9th Class Math Chapter 6 Notes

Q1(i) — Quadrant of $65^{\circ }$

Find the quadrant and a co-terminal angle.

1. $0^{\circ }<65^{\circ }<90^{\circ }$: Quadrant I.
2. Add $360^{\circ }$ for a co-terminal angle.

Answer: Quadrant I; co-terminal $=425^{\circ }$

Q1(ii) — Quadrant of $135^{\circ }$

Find the quadrant and a co-terminal angle.

1. $90^{\circ }<135^{\circ }<180^{\circ }$: Quadrant II.

Answer: Quadrant II; co-terminal $=495^{\circ }$

Q1(iii) — Quadrant of $-40^{\circ }$

Find the quadrant and a co-terminal angle.

1. $-40^{\circ }+360^{\circ }=320^{\circ }$, which is between $270^{\circ }$ and $360^{\circ }$: Quadrant IV.

Answer: Quadrant IV; co-terminal $=320^{\circ }$

Q1(iv) — Quadrant of $210^{\circ }$

Find the quadrant and a co-terminal angle.

1. $180^{\circ }<210^{\circ }<270^{\circ }$: Quadrant III.

Answer: Quadrant III; co-terminal $=570^{\circ }$

Q1(v) — Quadrant of $-150^{\circ }$

Find the quadrant and a co-terminal angle.

1. $-150^{\circ }+360^{\circ }=210^{\circ }$, between $180^{\circ }$ and $270^{\circ }$: Quadrant III.

Answer: Quadrant III; co-terminal $=210^{\circ }$

Q2(i) — Convert $123.456^{\circ }$ to D°M'S"

Convert to degrees, minutes, seconds.

1. Integer part: $123^{\circ }$.
2. $0.456\times 60=27.36^{\prime }$; integer part $27^{\prime }$.
3. $0.36\times 60=21.6^{\prime \prime }$.

Answer: $123^{\circ }27^{\prime }21.6^{\prime \prime }$

Q2(ii) — Convert $58.7891^{\circ }$

Convert to degrees, minutes, seconds.

1. $58^{\circ }$; $0.7891\times 60=47.346^{\prime }\to 47^{\prime }$; $0.346\times 60=20.76^{\prime \prime }$.

Answer: $58^{\circ }47^{\prime }20.76^{\prime \prime }$

Q2(iii) — Convert $90.5678^{\circ }$

Convert to degrees, minutes, seconds.

1. $90^{\circ }$; $0.5678\times 60=34.068^{\prime }\to 34^{\prime }$; $0.068\times 60=4.08^{\prime \prime }$.

Answer: $90^{\circ }34^{\prime }4.08^{\prime \prime }$

Q3(i) — Convert $65^{\circ }32^{\prime }15^{\prime \prime }$ to decimal degrees

Convert to decimal degrees.

1. $65+\frac{32}{60}+\frac{15}{3600}=65+0.5333+0.004167$.

Answer: $65.5375^{\circ }$

Q3(ii) — Convert $42^{\circ }18^{\prime }45^{\prime \prime }$

Convert to decimal degrees.

1. $42+\frac{18}{60}+\frac{45}{3600}=42+0.3+0.0125$.

Answer: $42.3125^{\circ }$

Q3(iii) — Convert $78^{\circ }45^{\prime }36^{\prime \prime }$

Convert to decimal degrees.

1. $78+\frac{45}{60}+\frac{36}{3600}=78+0.75+0.01$.

Answer: $78.76^{\circ }$

Q4(i) — Convert $36^{\circ }$ to radians

Convert to radians.

1. Multiply by $\frac{\pi }{180}$: $\frac{36\pi }{180}=\frac{\pi }{5}$.

Answer: $\frac{\pi }{5}\approx 0.6283$ rad

Q4(ii) — Convert $22.5^{\circ }$

Convert to radians.

1. $\frac{22.5\pi }{180}=\frac{\pi }{8}$.

Answer: $\frac{\pi }{8}\approx 0.3927$ rad

Q4(iii) — Convert $67.5^{\circ }$

Convert to radians.

1. $\frac{67.5\pi }{180}=\frac{3\pi }{8}$.

Answer: $\frac{3\pi }{8}\approx 1.1781$ rad

Q5(i) — Convert $\frac{\pi }{16}$ rad to degrees

Convert to degrees.

1. Multiply by $\frac{180}{\pi }$: $\frac{180}{16}=11.25^{\circ }$.

Answer: $11.25^{\circ }=11^{\circ }15^{\prime }$

Q5(ii) — Convert $\frac{11\pi }{5}$ rad

Convert to degrees.

1. $\frac{11\times 180}{5}=11\times 36$.

Answer: $396^{\circ }$

Q5(iii) — Convert $\frac{7\pi }{6}$ rad

Convert to degrees.

1. $\frac{7\times 180}{6}=7\times 30$.

Answer: $210^{\circ }$

Q6(i) — Arc length/area: $r=6$ cm, $\theta =\frac{\pi }{3}$

Find the arc length and sector area.

1. Arc length $s=r\theta =6\times \frac{\pi }{3}=2\pi$ cm.
2. Area $=\frac{1}{2}{r}^2\theta =\frac{1}{2}(36)\frac{\pi }{3}=6\pi$ cm${}^2$.

Answer: $s=2\pi \approx 6.28$ cm; Area $=6\pi \approx 18.85$ cm${}^2$

Q6(ii) — Arc length/area: $r=\frac{4.8}{\pi }$ cm, $\theta =\frac{5\pi }{6}$

Find the arc length and sector area.

1. $s=r\theta =\frac{4.8}{\pi }\times \frac{5\pi }{6}=\frac{4.8\times 5}{6}=4$ cm.
2. Area $=\frac{1}{2}{r}^2\theta =\frac{1}{2}{\left(\frac{4.8}{\pi }\right)}^2\frac{5\pi }{6}=\frac{9.6}{\pi }$ cm${}^2$.

Answer: $s=4$ cm; Area $=\frac{9.6}{\pi }\approx 3.06$ cm${}^2$

WP1 — Sector area and percentage: $\theta =60^{\circ }$, $r=12$ cm

Find the sector area and what percentage of the full circle it is.

1. Area $=\frac{60}{360}\pi r^2=\frac{1}{6}\times \pi \times 144=24\pi$ cm${}^2$.
2. Percentage $=\frac{60}{360}\times 100\%$.

Answer: Area $=24\pi \approx 75.4$ cm${}^2$; $16.67\%$ of the circle

Q7 — Percentage of area for $\theta =\frac{\pi }{8}$ rad

Find the percentage of the circle's area.

1. Convert to degrees: $\frac{\pi }{8}\times \frac{180}{\pi }=22.5^{\circ }$.
2. Percentage $=\frac{22.5}{360}\times 100\%$.

Answer: $6.25\%$

WP2 — Sector rolled into a cone: $r=12$ cm, angle $150^{\circ }$

Find the slant height and base radius of the resulting cone.

1. Arc length $=r\theta =12\times \left(150\times \frac{\pi }{180}\right)=12\times \frac{5\pi }{6}=10\pi$ cm.
2. This arc becomes the cone's base circumference: $2\pi R=10\pi \Rightarrow R=5$ cm.
3. The sector's radius becomes the cone's slant height.

Answer: Slant height $=12$ cm; base radius $=5$ cm

Q8(i) — Ratios: $BC=4,AB=3,CA=5$ (right angle at $B$)

Find all ten trig ratios for $\theta =\angle A$ and $\varphi =\angle C$.

1. Hypotenuse $CA=5$ (check: $3^2+4^2=5^2$ ✓).
2. $\theta$: opposite $=BC=4$, adjacent $=AB=3$, hyp $=5$.
3. $\varphi$: opposite $=AB=3$, adjacent $=BC=4$, hyp $=5$.

Answer: $\sin \theta =\frac{4}{5},\cos \theta =\frac{3}{5},\tan \theta =\frac{4}{3},\sec \theta =\frac{5}{3},\csc \theta =\frac{5}{4}$; $\cot \varphi =\frac{4}{3},\tan \varphi =\frac{3}{4},\csc \varphi =\frac{5}{3},\sec \varphi =\frac{5}{4},\cos \varphi =\frac{4}{5}$

Q8(ii) — Ratios: $BC=8,AB=15,CA=17$

Find all ten trig ratios for $\theta =\angle A$ and $\varphi =\angle C$.

1. Hypotenuse $CA=17$ (check: $8^2+15^2=17^2$ ✓).

Answer: $\sin \theta =\frac{8}{17},\cos \theta =\frac{15}{17},\tan \theta =\frac{8}{15},\sec \theta =\frac{17}{15},\csc \theta =\frac{17}{8}$; $\cot \varphi =\frac{8}{15},\tan \varphi =\frac{15}{8},\csc \varphi =\frac{17}{15},\sec \varphi =\frac{17}{8},\cos \varphi =\frac{8}{17}$

Q8(iii) — Ratios: $BC=5,AB=12,CA=13$

Find all ten trig ratios for $\theta =\angle A$ and $\varphi =\angle C$.

1. Hypotenuse $CA=13$ (check: $5^2+12^2=13^2$ ✓).

Answer: $\sin \theta =\frac{5}{13},\cos \theta =\frac{12}{13},\tan \theta =\frac{5}{12},\sec \theta =\frac{13}{12},\csc \theta =\frac{13}{5}$; $\cot \varphi =\frac{5}{12},\tan \varphi =\frac{12}{5},\csc \varphi =\frac{13}{12},\sec \varphi =\frac{13}{5},\cos \varphi =\frac{5}{13}$

Q9 — General ratios: right angle at $B$, $a=BC,b=AC$(hyp)$,c=AB$

Find $\sin \theta ,\cos \theta ,\tan \theta$ ($\theta =\angle C$) and $\sin \varphi ,\cos \varphi ,\tan \varphi$ ($\varphi =\angle A$).

1. For $\theta =\angle C$: opposite $=AB=c$, adjacent $=BC=a$, hyp $=b$.
2. For $\varphi =\angle A$: opposite $=BC=a$, adjacent $=AB=c$, hyp $=b$.

Answer: $\sin \theta =\frac{c}{b},\cos \theta =\frac{a}{b},\tan \theta =\frac{c}{a}$; $\sin \varphi =\frac{a}{b},\cos \varphi =\frac{c}{b},\tan \varphi =\frac{a}{c}$

Q10(i) — Verify $\sin \theta \csc \theta =1$

Triangle $ABC$: $AB=5$ (hyp), $BC=3$, $AC=4$, right angle at $C$, $\theta =\angle A$. Verify.

1. $\sin \theta =\frac{BC}{AB}=\frac{3}{5}$, $\csc \theta =\frac{5}{3}$.
2. $\sin \theta \csc \theta =\frac{3}{5}\times \frac{5}{3}=1$.

Answer: Verified: $1$

Q10(ii) — Verify $\cos \theta \sec \theta =1$

Using the same triangle, verify.

1. $\cos \theta =\frac{AC}{AB}=\frac{4}{5}$, $\sec \theta =\frac{5}{4}$.
2. $\cos \theta \sec \theta =\frac{4}{5}\times \frac{5}{4}=1$.

Answer: Verified: $1$

Q10(iii) — Verify $\tan \theta \cot \theta =1$

Using the same triangle, verify.

1. $\tan \theta =\frac{BC}{AC}=\frac{3}{4}$, $\cot \theta =\frac{4}{3}$.
2. $\tan \theta \cot \theta =\frac{3}{4}\times \frac{4}{3}=1$.

Answer: Verified: $1$

Q11(i) — $\sin 30^{\circ }=\sin (90^{\circ }-60^{\circ })=$?

Fill in the blank using the co-function identity.

1. $\sin (90^{\circ }-60^{\circ })=\cos 60^{\circ }$.

Answer: $\cos 60^{\circ }=\frac{1}{2}$

Q11(ii) — $\cos 30^{\circ }=\cos (90^{\circ }-60^{\circ })=$?

Fill in the blank.

1. $\cos (90^{\circ }-60^{\circ })=\sin 60^{\circ }$.

Answer: $\sin 60^{\circ }=\frac{\sqrt{3}}{2}$

Q11(iii) — $\tan 30^{\circ }=\tan (90^{\circ }-60^{\circ })=$?

Fill in the blank.

1. $\tan (90^{\circ }-60^{\circ })=\cot 60^{\circ }$.

Answer: $\cot 60^{\circ }=\frac{1}{\sqrt{3}}$

Q11(iv) — $\tan 60^{\circ }=\tan (90^{\circ }-30^{\circ })=$?

Fill in the blank.

1. $\tan (90^{\circ }-30^{\circ })=\cot 30^{\circ }$.

Answer: $\cot 30^{\circ }=\sqrt{3}$

Q11(v) — $\sin 60^{\circ }=\sin (90^{\circ }-30^{\circ })=$?

Fill in the blank.

1. $\sin (90^{\circ }-30^{\circ })=\cos 30^{\circ }$.

Answer: $\cos 30^{\circ }=\frac{\sqrt{3}}{2}$

Q11(vi) — $\cos 60^{\circ }=\cos (90^{\circ }-30^{\circ })=$?

Fill in the blank.

1. $\cos (90^{\circ }-30^{\circ })=\sin 30^{\circ }$.

Answer: $\sin 30^{\circ }=\frac{1}{2}$

Q11(vii) — $\sin 45^{\circ }=\sin (90^{\circ }-45^{\circ })=$?

Fill in the blank.

1. $\sin (90^{\circ }-45^{\circ })=\cos 45^{\circ }$.

Answer: $\cos 45^{\circ }=\frac{1}{\sqrt{2}}$

Q11(viii) — $\tan 45^{\circ }=\tan (90^{\circ }-45^{\circ })=$?

Fill in the blank.

1. $\tan (90^{\circ }-45^{\circ })=\cot 45^{\circ }$.

Answer: $\cot 45^{\circ }=1$

Q11(ix) — $\cos 45^{\circ }=\cos (90^{\circ }-45^{\circ })=$?

Fill in the blank.

1. $\cos (90^{\circ }-45^{\circ })=\sin 45^{\circ }$.

Answer: $\sin 45^{\circ }=\frac{1}{\sqrt{2}}$

Q12(i) — Find $\frac{BC}{AB}$ (triangle with $\angle A=30^{\circ },\angle C=60^{\circ },\angle B=90^{\circ }$)

Find $\frac{m\overline{BC}}{m\overline{AB}}$.

1. $BC$ is opposite $\angle A$, $AB$ is adjacent to $\angle A$, so this ratio is $\tan A=\tan 30^{\circ }$.

Answer: $\frac{1}{\sqrt{3}}$

Q12(ii) — $\cos 60^{\circ }$

Find $\cos 60^{\circ }$.

1. Standard value.

Answer: $\frac{1}{2}$

Q12(iii) — $\tan 60^{\circ }$

Find $\tan 60^{\circ }$.

1. Standard value.

Answer: $\sqrt{3}$

Q12(iv) — $\csc \frac{\pi }{3}$

Find $\csc \frac{\pi }{3}=\csc 60^{\circ }$.

1. Reciprocal of $\sin 60^{\circ }=\frac{\sqrt{3}}{2}$.

Answer: $\frac{2}{\sqrt{3}}$

Q12(v) — $\cot 60^{\circ }$

Find $\cot 60^{\circ }$.

1. Reciprocal of $\tan 60^{\circ }=\sqrt{3}$.

Answer: $\frac{1}{\sqrt{3}}$

Q12(vi) — $\sin 30^{\circ }$

Find $\sin 30^{\circ }$.

1. Standard value.

Answer: $\frac{1}{2}$

Q12(vii) — $\cos 30^{\circ }$

Find $\cos 30^{\circ }$.

1. Standard value.

Answer: $\frac{\sqrt{3}}{2}$

Q12(viii) — $\tan \frac{\pi }{6}$

Find $\tan \frac{\pi }{6}=\tan 30^{\circ }$.

1. Standard value.

Answer: $\frac{1}{\sqrt{3}}$

Q12(ix) — $\sec 30^{\circ }$

Find $\sec 30^{\circ }$.

1. Reciprocal of $\cos 30^{\circ }=\frac{\sqrt{3}}{2}$.

Answer: $\frac{2}{\sqrt{3}}$

Q12(x) — $\cot 30^{\circ }$

Find $\cot 30^{\circ }$.

1. Reciprocal of $\tan 30^{\circ }=\frac{1}{\sqrt{3}}$.

Answer: $\sqrt{3}$

Q13(i) — $\sin \theta =\frac{2}{3}$

Find the remaining trig ratios of $\theta$ (Quadrant I).

1. opp $=2$, hyp $=3$, adj $=\sqrt{9-4}=\sqrt{5}$.

Answer: $\cos \theta =\frac{\sqrt{5}}{3},\tan \theta =\frac{2}{\sqrt{5}},\csc \theta =\frac{3}{2},\sec \theta =\frac{3}{\sqrt{5}},\cot \theta =\frac{\sqrt{5}}{2}$

Q13(ii) — $\cos \theta =\frac{3}{4}$

Find the remaining trig ratios.

1. adj $=3$, hyp $=4$, opp $=\sqrt{16-9}=\sqrt{7}$.

Answer: $\sin \theta =\frac{\sqrt{7}}{4},\tan \theta =\frac{\sqrt{7}}{3},\csc \theta =\frac{4}{\sqrt{7}},\sec \theta =\frac{4}{3},\cot \theta =\frac{3}{\sqrt{7}}$

Q13(iii) — $\tan \theta =\frac{1}{2}$

Find the remaining trig ratios.

1. opp $=1$, adj $=2$, hyp $=\sqrt{5}$.

Answer: $\sin \theta =\frac{1}{\sqrt{5}},\cos \theta =\frac{2}{\sqrt{5}},\csc \theta =\sqrt{5},\sec \theta =\frac{\sqrt{5}}{2},\cot \theta =2$

Q13(iv) — $\sec \theta =3$

Find the remaining trig ratios.

1. $\cos \theta =\frac{1}{3}$; adj $=1$, hyp $=3$, opp $=\sqrt{9-1}=2\sqrt{2}$.

Answer: $\sin \theta =\frac{2\sqrt{2}}{3},\tan \theta =2\sqrt{2},\csc \theta =\frac{3\sqrt{2}}{4},\cot \theta =\frac{\sqrt{2}}{4}$

Q13(v) — $\cot \theta =\sqrt{\frac{3}{2}}$

Find the remaining trig ratios.

1. adj$:$opp $=\sqrt{3}:\sqrt{2}$; hyp $=\sqrt{3+2}=\sqrt{5}$.

Answer: $\sin \theta =\sqrt{\frac{2}{5}},\cos \theta =\sqrt{\frac{3}{5}},\tan \theta =\sqrt{\frac{2}{3}},\csc \theta =\sqrt{\frac{5}{2}},\sec \theta =\sqrt{\frac{5}{3}}$

Q14(i) — $\sin 30^{\circ }$

Find the value.

1. Standard angle value.

Answer: $\frac{1}{2}$

Q14(ii) — $\cos 30^{\circ }$

Find the value.

1. Standard angle value.

Answer: $\frac{\sqrt{3}}{2}$

Q14(iii) — $\tan \frac{\pi }{6}$

Find the value.

1. $=\tan 30^{\circ }$.

Answer: $\frac{1}{\sqrt{3}}$

Q14(iv) — $\tan 60^{\circ }$

Find the value.

1. Standard angle value.

Answer: $\sqrt{3}$

Q14(v) — $\sec 60^{\circ }$

Find the value.

1. Reciprocal of $\cos 60^{\circ }=\frac{1}{2}$.

Answer: $2$

Q14(vi) — $\cos \frac{\pi }{3}$

Find the value.

1. $=\cos 60^{\circ }$.

Answer: $\frac{1}{2}$

Q14(vii) — $\cot 60^{\circ }$

Find the value.

1. Reciprocal of $\tan 60^{\circ }=\sqrt{3}$.

Answer: $\frac{1}{\sqrt{3}}$

Q14(viii) — $\sin 60^{\circ }$

Find the value.

1. Standard angle value.

Answer: $\frac{\sqrt{3}}{2}$

Q14(ix) — $\sec 30^{\circ }$

Find the value.

1. Reciprocal of $\cos 30^{\circ }=\frac{\sqrt{3}}{2}$.

Answer: $\frac{2}{\sqrt{3}}$

Q14(x) — $\csc 30^{\circ }$

Find the value.

1. Reciprocal of $\sin 30^{\circ }=\frac{1}{2}$.

Answer: $2$

Q14(xi) — $\sin 45^{\circ }$

Find the value.

1. Standard angle value.

Answer: $\frac{1}{\sqrt{2}}$

Q14(xii) — $\cos \frac{\pi }{4}$

Find the value.

1. $=\cos 45^{\circ }$.

Answer: $\frac{1}{\sqrt{2}}$

Q15(i) — Evaluate: $2\sin 60^{\circ }\cos 60^{\circ }$

Evaluate.

1. $2\left(\frac{\sqrt{3}}{2}\right)\left(\frac{1}{2}\right)$.

Answer: $\frac{\sqrt{3}}{2}$

Q15(ii) — Evaluate: $2\cos \frac{\pi }{6}\sin \frac{\pi }{6}$

Evaluate.

1. $2\left(\frac{\sqrt{3}}{2}\right)\left(\frac{1}{2}\right)$.

Answer: $\frac{\sqrt{3}}{2}$

Q15(iii) — Evaluate: $2\sin 45^{\circ }+2\cos 45^{\circ }$

Evaluate.

1. $2\left(\frac{1}{\sqrt{2}}\right)+2\left(\frac{1}{\sqrt{2}}\right)=\frac{4}{\sqrt{2}}$.

Answer: $2\sqrt{2}$

Q15(iv) — Evaluate: $\sin 60^{\circ }\cos 30^{\circ }+\cos 60^{\circ }\sin 30^{\circ }$

Evaluate.

1. This is $\sin (60^{\circ }+30^{\circ })=\sin 90^{\circ }$.

Answer: $1$

Q15(v) — Evaluate: $\cos 60^{\circ }\cos 30^{\circ }-\sin 60^{\circ }\sin 30^{\circ }$

Evaluate.

1. This is $\cos (60^{\circ }+30^{\circ })=\cos 90^{\circ }$.

Answer: $0$

Q15(vi) — Evaluate: $\sin 60^{\circ }\cos 30^{\circ }-\cos 60^{\circ }\sin 30^{\circ }$

Evaluate.

1. This is $\sin (60^{\circ }-30^{\circ })=\sin 30^{\circ }$.

Answer: $\frac{1}{2}$

Q15(vii) — Evaluate: $\cos 60^{\circ }\cos 30^{\circ }+\sin 60^{\circ }\sin 30^{\circ }$

Evaluate.

1. This is $\cos (60^{\circ }-30^{\circ })=\cos 30^{\circ }$.

Answer: $\frac{\sqrt{3}}{2}$

Q15(viii) — Evaluate: $\tan \frac{\pi }{6}\cot \frac{\pi }{6}+1$

Evaluate.

1. $\tan \theta \cot \theta =1$ always, so this is $1+1$.

Answer: $2$

Q16(i) — Evaluate: $2\sin 45^{\circ }-2\cos 45^{\circ }$

Given $\sin 45^{\circ }=\cos 45^{\circ }=\frac{1}{\sqrt{2}}$, evaluate.

1. $2\left(\frac{1}{\sqrt{2}}\right)-2\left(\frac{1}{\sqrt{2}}\right)$.

Answer: $0$

Q16(ii) — Evaluate: $3\cos 45^{\circ }+4\sin 45^{\circ }$

Evaluate.

1. $3\left(\frac{1}{\sqrt{2}}\right)+4\left(\frac{1}{\sqrt{2}}\right)=\frac{7}{\sqrt{2}}$.

Answer: $\frac{7\sqrt{2}}{2}$

Q16(iii) — Evaluate: $5\cos 45^{\circ }-3\sin 45^{\circ }$

Evaluate.

1. $5\left(\frac{1}{\sqrt{2}}\right)-3\left(\frac{1}{\sqrt{2}}\right)=\frac{2}{\sqrt{2}}$.

Answer: $\sqrt{2}$

Q17(i) — $\angle B=90^{\circ },\angle A=30^{\circ },y=AB=4$: find $x=BC,z=AC$

Find $x$ and $z$.

1. $\tan 30^{\circ }=\frac{x}{y}\Rightarrow x=4\times \frac{1}{\sqrt{3}}$.
2. $\cos 30^{\circ }=\frac{y}{z}\Rightarrow z=\frac{4}{\sqrt{3}/2}=\frac{8}{\sqrt{3}}$.

Answer: $x=\frac{4\sqrt{3}}{3}$ cm; $z=\frac{8\sqrt{3}}{3}$ cm

Q17(ii) — $\angle B=90^{\circ },\angle A=\frac{\pi }{4},y=BC=\sqrt{3}$: find $x=AB,z=AC$

Find $x$ and $z$.

1. $\tan 45^{\circ }=\frac{y}{x}=1\Rightarrow x=y=\sqrt{3}$.
2. $\sin 45^{\circ }=\frac{y}{z}\Rightarrow z=\sqrt{3}\times \sqrt{2}$.

Answer: $x=\sqrt{3}$ cm; $z=\sqrt{6}$ cm

Q17(iii) — $\angle B=90^{\circ },\angle C=60^{\circ },z=CA=2$(hyp): find $x=CB,y=AB$

Find $x$ and $y$.

1. $\cos 60^{\circ }=\frac{x}{z}\Rightarrow x=2\times \frac{1}{2}$.
2. $\sin 60^{\circ }=\frac{y}{z}\Rightarrow y=2\times \frac{\sqrt{3}}{2}$.

Answer: $x=1$ cm; $y=\sqrt{3}$ cm

Q17(iv) — $\angle B=90^{\circ },\angle A=45^{\circ },y=AB=4$: find $x=BC,z=AC$

Find $x$ and $z$.

1. $\tan 45^{\circ }=1\Rightarrow x=y=4$.
2. $\cos 45^{\circ }=\frac{y}{z}\Rightarrow z=4\sqrt{2}$.

Answer: $x=4$ cm; $z=4\sqrt{2}$ cm

Q18(i) — $a=BC=\sqrt{5},c=AB=\sqrt{13}$: find $b=AC$ and the angles

Find the hypotenuse and both acute angles.

1. $b=\sqrt{a^2+c^2}=\sqrt{5+13}=\sqrt{18}=3\sqrt{2}$.
2. $\tan A=\frac{a}{c}=\sqrt{\frac{5}{13}}\approx 0.620\Rightarrow A\approx 31.8^{\circ }$; $C=90^{\circ }-A\approx 58.2^{\circ }$.

Answer: $b=3\sqrt{2}$ cm; $A\approx 31.8^{\circ }$, $C\approx 58.2^{\circ }$

Q18(ii) — $a=BC=4,c=AB=4$: find $b=AC$ and the angles

Find the hypotenuse and both acute angles.

1. $b=\sqrt{16+16}=\sqrt{32}=4\sqrt{2}$.
2. Since $a=c$, the triangle is isosceles: $A=C=45^{\circ }$.

Answer: $b=4\sqrt{2}$ cm; $A=C=45^{\circ }$

WP3 — Diagonal of a $60$ m square field

Find the length of the diagonal.

1. Diagonal $=$ side$\times \sqrt{2}=60\sqrt{2}$.

Answer: $60\sqrt{2}\approx 84.85$ m

Q19(i) — Solve: $\angle B=90^{\circ },\angle C=60^{\circ },c=3\sqrt{3}$

Solve the triangle.

1. $\angle A=30^{\circ }$.
2. $\tan C=\frac{c}{a}\Rightarrow a=\frac{c}{\tan 60^{\circ }}=\frac{3\sqrt{3}}{\sqrt{3}}=3$.
3. $b=\frac{c}{\sin C}=\frac{3\sqrt{3}}{\sqrt{3}/2}=6$.

Answer: $a=3$ cm, $b=6$ cm, $\angle A=30^{\circ }$

Q19(ii) — Solve: $\angle B=90^{\circ },\angle C=45^{\circ },a=8$

Solve the triangle.

1. $\angle A=45^{\circ }$ (since $B=90°,C=45°$), so the triangle is isosceles: $c=a=8$.
2. $b=a\sqrt{2}=8\sqrt{2}$.

Answer: $c=8$ cm, $b=8\sqrt{2}$ cm, $\angle A=45^{\circ }$

Q19(iii) — Solve: $\angle B=90^{\circ },a=12,c=6$

Solve the triangle.

1. $b=\sqrt{144+36}=\sqrt{180}=6\sqrt{5}$.
2. $\tan A=\frac{a}{c}=2\Rightarrow A\approx 63.43^{\circ }$; $C\approx 26.57^{\circ }$.

Answer: $b=6\sqrt{5}$ cm; $A\approx 63.43^{\circ }$, $C\approx 26.57^{\circ }$

Q19(iv) — Solve: $\angle B=90^{\circ },\angle A=60^{\circ },c=4$

Solve the triangle.

1. $\angle C=30^{\circ }$.
2. $a=c\tan 60^{\circ }=4\sqrt{3}$; $b=\frac{c}{\cos 60^{\circ }}=8$.

Answer: $a=4\sqrt{3}$ cm, $b=8$ cm, $\angle C=30^{\circ }$

Q19(v) — Solve: $\angle B=90^{\circ },\angle A=30^{\circ },c=4$

Solve the triangle.

1. $\angle C=60^{\circ }$.
2. $a=c\tan 30^{\circ }=\frac{4}{\sqrt{3}}$; $b=\frac{c}{\cos 30^{\circ }}=\frac{8}{\sqrt{3}}$.

Answer: $a=\frac{4\sqrt{3}}{3}$ cm, $b=\frac{8\sqrt{3}}{3}$ cm, $\angle C=60^{\circ }$

Q19(vi) — Solve: $\angle B=90^{\circ },b=10,a=6$

Solve the triangle.

1. $c=\sqrt{100-36}=\sqrt{64}=8$.
2. $\sin A=\frac{a}{b}=0.6\Rightarrow A\approx 36.87^{\circ }$; $C\approx 53.13^{\circ }$.

Answer: $c=8$ cm; $A\approx 36.87^{\circ }$, $C\approx 53.13^{\circ }$

WP4 — Canal width and angle $PQR$

$PQ=13$ m (hyp), $QR=5$ m, right angle at $R$. Find the canal's width and $\angle PQR$ in radians.

1. Width $PR=\sqrt{13^2-5^2}=\sqrt{144}=12$ m.
2. $\tan (\angle PQR)=\frac{PR}{QR}=\frac{12}{5}=2.4\Rightarrow \angle PQR\approx 67.38^{\circ }$.
3. Convert to radians: $67.38\times \frac{\pi }{180}\approx 1.176$.

Answer: Width $=12$ m; $\angle PQR\approx 1.176$ rad ($\approx 67.4^{\circ }$)

Q20 — Length $x$ in quadrilateral $ABCD$

Right angles at $B,C$; $AB=10$, $CD=8$, $AD=17$, $BC=x$. Find $x$.

1. Drop perpendiculars so $AB$ and $CD$ are parallel; the difference in their lengths ($10-8=2$) and $BC=x$ form the legs of a right triangle with hypotenuse $AD=17$.
2. $x^2+2^2=17^2\Rightarrow x^2=289-4=285$.

Answer: $x=\sqrt{285}\approx 16.88$ cm

WP5 — Ladder against a wall

Ladder $=8$ m, foot $2$ m from the wall. Find the height reached.

1. Height $=\sqrt{8^2-2^2}=\sqrt{60}=2\sqrt{15}$.

Answer: $2\sqrt{15}\approx 7.75$ m

WP6 — Rectangular field diagonal

Diagonal $=(x+9)$ m, sides $(x+7)$ m and $x$ m. Find $x$.

1. $x^2+(x+7{)}^2=(x+9{)}^2$.
2. $x^2+x^2+14x+49=x^2+18x+81\Rightarrow x^2-4x-32=0$.
3. $(x-8)(x+4)=0\Rightarrow x=8$ (rejecting the negative root).

Answer: $x=8$

Q21(i) — Calculate $x$: right angle at $B$, $AB=12$, $AC=20$(hyp)

Find $x=BC$.

1. $x=\sqrt{20^2-12^2}=\sqrt{256}$.

Answer: $x=16$ cm

Q21(ii) — Calculate $x$: isosceles triangle with altitude foot $B$

$D$ at the apex, $B$ the foot of the altitude with $AB=BC=4$; $AD=x$, $DC=5$. Find $x$.

1. In right triangle $DBC$: $DB=\sqrt{D{C}^2-B{C}^2}=\sqrt{25-16}=3$.
2. In right triangle $DBA$: $x=DA=\sqrt{D{B}^2+A{B}^2}=\sqrt{9+16}=\sqrt{25}$.

Answer: $x=5$ cm

WP7 — Flag post height

Angle of elevation $=60^{\circ }$ from $40$ m away. Find the height.

1. Height $=40\tan 60^{\circ }=40\sqrt{3}$.

Answer: $40\sqrt{3}\approx 69.28$ m

WP8 — Altitude of an isosceles triangle

Vertical angle $=120^{\circ }$, base $=10$ cm. Find the altitude.

1. The altitude splits the triangle into two right triangles with a $30^{\circ }$ base angle and half-base $5$ cm.
2. $\tan 30^{\circ }=\frac{h}{5}\Rightarrow h=5\tan 30^{\circ }=\frac{5}{\sqrt{3}}$.

Answer: $h=\frac{5\sqrt{3}}{3}\approx 2.89$ cm

WP9 — Angle of elevation of a tree

Tree $=72$ m, distance $=100$ m. Find the angle of elevation.

1. $\theta =\arctan \left(\frac{72}{100}\right)=\arctan (0.72)$.

Answer: $\theta \approx 35.75^{\circ }\approx 35^{\circ }45^{\prime }$