3D geometry, sine/cosine rule, radians, arcs & sectors – IB Maths exam questions with mark scheme

1CoreCalculator4 marks

The diagram shows a sector of a circle with centre $O$, radius $12$ cm and angle $1.4$ radians.

(a) Find the length of the arc. [2]

(b) Find the area of the sector. [2]

Uses $s=r\theta$: $12\times1.4$M1
$16.8$ cmA1
Uses $A=\tfrac12r^2\theta$: $\tfrac12\times12^2\times1.4$M1
$100.8$ cm$^2$ (accept $101$)A1
Answer: (a) $16.8$ cm (b) $100.8$ cm$^2$
Examiner tip: Both formulas only work in radians; check the calculator mode and never use $\tfrac12r^2\theta$ with degrees.

2StandardCalculator5 marks

In triangle $ABC$, $BC=7.2$ cm, $AC=9.5$ cm and angle $ACB=64^\circ$.

(a) Find the length of $AB$. [3]

(b) Find the area of triangle $ABC$. [2]

Substitutes into the cosine rule: $AB^2=7.2^2+9.5^2-2(7.2)(9.5)\cos64^\circ$M1
$AB^2=82.1$ (or better)A1
$AB=9.06$ cm (accept $9.062$)A1
Uses $\tfrac12ab\sin C$: $\tfrac12(7.2)(9.5)\sin64^\circ$M1
$30.7$ cm$^2$A1
Answer: (a) $9.06$ cm (b) $30.7$ cm$^2$
Examiner tip: The cosine rule needs the angle between the two known sides; here that is $C$, opposite $AB$.

3HardCalculator6 marks

Triangle $ABC$ has $BC=6$ cm, $AC=9$ cm and angle $BAC=35^\circ$. It is possible to draw two different triangles satisfying these conditions.

(a) Find the two possible sizes of angle $ABC$. [4]

(b) Find the smaller of the two possible lengths of $AB$. [2]

Sine rule: $\dfrac{\sin B}{9}=\dfrac{\sin35^\circ}{6}$M1
$\sin B=0.8604$ (accept $0.860$)A1
$B=59.4^\circ$ (accept $59.36^\circ$)A1
$B=120.6^\circ$ (obtuse, from $180^\circ-59.4^\circ$; accept $120.64^\circ$ or $121^\circ$ (3 s.f.))A1
Uses the larger $B$: $C=180-35-120.6=24.4^\circ$ (allow unrounded $C=24.36^\circ$), then $\dfrac{AB}{\sin C}=\dfrac{6}{\sin35^\circ}$ (or cosine rule: $6^2=AB^2+9^2-2(9)(AB)\cos35^\circ$, solving the quadratic and choosing the smaller root)M1
$AB=4.31$ cm (allow $4.32$ from $C=24.4^\circ$; do not accept $4.25$ from prematurely rounded $C=24^\circ$)A1
Answer: (a) $B=59.4^\circ$ or $120.6^\circ$ (b) $4.31$ cm
Examiner tip: In the ambiguous case always ask whether $180^\circ-B$ also fits; the smaller $AB$ comes from the obtuse $B$.

4StandardCalculator6 marks

The cuboid $ABCDEFGH$ has a horizontal rectangular base $ABCD$ with $AB=8$ cm and $BC=6$ cm. The height $CG=5$ cm.

(a) Find the length of $AC$. [2]

(b) Find the length of $AG$. [2]

Pythagoras in the base: $AC^2=8^2+6^2$M1
$AC=10$ cmA1
Pythagoras in triangle $ACG$: $AG^2=10^2+5^2$M1
$AG=\sqrt{125}=11.2$ cm (accept $5\sqrt5$)A1
Angle is $\angle GAC$: $\tan\theta=\dfrac{5}{10}$ (or $\sin\theta=\dfrac{5}{\sqrt{125}}$)M1
$\theta=26.6^\circ$A1
Answer: (a) $10$ cm (b) $11.2$ cm (c) $26.6^\circ$
Examiner tip: Name the right-angled triangle first: the angle between a line and a plane uses the line's projection on the plane ($AC$), not the edge $AB$.

5HardCalculator6 marks

A circle has centre $O$ and radius $9$ cm. The chord $AB$ subtends an angle of $2.2$ radians at $O$. The shaded segment is the region between the chord $AB$ and the minor arc $AB$.

(a) Find the length of the minor arc $AB$. [2]

(b) Find the length of the chord $AB$. [2]

$s=9\times2.2$M1
$19.8$ cmA1
Chord via cosine rule or $2r\sin\frac\theta2$: $2(9)\sin1.1$M1
$AB=16.0$ cm (3 s.f.)A1
Segment = sector $-$ triangle: $\tfrac12(9)^2(2.2)-\tfrac12(9)^2\sin2.2$M1
$56.4$ cm$^2$A1
Answer: (a) $19.8$ cm (b) $16.0$ cm (c) $56.4$ cm$^2$
Examiner tip: For the segment use $\tfrac12r^2(\theta-\sin\theta)$ and evaluate $\sin\theta$ in radian mode.

6HardCalculator6 marks

A sector of a circle has radius $r$ cm and angle $\theta$ radians. The perimeter of the sector is $24$ cm and its area is $32$ cm$^2$.

(a) Show that $r^2-12r+32=0$. [3]

(b) Hence find the two possible pairs of values of $r$ and $\theta$. [3]

Perimeter: $2r+r\theta=24$A1
Area: $\tfrac12r^2\theta=32$A1
Eliminates $\theta$, e.g. $r\theta=24-2r$ so $\tfrac12r(24-2r)=32$, leading to $12r-r^2=32\Rightarrow r^2-12r+32=0$ AG (must see the substitution step)M1
Factorises/solves: $(r-4)(r-8)=0$M1
$r=4,\ \theta=4$A1
$r=8,\ \theta=1$A1
Answer: $r=4,\theta=4$ or $r=8,\theta=1$
Examiner tip: Do not forget the two straight edges ($2r$) in the perimeter of a sector.

7HardCalculator7 marks

$VABCD$ is a right pyramid. The base $ABCD$ is a horizontal square of side $10$ cm. The apex $V$ is vertically above the centre $M$ of the base, and $VM=12$ cm.

(a) Find the length of $AM$. [2]

(b) Find the angle between the edge $VA$ and the base. [2]

(d) Find the volume of the pyramid. [1]

$AM=\tfrac12\sqrt{10^2+10^2}$M1
$AM=5\sqrt2=7.07$ cmA1
$\tan\theta=\dfrac{12}{5\sqrt2}$M1
$59.5^\circ$A1
Uses midpoint $N$ of $AB$, $MN=5$: $\tan\phi=\dfrac{12}{5}$M1
$67.4^\circ$A1
$V=\tfrac13\times10^2\times12=400$ cm$^3$A1
Answer: (a) $7.07$ cm (b) $59.5^\circ$ (c) $67.4^\circ$ (d) $400$ cm$^3$
Examiner tip: The angle between a face and the base is measured along a line at right angles to the shared edge, so use the midpoint of $AB$, not the corner $A$.

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