Trigonometry — Foundation

Question 1

(a) In triangle $ABC$ with altitude $CD$ to hypotenuse:

• $\triangle ACD \sim \triangle CBD$ (both similar to $\triangle ABC$; equal angles since $\angle ADC=\angle CDB=90°$ and $\angle DAC=\angle DCB$ as complements of $\angle DCA$).
• From similarity $\dfrac{AD}{CD} = \dfrac{CD}{DB}$, so $CD^2 = AD\cdot DB$. (2 marks: similarity + ratio)
• $CD^2 = 9\times 4 = 36 \Rightarrow CD = 6$. (2 marks)

(b) $AC^2 = AD\cdot AB = 9\times 13 = 117 \Rightarrow AC = \sqrt{117}=3\sqrt{13}$. $BC^2 = DB\cdot AB = 4\times 13 = 52 \Rightarrow BC = \sqrt{52}=2\sqrt{13}$. (2 marks) Check: $AC^2+BC^2 = 117 + 52 = 169 = 13^2 = AB^2$. ✓ (2 marks)

(c) $\sin(\angle CAB) = \dfrac{BC}{AB} = \dfrac{2\sqrt{13}}{13}$; $\cos(\angle CBA) = \dfrac{BC}{AB} = \dfrac{2\sqrt{13}}{13}$. (1 mark each value / 1 for pair) They are equal because $\angle CAB$ and $\angle CBA$ are complementary ($\sin\theta=\cos(90°-\theta)$), and both equal the same ratio $BC/AB$. (1 mark)

Question 2

(a) Equilateral side $2$, drop altitude → splits base into $1,1$; altitude $=\sqrt{2^2-1^2}=\sqrt3$. The half-triangle is right-angled with angles $30°$ (top) and $60°$ (base). (2 marks)

• $\sin 30° = \dfrac{\text{opp}}{\text{hyp}} = \dfrac{1}{2}$. (1)
• $\cos 30° = \dfrac{\sqrt3}{2}$. (1)
• $\tan 60° = \dfrac{\text{opp}}{\text{adj}} = \dfrac{\sqrt3}{1} = \sqrt3$. (1)

(b) Legs $1,1$; hypotenuse $=\sqrt2$; angles $45°$.

• $\sin 45° = \dfrac{1}{\sqrt2} = \dfrac{\sqrt2}{2}$. (2)
• $\tan 45° = \dfrac{1}{1} = 1$. (1)

(c) $\tan 60°=\sqrt3$, $\tan 30°=\dfrac{1}{\sqrt3}$. $\frac{\sqrt3 - \tfrac{1}{\sqrt3}}{1 + \sqrt3\cdot\tfrac{1}{\sqrt3}} = \frac{\tfrac{3-1}{\sqrt3}}{1+1} = \frac{2/\sqrt3}{2} = \frac{1}{\sqrt3} = \frac{\sqrt3}{3}.$ (2 marks) (This equals $\tan 30°$.)

Question 3

(a) By definition $\sec(90°-\theta) = \dfrac{1}{\cos(90°-\theta)}$. (1) Complementary relation: $\cos(90°-\theta) = \sin\theta$ (in a right triangle the adjacent to $(90°-\theta)$ is the opposite of $\theta$). (2) So $\sin\theta\cdot\sec(90°-\theta) = \sin\theta\cdot\dfrac{1}{\sin\theta} = 1$. (1)

(b) $\sin\theta = \dfrac{3}{5}$. Right triangle: opp $=3$, hyp $=5$, adj $=\sqrt{25-9}=4$. (1)

• $\cos\theta = \dfrac{4}{5}$ (1)
• $\tan\theta = \dfrac{3}{4}$ (0.5), $\cot\theta = \dfrac{4}{3}$ (0.5)
• $\csc\theta = \dfrac{5}{3}$ (1)

(c) $\csc^2\theta - \cot^2\theta = \dfrac{25}{9} - \dfrac{16}{9} = \dfrac{9}{9} = 1$. (1) Comment: confirms the identity $\csc^2\theta - \cot^2\theta = 1$ (Pythagorean identity). (1)

Question 4

Let first position $A$, second position $B$ (closer), foot $O$, top $T$, $BO = x$, $AO = x+50$.

(a) $\tan 30° = \dfrac{h}{x+50}$ and $\tan 60° = \dfrac{h}{x}$. (3 marks)

(b) From these: $h = (x+50)\tan30° = \dfrac{x+50}{\sqrt3}$ and $h = x\tan60° = x\sqrt3$. (1) Set equal: $x\sqrt3 = \dfrac{x+50}{\sqrt3} \Rightarrow 3x = x+50 \Rightarrow 2x = 50 \Rightarrow x = 25\text{ m}$. (2) $h = 25\sqrt3 \approx 43.30\text{ m}$ (exact $25\sqrt3$). (1)

(c) Distance from first position to top $AT$: hypotenuse where elevation $=30°$, so $AT = \dfrac{h}{\sin 30°} = \dfrac{25\sqrt3}{1/2} = 50\sqrt3 \approx 86.60\text{ m}$. (3 marks)

Question 5

Let $PF = a$, $QF = b$, $a+b = 60$, pole height $h$.

(a) From $P$: $\tan 45° = \dfrac{h}{a} \Rightarrow a = h$ (since $\tan45°=1$). (2) From $Q$: $\tan 30° = \dfrac{h}{b} \Rightarrow b = \dfrac{h}{\tan30°} = h\sqrt3$. (2)

(b) $a + b = 60 \Rightarrow h + h\sqrt3 = 60 \Rightarrow h(1+\sqrt3) = 60$. (2) $h = \frac{60}{1+\sqrt3} = \frac{60(\sqrt3-1)}{(\sqrt3+1)(\sqrt3-1)} = \frac{60(\sqrt3-1)}{2} = 30(\sqrt3-1)\text{ m}.$ (2 marks) Numerically $h \approx 30(0.7320508) \approx 21.96\text{ m}$.

(c) $PF = a = h = 30(\sqrt3-1) \approx 21.96\text{ m}$. (2 marks)

[
  {"claim":"Q1: AC^2+BC^2 = AB^2 with AC^2=117, BC^2=52, AB=13","code":"result = (117 + 52 == 13**2)"},
  {"claim":"Q2c: (tan60-tan30)/(1+tan60*tan30) = 1/sqrt3","code":"expr = (tan(pi/3)-tan(pi/6))/(1+tan(pi/3)*tan(pi/6)); result = simplify(expr - 1/sqrt(3)) == 0"},
  {"claim":"Q3b/c: with sin=3/5, csc^2 - cot^2 = 1","code":"s=Rational(3,5); c=Rational(4,5); csc=1/s; cot=c/s; result = simplify(csc**2 - cot**2) == 1"},
  {"claim":"Q4: x=25, h=25*sqrt3, AT=50*sqrt3","code":"x=25; h=x*sqrt(3); AT=h/sin(pi/6); result = (simplify(h-25*sqrt(3))==0) and (simplify(AT-50*sqrt(3))==0) and (3*x==x+50)"},
  {"claim":"Q5: h = 30(sqrt3-1) satisfies h + h*sqrt3 = 60","code":"h=30*(sqrt(3)-1); result = simplify(h + h*sqrt(3) - 60) == 0"}
]