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Cambridge IGCSE Maths · Topic 6 · 28 cards
Trigonometry flashcards
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- 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).
- 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.
- 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.
- 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.).
- How do you find an unknown angle in a right-angled triangle?Use the inverse ratio. E.g. if opposite \(=7\) and hypotenuse \(=10\): \(\theta=\sin^{-1}\left(\frac{7}{10}\right)=44.4^\circ\).
- 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 \(\theta=\cos^{-1}0.3=72.5^\circ\).
- 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).
- 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.
- (Extended) What are the exact values of \(\sin\), \(\cos\) and \(\tan\) for \(30^\circ\), \(45^\circ\) and \(60^\circ\)?\(\sin30=\frac12,\cos30=\frac{\sqrt3}{2},\tan30=\frac{1}{\sqrt3}\); \(\sin45=\cos45=\frac{\sqrt2}{2},\tan45=1\); \(\sin60=\frac{\sqrt3}{2},\cos60=\frac12,\tan60=\sqrt3\).
- (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.
- (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.
- (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 \(b\).
- (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.).
- (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.).
- (Extended) In triangle \(ABC\), \(A=40^\circ\), \(a=7\) and \(B=65^\circ\). Find \(b\).\(\frac{b}{\sin65^\circ}=\frac{7}{\sin40^\circ}\), so \(b=\frac{7\sin65^\circ}{\sin40^\circ}=9.87\).
- (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 \(360^\circ\).
- (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 \(270^\circ\).
- (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.
- (Extended) State the amplitude and period of \(y=3\sin2x\).Amplitude 3 (maximum value). Period \(=\frac{360^\circ}{2}=180^\circ\).
- (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: \(x=180^\circ-30^\circ=150^\circ\).
- (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 \(360^\circ-120^\circ=240^\circ\).
- 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.
- (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.
- (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.
- (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).
- What is the value of \(\tan\theta\) in terms of \(\sin\theta\) and \(\cos\theta\)?\(\tan\theta=\frac{\sin\theta}{\cos\theta}\). Also \(\sin^2\theta+\cos^2\theta=1\).
- 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.
- 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.