Trigonometry — Foundation

Question 1

(a) Similar-triangles proof (6) Right triangle $ABC$, right angle at $C$, legs $a=BC$, $b=CA$, hypotenuse $c=AB$. Drop altitude $CD\perp AB$, foot $D$ splits hypotenuse into $AD=p$, $DB=q$, $p+q=c$. (1, setup)

• $\triangle ACD \sim \triangle ABC$ (share $\angle A$, both right-angled) ⇒ $\dfrac{b}{c}=\dfrac{p}{b}\Rightarrow b^2=pc$. (2)
• $\triangle BCD \sim \triangle BAC$ (share $\angle B$, both right-angled) ⇒ $\dfrac{a}{c}=\dfrac{q}{a}\Rightarrow a^2=qc$. (2)
• Add: $a^2+b^2 = qc+pc = c(p+q)=c\cdot c = c^2$. ∎ (1)

(b) Standard angles (6) Equilateral triangle side $2$; altitude bisects base into $1$ and, by Pythagoras, altitude $=\sqrt{2^2-1^2}=\sqrt3$. The half-triangle is right-angled with angles $30°$ and $60°$. (2 for construction/altitude)

• Opposite $30°$ is the side $1$, hypotenuse $2$: $\sin30°=\tfrac12$, $\cos30°=\tfrac{\sqrt3}{2}$, $\tan30°=\tfrac{1}{\sqrt3}$. (2)
• Opposite $60°$ is $\sqrt3$: $\sin60°=\tfrac{\sqrt3}{2}$, $\cos60°=\tfrac12$, $\tan60°=\sqrt3$. (2)

(c) Complementary identity (4) In right triangle with acute angles $\theta$ and $90°-\theta$: the side opposite $\theta$ is adjacent to $(90°-\theta)$. (1) $\sin(90°-\theta)=\dfrac{\text{opp of }(90°-\theta)}{\text{hyp}}=\dfrac{\text{adj of }\theta}{\text{hyp}}=\cos\theta$. (2) Corresponding: $\tan\theta = \cot(90°-\theta)$. (1)

Question 2

(a) (6) $\sin30°=\tfrac12$: $T=\dfrac{2(20)(\tfrac12)}{10}=\dfrac{20}{10}=2\ \text{s}$. (2) $\sin2\theta=2\sin30°\cos30°=2\cdot\tfrac12\cdot\tfrac{\sqrt3}{2}=\tfrac{\sqrt3}{2}$. (2) $R=\dfrac{20^2\cdot(\sqrt3/2)}{10}=\dfrac{400}{10}\cdot\tfrac{\sqrt3}{2}=40\cdot\tfrac{\sqrt3}{2}=20\sqrt3\ \text{m}\approx 34.6\ \text{m}$. (2)

(b) (7) $H=\dfrac{u^2\sin^2\theta}{2g}=\dfrac{400\cdot(1/2)^2}{20}=\dfrac{400\cdot\tfrac14}{20}=\dfrac{100}{20}=5\ \text{m}$. (3) Horizontal distance to point below peak $=R/2=10\sqrt3$ m. (1) $\tan\phi=\dfrac{H}{R/2}=\dfrac{5}{10\sqrt3}=\dfrac{1}{2\sqrt3}=\dfrac{\sqrt3}{6}$. (2) $\phi=\arctan(0.2887)\approx 16°$. (1)

(c) (5) $R(\theta)=\dfrac{u^2\sin2\theta}{g}$. Since $\sin2\theta\le1$ for all $\theta$, $R(\theta)\le\dfrac{u^2}{g}$. (2) Equality holds when $\sin2\theta=1\Rightarrow 2\theta=90°\Rightarrow\theta=45°$. So $R(45°)=\dfrac{u^2}{g}$ is the maximum and $R(\theta)\le R(45°)$. (2) $R(45°)=\dfrac{400}{10}=40\ \text{m}$. (1)

Question 3

(a) (8) Sample Python:

import math
def solve_right_triangle(known):
    a = known.get('a'); b = known.get('b'); c = known.get('c')
    if c is None:
        c = math.hypot(a, b)                 # Pythagoras
    elif a is None:
        a = math.sqrt(c*c - b*b)
    else:
        b = math.sqrt(c*c - a*a)
    A = math.degrees(math.atan2(a, b))       # angle opposite a
    B = 90.0 - A
    # reciprocal-identity self-test
    s = math.sin(math.radians(A))
    assert abs((1/s) - c/a) < 1e-9           # cosec A == c/a
    return dict(a=a, b=b, c=c, A=A, B=B)

Marks: missing-side via Pythagoras (2); angle via atan/atan2 (2); second angle as complement (1); reciprocal-identity assertion $\csc A=1/\sin A=c/a$ (2); correct return structure/handles all three input cases (1).

(b) (6) Let horizontal distance from $B$ to tower base be $x$; height $h$. At $B$: $\tan60°=\dfrac{h}{x}\Rightarrow h=x\sqrt3$. (2) At $A$: $\tan30°=\dfrac{h}{x+40}\Rightarrow \dfrac{1}{\sqrt3}=\dfrac{h}{x+40}\Rightarrow x+40=h\sqrt3$. (2) Substitute $x=h/\sqrt3$: $\dfrac{h}{\sqrt3}+40=h\sqrt3\Rightarrow 40=h\sqrt3-\dfrac{h}{\sqrt3}=h\dfrac{3-1}{\sqrt3}=\dfrac{2h}{\sqrt3}$. $h=\dfrac{40\sqrt3}{2}=20\sqrt3\approx34.64\ \text{m}$. (2)

(c) (2) Converse: if in a triangle with sides $a\le b\le c$ we have $a^2+b^2=c^2$, then the angle opposite $c$ is a right angle. (1) Test $(9,40,41)$: $9^2+40^2=81+1600=1681=41^2$. ✓ So it is right-angled. (1)

[
  {"claim":"Range at 30deg is 20*sqrt(3) m","code":"u=20; g=10; R=Rational(u**2)/g*sin(2*pi/6); result = simplify(R-20*sqrt(3))==0"},
  {"claim":"Max height at 30deg is 5 m","code":"u=20; g=10; H=Rational(u**2)*sin(pi/6)**2/(2*g); result = simplify(H-5)==0"},
  {"claim":"tan(phi)=H/(R/2)=1/(2*sqrt(3)) for 30deg launch","code":"H=5; R=20*sqrt(3); t=H/(R/2); result = simplify(t-1/(2*sqrt(3)))==0"},
  {"claim":"Tower height h = 20*sqrt(3)","code":"h=symbols('h',positive=True); sol=solve(Eq(h/sqrt(3)+40, h*sqrt(3)), h); result = simplify(sol[0]-20*sqrt(3))==0"},
  {"claim":"(9,40,41) satisfies Pythagoras converse","code":"result = (9**2+40**2==41**2)"}
]