Sec 2 Maths: Pythagoras’ Theorem & Trigonometry practice questions with worked solutions

(a) Value of $\theta$

Divide both sides by 2, then use the inverse sine.

$$\begin{aligned}\sin\theta &= \frac{16}{21}\div 2 = \frac{8}{21}\\ \theta &= \sin^{-1}\left(\frac{8}{21}\right)\\ &= 22.39^\circ\\ &= 22.4^\circ \text{ (1 d.p.)}\end{aligned}$$

(b) Value of $x$

Multiply both sides by 18.5.

$$\begin{aligned}x &= 18.5\tan 72^\circ\\ &= 56.94\\ &= 56.9 \text{ (3 s.f.)}\end{aligned}$$

The longest side is $DE$. Compare its square with the sum of the squares of the other two sides.

$$\begin{aligned}DE^2 &= 2.4^2 = 5.76\\ EF^2 + DF^2 &= 2^2 + 1.5^2 = 4 + 2.25 = 6.25\end{aligned}$$

Since $DE^2 \ne EF^2 + DF^2$, the converse of Pythagoras’ theorem does not apply, so triangle $DEF$ is not a right-angled triangle.

Expand the left side using the difference of two squares.

$$\begin{aligned}(x+y)(x-y) &= 36\\ x^2 – y^2 &= 6^2\\ x^2 &= 6^2 + y^2\end{aligned}$$

The square of the longest side equals the sum of the squares of the other two sides, so by the converse of Pythagoras’ theorem the triangle is right-angled (the right angle is opposite the side of length $x$ cm).

(a) Length of $AC$

In right-angled triangle $AMC$, $MC$ is opposite angle $MAC$ and $AC$ is the hypotenuse.

$$\begin{aligned}\sin\angle MAC &= \frac{MC}{AC}\\ \frac{30}{AC} &= \frac35\\ AC &= 50 \text{ cm}\end{aligned}$$

(b) $\triangle PBC$ is right-angled

Since $AB:AC = 4:5$, $BC$ is $\tfrac15$ of $AC$.

$$\begin{aligned}BC &= \frac15 \times 50 = 10 \text{ cm}\\ BC^2 + PC^2 &= 10^2 + 24^2 = 676\\ PB^2 &= 26^2 = 676\end{aligned}$$

Since $BC^2 + PC^2 = PB^2$, by the converse of Pythagoras’ theorem $\triangle PBC$ is right-angled, with the right angle at $C$.

In right-angled triangle $ABC$, $BC = 0.4$ m is adjacent to angle $ACB$ and $AC = 2$ m is the hypotenuse.

$$\begin{aligned}\cos\angle ACB &= \frac{0.4}{2}\\ \angle ACB &= \cos^{-1}(0.2)\\ &= 78.46^\circ\end{aligned}$$

Since $78.46^\circ$ is not between $74^\circ$ and $76^\circ$, the ladder is not in a safe position (it is too steep).

(a) Length of $PQ$

In right-angled triangle $PQS$, $PQ$ is adjacent to the $36^\circ$ angle and $QS$ is the hypotenuse.

$$\begin{aligned}\cos 36^\circ &= \frac{PQ}{6}\\ PQ &= 6\cos 36^\circ\\ &= 4.8541\\ &= 4.85 \text{ cm (3 s.f.)}\end{aligned}$$

(b) Angle $PRQ$

In right-angled triangle $PQR$, $PQ$ is opposite angle $PRQ$ and $QR$ is the hypotenuse.

$$\begin{aligned}\sin\angle PRQ &= \frac{4.8541}{10}\\ \angle PRQ &= \sin^{-1}(0.48541)\\ &= 29.04^\circ\\ &= 29.0^\circ \text{ (1 d.p.)}\end{aligned}$$

(a) Distance $AC$

Distance = speed × time, with half a minute = 30 s.

$$\begin{aligned}AC &= 78 \times 30\\ &= 2340 \text{ m}\\ &= 2.34 \text{ km}\end{aligned}$$

(b) Height $BC$

In right-angled triangle $ABC$, $BC$ is opposite the $15^\circ$ angle and $AC$ is the hypotenuse.

$$\begin{aligned}\sin 15^\circ &= \frac{BC}{2340}\\ BC &= 2340\sin 15^\circ\\ &= 605.64\\ &= 606 \text{ m (3 s.f.)}\end{aligned}$$

(c) Cheapest airline

Airline A: 5% off the full cost.

$$\begin{aligned}2(500) + 3(350) &= 2050\\ \tfrac{95}{100} \times 2050 &= 1947.50\end{aligned}$$

Airline B: Manila is more than 2000 km away, so the 2% surcharge applies.

$$\begin{aligned}5(380) &= 1900\\ \tfrac{102}{100} \times 1900 &= 1938\end{aligned}$$

Airline C: each of the 2 adult tickets gives 1 free child ticket, so only 1 child ticket is paid for.

$$\begin{aligned}2(700) + 1(600) &= 2000\end{aligned}$$

The totals are $1947.50, $1938 and $2000, so Jane should choose Airline B.

(a) Angle $ADB$

In right-angled triangle $ABC$, find $BC$ first.

$$\begin{aligned}\tan 54^\circ &= \frac{15}{BC}\\ BC &= \frac{15}{\tan 54^\circ} = 10.898 \text{ m}\\ BD &= 10.898 + 12 = 22.898 \text{ m}\end{aligned}$$

In right-angled triangle $ABD$:

$$\begin{aligned}\tan\angle ADB &= \frac{15}{22.898}\\ \angle ADB &= 33.22^\circ\\ &= 33.2^\circ \text{ (1 d.p.)}\end{aligned}$$

(b) Length $AD$

By Pythagoras’ theorem in triangle $ABD$:

$$\begin{aligned}AD^2 &= 15^2 + 22.898^2 = 749.32\\ AD &= 27.37 \text{ m}\end{aligned}$$

Since $27.37 \text{ m} \lt 28 \text{ m}$, the wire can be stretched from $A$ to $D$ without snapping.

In right-angled triangle $ALB$, $AB$ is opposite the $68.5^\circ$ angle at $L$ and $AL = 20$ m is adjacent.

$$\begin{aligned}\tan 68.5^\circ &= \frac{AB}{20}\\ AB &= 20\tan 68.5^\circ = 50.773 \text{ m}\end{aligned}$$

In right-angled triangle $AUB$, $AU = h + 20$ is opposite the $25.5^\circ$ angle at $B$.

$$\begin{aligned}\tan 25.5^\circ &= \frac{h+20}{50.773}\\ h + 20 &= 50.773\tan 25.5^\circ = 24.217\\ h &= 4.217\\ &= 4.22 \text{ m (3 s.f.)}\end{aligned}$$

(a) Distance between the ships

East and south are perpendicular, so the two distances are the shorter sides of a right-angled triangle.

$$\begin{aligned}\text{Ship A: } 550 \times 3 &= 1650 \text{ km}\\ \text{Ship B: } 420 \times 3 &= 1260 \text{ km}\\ d^2 &= 1650^2 + 1260^2 = 4\,310\,100\\ d &= 2076.1\\ &= 2080 \text{ km (3 s.f.)}\end{aligned}$$

(b)(i) Length of $BM$

$BCDM$ is a rectangle, so $MD = BC = 1.6$ cm and $AM = 3.6 – 1.6 = 2$ cm. In right-angled triangle $ABM$ (right angle at $M$):

$$\begin{aligned}\tan 35^\circ &= \frac{AM}{BM} = \frac{2}{BM}\\ BM &= \frac{2}{\tan 35^\circ} = 2.8563\\ &= 2.86 \text{ cm (3 s.f.)}\end{aligned}$$

(b)(ii) Angle $NAB$

Angle sum of triangle $ABM$:

$$\begin{aligned}\angle BAM &= 180^\circ – 90^\circ – 35^\circ = 55^\circ\end{aligned}$$

In right-angled triangle $ADN$, $AD$ is adjacent to angle $DAN$ and $AN$ is the hypotenuse.

$$\begin{aligned}\cos\angle DAN &= \frac{3.6}{4.5}\\ \angle DAN &= 36.870^\circ\\ \angle NAB &= 55^\circ – 36.870^\circ\\ &= 18.13^\circ = 18.1^\circ \text{ (1 d.p.)}\end{aligned}$$

(b)(iii) Length of $CN$

By Pythagoras’ theorem in triangle $ADN$, and $CD = BM$ (opposite sides of rectangle $BCDM$):

$$\begin{aligned}ND &= \sqrt{4.5^2 – 3.6^2} = 2.7 \text{ cm}\\ CN &= CD – ND = 2.8563 – 2.7\\ &= 0.156 \text{ cm (3 s.f.)}\end{aligned}$$

(a) Length of $BC$

In right-angled triangle $ACD$, $AD$ is opposite the $35^\circ$ angle and $AC$ is adjacent.

$$\begin{aligned}\tan 35^\circ &= \frac{5}{AC}\\ AC &= \frac{5}{\tan 35^\circ} = 7.1407 \text{ cm}\\ BC &= 7.1407 – 2 = 5.1407\\ &= 5.14 \text{ cm (3 s.f.)}\end{aligned}$$

(b) Length of $BE$

In right-angled triangle $BEC$ (right angle at $E$), $BE$ is opposite the $35^\circ$ angle and $BC$ is the hypotenuse.

$$\begin{aligned}\sin 35^\circ &= \frac{BE}{5.1407}\\ BE &= 5.1407\sin 35^\circ = 2.9486\\ &= 2.95 \text{ cm (3 s.f.)}\end{aligned}$$

(c) Angle $DBE$

By Pythagoras’ theorem in triangle $ABD$, then use triangle $BED$ (right angle at $E$), where $BE$ is adjacent to angle $DBE$ and $BD$ is the hypotenuse.

$$\begin{aligned}BD &= \sqrt{2^2 + 5^2} = \sqrt{29} = 5.3852 \text{ cm}\\ \cos\angle DBE &= \frac{2.9486}{5.3852}\\ \angle DBE &= 56.80^\circ\\ &= 56.8^\circ \text{ (1 d.p.)}\end{aligned}$$

(a) $PQ = 13$ cm

Triangle $PTQ$ is right-angled at $T$, so its area is $\tfrac12 \times PT \times TQ$.

$$\begin{aligned}\tfrac12 \times 5 \times TQ &= 30\\ TQ &= 12 \text{ cm}\\ PQ^2 &= 5^2 + 12^2 = 169\\ PQ &= 13 \text{ cm (shown)}\end{aligned}$$

(b) Value of $x$

$T$ is level with $Q$, so $TS = QR = x$ and the large radius is $PS = 5 + x$. The circles touch at $E$ on $PQ$, so $PQ = PE + EQ$.

$$\begin{aligned}(5 + x) + x &= 13\\ 2x &= 8\\ x &= 4\end{aligned}$$

(c) Angle $PQR$

$\angle PQR = \angle PQT + \angle TQR$, and $\angle TQR = 90^\circ$ because $TQ$ is horizontal and $QR$ is vertical.

$$\begin{aligned}\tan\angle PQT &= \frac{PT}{TQ} = \frac{5}{12}\\ \angle PQT &= 22.620^\circ\\ \angle PQR &= 90^\circ + 22.620^\circ\\ &= 112.6^\circ \text{ (1 d.p.)}\end{aligned}$$

(a)

In right-angled triangle $TBM$, $TB = y$ is opposite the $57^\circ$ angle and $BM = x$ is adjacent.

$$\tan 57^\circ = \frac{y}{x}$$

(b)

In right-angled triangle $TBN$, the adjacent side is $BN = x + 4$.

$$\tan 50^\circ = \frac{y}{x+4}$$

(c)

Make $y$ the subject of both equations and equate them.

$$\begin{aligned}y &= x\tan 57^\circ \quad\text{(1)}\\ y &= (x+4)\tan 50^\circ \quad\text{(2)}\\ x\tan 57^\circ &= (x+4)\tan 50^\circ\\ x\tan 57^\circ &= x\tan 50^\circ + 4\tan 50^\circ \text{ (shown)}\end{aligned}$$

(d) Height of the building

$$\begin{aligned}x\tan 57^\circ – x\tan 50^\circ &= 4\tan 50^\circ\\ x(\tan 57^\circ – \tan 50^\circ) &= 4\tan 50^\circ\\ x &= \frac{4\tan 50^\circ}{\tan 57^\circ – \tan 50^\circ} = 13.694\\ y &= 13.694\tan 57^\circ = 21.087\\ &= 21.1 \text{ m (3 s.f.)}\end{aligned}$$

(a) Angle $SPQ$

Triangle $SPR$ is isosceles ($SP = SR$), so its base angles are equal. In right-angled triangle $PQR$, $QR$ is opposite angle $RPQ$ and $PQ$ is adjacent.

$$\begin{aligned}\angle SPR &= \frac{180^\circ – 73.8^\circ}{2} = 53.1^\circ\\ \angle RPQ &= \tan^{-1}\left(\frac{80}{60}\right) = 53.130^\circ\\ \angle SPQ &= 53.1^\circ + 53.130^\circ\\ &= 106.2^\circ \text{ (1 d.p., shown)}\end{aligned}$$

(b) Length $SR$

By Pythagoras’ theorem, $PR = \sqrt{60^2 + 80^2} = 100$ m. The line from $S$ to the midpoint $X$ of $PR$ is perpendicular to $PR$ and bisects angle $PSR$ (isosceles triangle), so $\angle XSR = 36.9^\circ$ and $XR = 50$ m.

$$\begin{aligned}\sin 36.9^\circ &= \frac{50}{SR}\\ SR &= \frac{50}{\sin 36.9^\circ} = 83.27\\ &= 83.3 \text{ m (3 s.f.)}\end{aligned}$$

Note: the school’s answer key lists 83.2 m, from a method that drops a perpendicular from $P$ to $SR$ and assumes $SR$ is parallel to $PQ$; the key’s other methods give 83.3 m, as here. The difference comes from rounding $73.8^\circ$.

(c) Shortest distance from $Q$ to $PR$

The shortest distance is the perpendicular height $h$ from $Q$ to $PR$. Work out the area of triangle $PQR$ in two ways.

$$\begin{aligned}\tfrac12 \times 60 \times 80 &= \tfrac12 \times 100 \times h\\ 2400 &= 50h\\ h &= 48 \text{ m}\end{aligned}$$