Edexcel IGCSE Math B Geometry and Trigonometry Question Bank

$$\angle OAC=\angle OBC=90^\circ$$$$\angle ACB=80^\circ$$
Reasons may use tangent-radius angles and angles in a quadrilateral, or split into two triangles using tangent, triangle angle sum, and the isosceles/bisector facts.

One correct method:
$$\tan40^\circ=\frac{10}{BC}$$
Equivalently:
$$\tan50^\circ=\frac{BC}{10}$$
Hence:
$$BC=10\tan50^\circ=11.9175\ldots$$
Final answer:
$$11.9\text{ cm}$$

O A C or O B C=0.5 x 10 x " 11.9 " (=59.58767 ) OR O A B=0.5 x 10 x 10 x 100^(=49.24038 ) OR
ACB=0.5 x " 11.9 " x " 11.9 " x " 80 "(=69.93497 )

M1ft their BC

For 2 x 0.5 x 10 x " 11.9 " oe OR
0.5 x 10 x 10 x 100^+0.5 x " 11.9 " x " 11.9 " x " 80 " oe

M1ft their BC

O A C B= awrt 1 1 9(c m^2)

A1cao

Sector area:
$$\frac{100}{360}\times\pi\times10^{2}=87.27\ldots$$
Shaded region:
$$119-87.27\ldots=31.7\ldots$$
Final answer:
$$31.7\text{ to }31.9\text{ cm}^{2}$$

$$\overrightarrow{AB}=\binom{9}{-12}$$

Marks: B1.

$$\overrightarrow{OP}=\overrightarrow{OA}+\overrightarrow{AP} =\binom{1}{8}+\frac13\binom{9}{-12} =\binom{4}{4}$$

Alternative:
$$\overrightarrow{OP}=\overrightarrow{OB}+\overrightarrow{BP} =\binom{10}{-4}-\frac23\binom{9}{-12} =\binom{4}{4}$$

Marks: M1, A1.

$$|\overrightarrow{AP}|=\sqrt{3^{2}+(-4)^{2}}=5$$
Final answer:
5

Follow through from $\overrightarrow{AP}$ in part (b), or from $\overrightarrow{AP}=\frac13\overrightarrow{AB}$, is allowed.

Area of trapezium BCHG:
$$\frac32(1.5+2.4)=5.85$$
Volume:
$$5.85\times4=23.4$$
Final answer:
$$23.4\text{ m}^{3}$$

Horizontal diagonal:
$$DB=\sqrt{4^{2}+3^{2}}=5$$
Vertical difference:
2.4-1.5=0.9$$\tan\theta=\frac{0.9}{5}$$
Final answer:
$$10.2^\circ$$
An answer of $79.8^\circ$ gains 2 marks.

$$AH=\sqrt{4^{2}+3^{2}+2.4^{2}} = \sqrt{30.76}$$
Final answer:
$$5.55\text{ m}$$

$$PQ=1.5+\frac23(2.4-1.5)=2.1$$$$AQ=\sqrt{4^{2}+2^{2}}=\sqrt{20}$$$$\tan\angle APQ=\frac{\sqrt{20}}{2.1}$$
Final answer:
$$64.8^\circ$$
Only allow $65^\circ$ from correct working.