {\text{M}}\limits^ \wedge  {\text{P}}$ in radians.

Find the area of the shaded region.

METHOD 1

PC $= \frac{{\sqrt 3 }}{2}$ or 0.8660       (M1)

PM $= \frac{1}{2}$PC $= \frac{{\sqrt 3 }}{4}$ or 0.4330     (A1)

AM $= \sqrt {\frac{1}{4} + \frac{3}{{16}}}$

$= \frac{{\sqrt 7 }}{4}$ or 0.661 (m)     A1

METHOD 2

using the cosine rule

AM² $= {1^2} + {\left( {\frac{{\sqrt 3 }}{4}} \right)^2} - 2 \times \frac{{\sqrt 3 }}{4} \times {\text{cos}}\left( {30^\circ } \right)$      M1A1

AM $= \frac{{\sqrt 7 }}{4}$ or 0.661 (m)     A1

[3 marks]

tan (${\text{A}}\mathop {\text{M}}\limits^ \wedge  {\text{P}}$) $= \frac{2}{{\sqrt 3 }}$ or equivalent      (M1)

= 0.857      A1

[2 marks]

EITHER

$\frac{1}{2}{\text{A}}{{\text{M}}^2}\left( {2\,{\text{A}}\mathop {\text{M}}\limits^ \wedge  {\text{P}} - {\text{sin}}\left( {2\,{\text{A}}\mathop {\text{M}}\limits^ \wedge  {\text{P}}} \right)} \right)$     (M1)A1

OR

$\frac{1}{2}{\text{A}}{{\text{M}}^2} \times 2\,{\text{A}}\mathop {\text{M}}\limits^ \wedge  {\text{P}} -  = \frac{{\sqrt 3 }}{8}$     (M1)A1

= 0.158(m²)      A1

Note: Award M1 for attempting to calculate area of a sector minus area of a triangle.

[3 marks]