Card 1 of 23
The component adjacent to the given angle uses [ ]; the component opposite the angle uses sine.
A
Tsin40∘=120T\sin40^\circ=120Tsin40∘=120, so T≈187 N\mathbf{T\approx187\text{ N}}T≈187 N.
B
The component adjacent to the given angle uses cosine; the component opposite the angle uses sine.
C
P=187cos40∘≈143 NP=187\cos40^\circ\approx\mathbf{143\text{ N}}P=187cos40∘≈143 N.
D
Both components are negative.
Card 1 of 23
1.8.9 Trigonometry in context Flashcards
23 flashcards on AQA A Level Maths 1.8.9 Trigonometry in context: the key formulae, methods and definitions you need to recall for Paper 1, Paper 2 and Paper 3.