Solutions — Y10 Core Trig 6B Worksheet

Warm-up

Q1. Missing parts:

a $60$ ($\cos^{-1}(0.5)=60^\circ$)

b $\tfrac{1}{2}$ ($\sin^{-1}(\tfrac{1}{2})=30^\circ$)

c $0.75$ ($\tan^{-1}(0.75)\approx 36.87^\circ$)

Q2. Inverse trig (2 d.p.):

a $23.58^\circ$

b $60^\circ$

c $11.31^\circ$

d $5.74^\circ$

Q3. Apply the inverse, 2 d.p.

a $36.87^\circ$ ($\sin^{-1}(\tfrac{3}{5})$)

b $61.93^\circ$ ($\cos^{-1}(\tfrac{8}{17})$)

c $16.26^\circ$ ($\tan^{-1}(\tfrac{7}{24})$)

d $36.87^\circ$ ($\tfrac{9}{15}=0.6$)

e $66.42^\circ$ ($\cos^{-1}(0.4)$)

f $22.62^\circ$ ($\tan^{-1}(\tfrac{5}{12})$)

Q4. Pick the ratio, then invert.

a opp + hyp → $\sin\theta=\tfrac{1}{2}\;\Rightarrow\;\theta=\sin^{-1}(0.5)=$ $30^\circ$

b adj + hyp → $\cos\theta=\tfrac{1.5}{2.5}=0.6\;\Rightarrow\;\theta=\cos^{-1}(0.6)\approx$ $53.13^\circ$

c opp + adj → $\tan\theta=\tfrac{3}{5}\;\Rightarrow\;\theta=\tan^{-1}(0.6)\approx$ $30.96^\circ$

Q5. Mine tunnel:

· opp = 120, adj = 100 → $\tan\theta=\tfrac{120}{100}\;\Rightarrow\;\theta=\tan^{-1}(1.2)\approx$ $50.2^\circ$

Q6. Ladder:

· adj $=1.2$, hyp $=4$ → $\cos\theta=\tfrac{1.2}{4}=0.3\;\Rightarrow\;\theta=\cos^{-1}(0.3)\approx$ $72.5^\circ$

Q7. Ramp:

· opp $=0.8$, adj $=5$ → $\tan\theta=\tfrac{0.8}{5}=0.16\;\Rightarrow\;\theta=\tan^{-1}(0.16)\approx$ $9.1^\circ$

Q8. Kite (string is hypotenuse $35$, height is opposite $28$):

· $\sin\theta=\tfrac{28}{35}=0.8\;\Rightarrow\;\theta=\sin^{-1}(0.8)\approx$ $53.1^\circ$