Trigonometry — Foundation

Q1. (3 marks)

• Theorem: In a right triangle, the square on the hypotenuse equals the sum of the squares on the other two sides: $h^2 = a^2 + b^2$. (1.5)
• Converse: If in a triangle the square of one side equals the sum of squares of the other two, then the angle opposite that side is a right angle. (1.5)

Q2. (3 marks) With opposite $a$, adjacent $b$, hyp $h$: $\sin A=\tfrac{a}{h},\ \cos A=\tfrac{b}{h},\ \tan A=\tfrac{a}{b}$ (1.5) $\csc A=\tfrac{h}{a},\ \sec A=\tfrac{h}{b},\ \cot A=\tfrac{b}{a}$ (1.5)

Q3. (4 marks)

• $h=\sqrt{9^2+12^2}=\sqrt{81+144}=\sqrt{225}=15\text{ cm}$. (2) — Pythagoras.
• $\theta$ opposite the 9 cm side: $\sin\theta=\dfrac{9}{15}=\dfrac{3}{5}=0.6$. (1)
• $\cos\theta=\dfrac{12}{15}=\dfrac{4}{5}=0.8$. (1)

Q4. (4 marks)

• The distance from wall (2 m) is adjacent to $60^\circ$; ladder is hypotenuse. Use CAH: $\cos 60^\circ=\dfrac{\text{adj}}{\text{hyp}}=\dfrac{2}{L}$. (2)
• $\cos 60^\circ=\tfrac12$, so $\tfrac12=\dfrac{2}{L}\Rightarrow L=4\text{ m}$. (2)

Q5. (5 marks)

• Equilateral triangle side 2; drop altitude splitting into two right triangles with base 1, hyp 2. (1)
• Altitude $=\sqrt{2^2-1^2}=\sqrt3$. (1)
• For $30^\circ$ angle: opposite $=1$, adjacent $=\sqrt3$, hyp $=2$. $\sin30^\circ=\dfrac{1}{2}$ (1), $\cos30^\circ=\dfrac{\sqrt3}{2}$ (1), $\tan30^\circ=\dfrac{1}{\sqrt3}$ (1).

Q6. (4 marks)

• $\sin^2 60^\circ+\cos^2 60^\circ=1$ (Pythagorean identity, or $\tfrac34+\tfrac14$). (2)
• $\tan 45^\circ=1$. (1)
• Total $=1+1=2$. (1)

Q7. (4 marks)

• $\cos 55^\circ=\sin(90^\circ-55^\circ)=\sin35^\circ$, so $\dfrac{\sin35^\circ}{\cos55^\circ}=1$. (1.5)
• $\csc 65^\circ=\sec(90^\circ-65^\circ)=\sec25^\circ$, so $\cos25^\circ\cdot\csc65^\circ=\cos25^\circ\cdot\sec25^\circ=1$. (1.5)
• Total $=1+1=2$. (1)

Q8. (5 marks)

• $\tan\theta=\tfrac34$: opposite 3, adjacent 4, hyp $=\sqrt{9+16}=5$. (2)
• $\sin\theta=\dfrac{3}{5}=0.6$ (1), $\cos\theta=\dfrac{4}{5}=0.8$. (1)
• $\csc\theta=\dfrac{h}{\text{opp}}=\dfrac{5}{3}$ and $\dfrac{1}{\sin\theta}=\dfrac{1}{3/5}=\dfrac53$; equal ✓. (1)

Q9. (4 marks)

• $\tan 30^\circ=\dfrac{\text{height}}{30}$. (2)
• Height $=30\tan30^\circ=30\cdot\dfrac{1}{\sqrt3}=\dfrac{30}{\sqrt3}=10\sqrt3\approx17.32\text{ m}$. (2)

Q10. (4 marks)

• $1+\cot^2\theta=1+\dfrac{\cos^2\theta}{\sin^2\theta}=\dfrac{\sin^2\theta+\cos^2\theta}{\sin^2\theta}=\dfrac{1}{\sin^2\theta}$. (2)
• Multiply by $\sin^2\theta$: $\dfrac{1}{\sin^2\theta}\cdot\sin^2\theta=1$. (2)

[
  {"claim":"Q3 hypotenuse=15 and sinθ=3/5, cosθ=4/5","code":"h=sqrt(9**2+12**2); result=(h==15) and (Rational(9,15)==Rational(3,5)) and (Rational(12,15)==Rational(4,5))"},
  {"claim":"Q4 ladder length = 4","code":"L=2/cos(rad(60)); result=simplify(L-4)==0"},
  {"claim":"Q6 expression equals 2","code":"e=sin(rad(60))**2+cos(rad(60))**2+tan(rad(45)); result=simplify(e-2)==0"},
  {"claim":"Q7 expression equals 2","code":"e=sin(rad(35))/cos(rad(55))+cos(rad(25))*(1/sin(rad(65))); result=simplify(e-2)==0"},
  {"claim":"Q9 tower height = 10*sqrt(3)","code":"H=30*tan(rad(30)); result=simplify(H-10*sqrt(3))==0"}
]