Geometry & Trigonometry · Paper 2
3D Trigonometry & Sectors
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Section 1 of 5
Seeing 3D — the corner-of-a-room idea
Every 3D problem becomes a 2D problem. Pick the right triangle inside the 3D shape, then use one of four tools. That's the whole game.
Look at any corner of a room: floor meets wall meets wall. Three edges, all at right angles. Anywhere you see a vertical line meeting a horizontal line, you've got a right-angled triangle to start with.
The 3D Toolkit — Must learn
Pythagoras (right-angled only) $a^{2} + b^{2} = c^{2}$
Sine rule $\dfrac{a}{\sin A} = \dfrac{b}{\sin B} = \dfrac{c}{\sin C}$
Cosine rule $a^{2} = b^{2} + c^{2} - 2bc \cos A$
Area $\text{Area} = \dfrac{1}{2} ab \sin C$
Pick the triangle. Label what you know. Choose the tool.
Worked example — diagonal of a box
A box has length $5$ m, width $4$ m, height $3$ m. Find the diagonal from one bottom corner to the opposite top corner.
Step 1. Flatten the floor first. Find the diagonal $d$ across the base ($5 \times 4$ rectangle).
$d^{2} = 5^{2} + 4^{2} = 25 + 16 = 41$
$d = \sqrt{41}$
Step 2. Now stand the diagonal up. The space diagonal $D$ is the hypotenuse of a right-angled triangle with horizontal leg $\sqrt{41}$ and vertical leg $3$.
$D^{2} = (\sqrt{41})^{2} + 3^{2} = 41 + 9 = 50$
$D = \sqrt{50} = 5\sqrt{2}$ m
Two right triangles, in sequence. That's all 3D is — finding the right 2D triangles and stringing them together.
Section 2 of 5
Two vertical poles, equal wires
$[sp]$ and $[tq]$ are vertical poles each of height $10$ m. The points $p$, $q$, $r$ lie on level ground. Two wires of equal length join $s$ and $t$ to $r$, so $|sr| = |tr|$.
Given $|pr| = 8$ m, $\angle pqr = 30°$, $\angle prq = 120°$, find
(i) $|pq|$ to the nearest metre (ii) $|sr|$ in surd form (iii) $\angle srt$ to the nearest degree.
Two vertical poles sp and tq, each height 10 m. Equal wires from s and t meet at r.
(i) Find $|pq|$
Look at the ground triangle $pqr$. The three angles sum to $180°$:
Ground triangle pqr (flat). Isoceles: 120° at r, 30° at p and q. Equal sides 8.
Vertical pole sp = 10, ground distance pr = 8. Right-angle at p. The wire sr is the hypotenuse, length x.
Top triangle srt. Base st = 14, equal slanted sides √164. Find angle A at r.
Right triangle psr (flattened). 60° at p, right angle at r. Vertical leg h, horizontal leg x.
Right triangle qsr (flattened). 30° at q, right angle at r. Vertical leg h, horizontal leg y.
Ground triangle pqr (flat). Apex at r with unknown angle C. Sides h/√3 and √3 h.
The three midpoints sit on the three edges of the corner: two are $\tfrac{1}{2}(8) = 4$ cm from the corner, and one is $\tfrac{1}{2}(4) = 2$ cm from the corner.
Each side of triangle $abc$ runs between two midpoints. Each is the hypotenuse of a little right-angled triangle hidden inside the block:
The two $\sqrt{20}$ sides ($4$ along one edge, $2$ along the perpendicular edge):
$|ab|^{2} = 4^{2} + 2^{2} = 20 \;\;\Rightarrow\;\; |ab| = \sqrt{20}$
$|bc|^{2} = 4^{2} + 2^{2} = 20 \;\;\Rightarrow\;\; |bc| = \sqrt{20}$
The $\sqrt{32}$ side ($4$ along one $8$-edge, $4$ along the other $8$-edge):
$|ac|^{2} = 4^{2} + 4^{2} = 32 \;\;\Rightarrow\;\; |ac| = \sqrt{32}$
Triangle $abc$ is isoceles: two sides $\sqrt{20}$ meeting at vertex $b$, third side $\sqrt{32}$ opposite $b$.
Triangle abc (flat). Isoceles: ab = √32, ac = bc = √20. Apex angle A at c.
(i) Slanted edge
Step A — half the diagonal of the base. The apex sits above the centre, so we need the distance from a base corner to the centre. Half of each side is $\tfrac{230}{2} = 115$. Pythagoras on the floor:
$t^{2} = 115^{2} + 115^{2} = 2(115)^{2}$
$t = 115\sqrt{2}$ (corner-to-centre distance)
Step B — stand the apex up. The slanted edge $y$ runs from base corner up to the apex. Right-angled triangle: horizontal leg $115\sqrt{2}$, vertical leg $146$.
Right triangle inside the pyramid: vertical 146, horizontal half-diagonal 115√2, hypotenuse y = slant edge.
One triangular face of the pyramid. Equal sides 219, base 230, angle A at apex.
Quick conversions
$180° = \pi$ rad $\Rightarrow$ $1° = \dfrac{\pi}{180}$ rad and $1$ rad $= \dfrac{180°}{\pi}$
$30° = \dfrac{\pi}{6}$ $45° = \dfrac{\pi}{4}$ $60° = \dfrac{\pi}{3}$ $90° = \dfrac{\pi}{2}$ $120° = \dfrac{2\pi}{3}$
Worked example
A sector has radius $6$ cm and central angle $\dfrac{\pi}{3}$ rad. Find the arc length and the sector area.
$\ell = r\theta = 6 \cdot \dfrac{\pi}{3} = 2\pi$ cm
$A = \dfrac{1}{2}r^{2}\theta = \dfrac{1}{2}(36)\dfrac{\pi}{3} = 6\pi$ cm²
If the question had said "$60°$" instead, the first move is always: convert to $\dfrac{\pi}{3}$ rad, then plug in.
Section 3 of 5
Two intersecting circles
$C_{1}$ is a circle with centre $a$ and radius $r$. $C_{2}$ is a circle with centre $b$ and radius $r$.
$C_{1}$ and $C_{2}$ intersect at $k$ and $p$, with $a \in C_{2}$ and $b \in C_{1}$.
(i) Find, in radians, the measure of $\angle kap$.
(ii) Calculate the area of the shaded region (the overlap). Give your answer in terms of $r$ and $\pi$.
(i) The angle $\angle kap$
$b \in C_{1}$ means $b$ is on $C_{1}$, so $|ab| = r$ (radius of $C_{1}$). Similarly $|ab| = r$ from the other condition (consistent).
Look at triangle $akb$: $|ak| = r$ (radius $C_{1}$), $|bk| = r$ (radius $C_{2}$), $|ab| = r$. All three sides equal — equilateral. So $\angle kab = 60° = \dfrac{\pi}{3}$.
By symmetry, triangle $apb$ is also equilateral, so $\angle pab = \dfrac{\pi}{3}$.
$\angle kap = \angle kab + \angle pab = \dfrac{\pi}{3} + \dfrac{\pi}{3}$
$\angle kap = \dfrac{2\pi}{3}$ rad ($= 120°$)
(ii) The shaded overlap
The overlap (a "lens") is symmetric — split it into two equal segments, one from each circle. Each segment is cut from its circle by the chord $kp$.
In $C_{1}$, the chord $kp$ subtends $\angle kap = \dfrac{2\pi}{3}$ at the centre. Segment formula:
$\text{Segment} = \dfrac{1}{2}r^{2}\left(\theta - \sin\theta\right) = \dfrac{1}{2}r^{2}\left(\dfrac{2\pi}{3} - \sin\dfrac{2\pi}{3}\right)$
$= \dfrac{1}{2}r^{2}\left(\dfrac{2\pi}{3} - \dfrac{\sqrt{3}}{2}\right)$
$= \dfrac{\pi r^{2}}{3} - \dfrac{\sqrt{3}\,r^{2}}{4}$
Overlap is two of these:
$\text{Shaded area} = 2\left(\dfrac{\pi r^{2}}{3} - \dfrac{\sqrt{3}\,r^{2}}{4}\right)$
$\text{Shaded area} = \dfrac{2\pi r^{2}}{3} - \dfrac{\sqrt{3}\,r^{2}}{2}$
Equilateral triangle $\Rightarrow$ angle $\dfrac{\pi}{3}$ at each corner. The whole problem hinges on spotting that.
Section 4 of 5
Arc, chord, and a hidden equilateral fact
$p$, $q$ and $r$ are points on a circle with centre $k$. The radius is $2$ cm and the length of the minor arc $pq$ is $\dfrac{5\pi}{3}$ cm.
(i) Find the length of the chord $[pq]$, correct to two decimal places.
(ii) If $|pq| = |pr|$, find $|rq|$.
(i) Chord $[pq]$
Step A — central angle from the arc.
$\ell = r\theta \;\;\Rightarrow\;\; \dfrac{5\pi}{3} = 2\theta$
$\theta = \dfrac{5\pi}{6}$ rad ($= 150°$)
Step B — chord via cosine rule. Triangle $kpq$ is isoceles with $|kp| = |kq| = 2$ and included angle $\dfrac{5\pi}{6}$:
$|pq|^{2} = 2^{2} + 2^{2} - 2(2)(2)\cos\dfrac{5\pi}{6}$
$= 8 - 8\left(-\dfrac{\sqrt{3}}{2}\right) = 8 + 4\sqrt{3}$
$|pq| = \sqrt{8 + 4\sqrt{3}} \approx 3.864$
$|pq| \approx 3.86$ cm
(ii) Find $|rq|$ given $|pq| = |pr|$
Key idea: equal chords subtend equal central angles. Since $|pq| = |pr|$, the central angle for chord $pr$ equals the central angle for chord $pq$:
$\angle pkr = \angle pkq = \dfrac{5\pi}{6}$
The three central angles around $k$ sum to $2\pi$:
$\angle pkq + \angle pkr + \angle qkr = 2\pi$
$\dfrac{5\pi}{6} + \dfrac{5\pi}{6} + \angle qkr = 2\pi$
$\angle qkr = 2\pi - \dfrac{5\pi}{3} = \dfrac{\pi}{3}$
So triangle $kqr$ is isoceles with $|kq| = |kr| = 2$ and included angle $\dfrac{\pi}{3} = 60°$. Two equal sides and a $60°$ apex angle ⇒ equilateral. Therefore
$|rq| = 2$ cm
Or by cosine rule: $|rq|^{2} = 4 + 4 - 8\cos(\pi/3) = 8 - 4 = 4$, $|rq| = 2$. Same answer.
Section 5 of 5
Sector circumscribed by a circle
A sector has two straight sides of length $k$ meeting at an angle of $60°$. A circle is drawn that passes through the apex of the sector and the two endpoints of its arc.
(i) Find the radius of the circumscribing circle in terms of $k$.
(ii) Show that the area enclosed by the circle is double the area of the sector.
(i) Radius of the circumscribing circle
Call the apex $A$ and the arc endpoints $B$, $C$. Triangle $ABC$ has $|AB| = |AC| = k$ and apex angle $\angle BAC = 60°$.
Two equal sides with a $60°$ angle between them ⇒ the triangle is equilateral, so $|BC| = k$ too, and all angles are $60°$.
Now find the circumradius $R$ of an equilateral triangle of side $k$. Drop the centre of the circle ($O$, say) into the middle. Side $|BC| = k$ subtends $\angle BOC = 2(60°) = 120°$ at the centre (angle at centre is twice angle at circumference).
Sine rule on triangle $OBC$ (sides $R$, $R$, $k$ — the two angles opposite the $R$'s are each $\tfrac{1}{2}(180° - 120°) = 30°$):
$\dfrac{R}{\sin 30°} = \dfrac{k}{\sin 120°}$
$R = \dfrac{k \sin 30°}{\sin 120°} = \dfrac{k \cdot \tfrac{1}{2}}{\tfrac{\sqrt{3}}{2}} = \dfrac{k}{\sqrt{3}}$
$R = \dfrac{k}{\sqrt{3}} = \dfrac{\sqrt{3}}{3}\,k$
(ii) Show that the circle's area is twice the sector's area
Sector area ($60° = \pi/3$ rad):
$A_{\text{sector}} = \dfrac{1}{2}k^{2}\theta = \dfrac{1}{2}k^{2}\cdot\dfrac{\pi}{3} = \dfrac{\pi k^{2}}{6}$
Circle area:
$A_{\text{circle}} = \pi R^{2} = \pi\left(\dfrac{k}{\sqrt{3}}\right)^{2} = \dfrac{\pi k^{2}}{3}$
Ratio:
$\dfrac{A_{\text{circle}}}{A_{\text{sector}}} = \dfrac{\pi k^{2}/3}{\pi k^{2}/6} = \dfrac{6}{3} = 2$ ✓
$A_{\text{circle}} = 2 \cdot A_{\text{sector}}$
"Show that" questions: produce both quantities separately, then take the ratio (or subtract, or whatever the relationship asks). The work itself is the answer.
That's 3D Trigonometry & Sectors.
Every 3D problem reduces to 2D triangles — find the right one, label it, then pick from Pythagoras / sine rule / cosine rule / $\tfrac{1}{2}ab\sin C$. For sectors, $\theta$ goes in radians and three formulas (arc, sector area, segment area) cover everything.