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Cambridge IGCSE Maths · Topic 6 · 28 cards

Trigonometry flashcards

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  1. State the three trigonometric ratios for a right-angled triangle.
    \(\sin\theta=\frac{\text{opp}}{\text{hyp}}\), \(\cos\theta=\frac{\text{adj}}{\text{hyp}}\), \(\tan\theta=\frac{\text{opp}}{\text{adj}}\) (SOH CAH TOA).
  2. How do you label the sides of a right-angled triangle for trigonometry?
    Hypotenuse: opposite the right angle (longest). Opposite: across from the angle you are using. Adjacent: the remaining side next to that angle.
  3. How do you choose which trig ratio to use?
    Mark the side you know and the side you want (relative to the angle). Pick the ratio containing exactly those two sides: opp/hyp is sin, adj/hyp is cos, opp/adj is tan.
  4. Find \(x\): a right-angled triangle has hypotenuse 10 cm and an angle \(35^\circ\); \(x\) is the side opposite the angle.
    \(\sin35^\circ=\frac{x}{10}\), so \(x=10\sin35^\circ=5.74\) cm (3 s.f.).
  5. How do you find an unknown angle in a right-angled triangle?
    Use the inverse ratio. E.g. if opposite \(=7\) and hypotenuse \(=10\): \(\theta=\sin^{-1}\left(\frac{7}{10}\right)=44.4^\circ\).
  6. A ladder 5 m long leans against a wall with its base 1.5 m from the wall. Find the angle with the ground.
    \(\cos\theta=\frac{1.5}{5}=0.3\), so \(\theta=\cos^{-1}0.3=72.5^\circ\).
  7. What are angles of elevation and depression?
    Both are measured from the horizontal. Elevation: looking up at an object. Depression: looking down. The angle of depression from \(A\) to \(B\) equals the angle of elevation from \(B\) to \(A\) (alternate angles).
  8. From a point 40 m from the base of a tower, the angle of elevation of the top is \(28^\circ\). Find the tower's height.
    \(\tan28^\circ=\frac{h}{40}\), so \(h=40\tan28^\circ=21.3\) m.
  9. (Extended) What are the exact values of \(\sin\), \(\cos\) and \(\tan\) for \(30^\circ\), \(45^\circ\) and \(60^\circ\)?
    \(\sin30=\frac12,\cos30=\frac{\sqrt3}{2},\tan30=\frac{1}{\sqrt3}\); \(\sin45=\cos45=\frac{\sqrt2}{2},\tan45=1\); \(\sin60=\frac{\sqrt3}{2},\cos60=\frac12,\tan60=\sqrt3\).
  10. (Extended) State the sine rule and when it is used.
    \(\frac{a}{\sin A}=\frac{b}{\sin B}=\frac{c}{\sin C}\). Use it in any triangle when you know a side and its opposite angle, plus one other side or angle.
  11. (Extended) State the cosine rule for a side and for an angle.
    \(a^2=b^2+c^2-2bc\cos A\), and rearranged \(\cos A=\frac{b^2+c^2-a^2}{2bc}\). Use it for two sides and the included angle, or for all three sides.
  12. (Extended) State the formula for the area of a triangle using two sides and the included angle.
    \(\text{Area}=\frac12ab\sin C\), where \(C\) is the angle between sides \(a\) and \(b\).
  13. (Extended) Find the area of a triangle with sides 8 cm and 11 cm enclosing an angle of \(40^\circ\).
    \(\frac12\times8\times11\times\sin40^\circ=28.3\) cm² (3 s.f.).
  14. (Extended) In triangle \(ABC\), \(b=9\), \(c=12\), \(A=50^\circ\). Find \(a\).
    \(a^2=81+144-2(9)(12)\cos50^\circ=225-138.84=86.16\), so \(a=9.28\) (3 s.f.).
  15. (Extended) In triangle \(ABC\), \(A=40^\circ\), \(a=7\) and \(B=65^\circ\). Find \(b\).
    \(\frac{b}{\sin65^\circ}=\frac{7}{\sin40^\circ}\), so \(b=\frac{7\sin65^\circ}{\sin40^\circ}=9.87\).
  16. (Extended) Describe the graph of \(y=\sin x\) for \(0^\circ\le x\le360^\circ\).
    A smooth wave starting at 0, rising to a maximum of 1 at \(90^\circ\), crossing zero at \(180^\circ\), reaching \(-1\) at \(270^\circ\) and returning to 0 at \(360^\circ\). The period is \(360^\circ\).
  17. (Extended) How does the graph of \(y=\cos x\) differ from \(y=\sin x\)?
    Same shape and period \(360^\circ\), but it starts at its maximum 1 when \(x=0^\circ\) (a sine curve shifted \(90^\circ\) left). It crosses zero at \(90^\circ\) and \(270^\circ\).
  18. (Extended) What is special about the graph of \(y=\tan x\)?
    Period \(180^\circ\), range is all real numbers, and it has vertical asymptotes at \(x=90^\circ,270^\circ,\ldots\) where \(\tan x\) is undefined.
  19. (Extended) State the amplitude and period of \(y=3\sin2x\).
    Amplitude 3 (maximum value). Period \(=\frac{360^\circ}{2}=180^\circ\).
  20. (Extended) Solve \(\sin x=0.5\) for \(0^\circ\le x\le360^\circ\).
    \(x=\sin^{-1}0.5=30^\circ\). Sine is also positive in the second quadrant: \(x=180^\circ-30^\circ=150^\circ\).
  21. (Extended) Solve \(\cos x=-0.5\) for \(0^\circ\le x\le360^\circ\).
    \(\cos^{-1}(-0.5)=120^\circ\). By symmetry the other solution is \(360^\circ-120^\circ=240^\circ\).
  22. A ship sails 12 km on a bearing of \(060^\circ\). How far north has it travelled?
    North component \(=12\cos60^\circ=6\) km (bearing is measured from north). East component \(=12\sin60^\circ=10.4\) km.
  23. (Extended) How do you find the angle between a line and a plane in 3D?
    Find the right-angled triangle formed by the line, its projection on the plane and the perpendicular height. Then use trigonometry in that triangle.
  24. (Extended) A cuboid is 6 cm by 8 cm by 5 cm. Find the length of the space diagonal.
    Base diagonal \(=\sqrt{6^2+8^2}=10\). Then \(\sqrt{10^2+5^2}=\sqrt{125}=11.2\) cm.
  25. (Extended) Why is \(\sin\theta=\sin(180^\circ-\theta)\) important with the sine rule?
    Finding an angle with \(\sin^{-1}\) gives an acute angle, but an obtuse angle has the same sine, so there may be two possible triangles (the ambiguous case).
  26. What is the value of \(\tan\theta\) in terms of \(\sin\theta\) and \(\cos\theta\)?
    \(\tan\theta=\frac{\sin\theta}{\cos\theta}\). Also \(\sin^2\theta+\cos^2\theta=1\).
  27. Why is the hypotenuse always the longest side, and what does this say about \(\sin\) and \(\cos\) of acute angles?
    It is opposite the largest angle (\(90^\circ\)). So opp/hyp and adj/hyp are always less than 1: \(\sin\theta\) and \(\cos\theta\) lie between 0 and 1 for acute angles.
  28. What mode should the calculator be in and what is the check for a trig answer?
    Degrees mode. Check sense: the hypotenuse is the longest side, and a bigger angle gives a bigger opposite side.