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AI SL · 12.10

Geometry & Trigonometry

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12.10.4

Mixed practice: rules of trigonometry

12.10.4

Mixed practice: rules of trigonometry

Target: I can choose between the sine rule, the cosine rule and right-angled trigonometry from what a triangle shows. (+5 more)

By the end of this section

Key points

Choose the rule from what you know

  • Right-angled triangle: sin, cos or tan. Not right-angled: the sine rule or the cosine rule.
  • Two sides and the angle between them, or all three sides: cosine rule. Cosine rule for a side: \(a^2=b^2+c^2-2bc\cos A\). For an angle: \(\cos A=\frac{b^2+c^2-a^2}{2bc}\).
  • A side and its opposite angle known, plus one more side or angle: sine rule, \(\frac{a}{\sin A}=\frac{b}{\sin B}=\frac{c}{\sin C}\). Put the unknown on top.
  • Area = \(\frac12ab\sin C\), where C is the angle between sides a and b. With three sides and no angle, find an angle first.

Warnings

  • The sine rule for an angle can give two answers: \(\theta\) and \(180^\circ-\theta\). Keep the second only if the angles still add to less than 180°.
  • The calculator must be in degree mode. Keep full accuracy until the last line; round to 3 significant figures at the end.
  • A negative cosine means an obtuse angle. A sine rule on its own cannot tell you that.

Compound figures

  • Draw the figure and split it into right-angled triangles or into triangles you can use a rule on.
  • Find a side shared by two triangles first. Then use it in the second triangle.
  • Two angles at one vertex add to give the angle of the big triangle. A straight line gives 180°.
  • Height problems: write each length with tan, such as \(PO=\frac{h}{\tan 32^\circ}\), then set two expressions for the same length equal, or solve the equation with technology.

Lines and angles

  • Axis intercepts: put y = 0 to find where the line meets the x-axis, and x = 0 for the y-axis.
  • A line with gradient m makes an angle \(\theta\) with the x-axis where \(\tan\theta=|m|\).
  • Angle between two lines: find each line's angle with the x-axis. Lines sloping the same way: subtract. Opposite ways: add, and if the total is over 90°, take 180° minus it.
  • Two lines meet on the x-axis if they have the same x-intercept.

On the calculator

Degree mode and inverse trig

  • Check the calculator is in degree mode before every trig question.
  • For an angle, use the inverse key: sin⁻¹, cos⁻¹ or tan⁻¹ of the ratio.
  • Type the whole fraction in brackets, such as (7²+9²−12²) ÷ (2×7×9), before taking cos⁻¹.

Screenshot and video to come

Keeping full accuracy

  • Use the previous answer, or store a value, instead of typing a rounded copy.
  • Round only the final answer, to 3 significant figures.

Screenshot and video to come

Solving an equation for h

  • Type the equation with h as the unknown, such as h ÷ tan(24°) + h ÷ tan(41°) = 60.
  • Use the solver: Solve for h.
  • Check the answer is sensible: a height must be positive.

Screenshot and video to come

Explore: which rule to use

Drag the corners to make any triangle. Tick the three measurements you are given, then choose what to find. The applet names the rule and writes the working.

Sides
Angles
Area

Tick three measurements.

  1. Tick b, c and A, then find a. Which rule is it? Now tick a, b and A instead. What changes?
  2. Tick all three sides. Why can you find an angle but not need the sine rule?
  3. Tick A, B and C. Why can you not find a side? Drag a corner and see what stays the same.
  4. Make one angle obtuse. Which rule can you trust to find it, and which could give the wrong angle?

Worked examples

Problems like the ones in class. Pick one, try it, then open the working.
Two angles at one vertex

In the diagram, angle ABC is a right angle and D lies on BC. BD = 5 cm, angle BAD = 40° and angle DAC = 15°. Find the length x = DC.

5 cmx40°, 15°ABDC
Working
  1. Find the shared side

    \(AB=\frac{5}{\tan 40^\circ}=5.959\) cm

    Triangle ABD is right-angled. AB is in both triangles.

  2. The big triangle

    \(BC=AB\tan 55^\circ=8.510\) cm

    Angle BAC = 40° + 15° = 55°. Keep the full value of AB.

  3. Subtract

    \(x=8.510-5=3.51\) cm

    x is only the part of BC beyond D.

Video walkthrough coming
Intercepts and the angle with the x-axis

The line \(L\) has equation \(7x+3y=21\). Find the coordinates of the points where \(L\) meets the axes, and the acute angle between \(L\) and the x-axis.

−1123452468xyL
Working
  1. Intercepts

    y = 0: \(7x=21\), x = 3. x = 0: \(3y=21\), y = 7.

    Set one letter to 0 and solve for the other. Points (3, 0) and (0, 7).

  2. The gradient

    \(y=-\frac73x+7\), gradient \(-\frac73\)

    The line slopes down. Use the size of the gradient for the acute angle.

  3. The angle

    \(\tan\theta=\frac73\), \(\theta=66.8^\circ\)

    From the calculator: inverse tangent of 7 ÷ 3.

Video walkthrough coming
Two lines meeting on the x-axis

The lines \(L_1: y=2x-6\) and \(L_2: x+3y=3\) are given. (a) Show that they meet on the x-axis. (b) Find the acute angle between them.

−1123456−6−4−22xyL1L2
Working
  1. Where each crosses the x-axis

    \(L_1\): y = 0 gives \(2x=6\), x = 3. \(L_2\): y = 0 gives x = 3.

    Both lines pass through (3, 0), so they meet on the x-axis.

  2. Angle of each line

    \(L_1\): gradient 2, \(\tan\theta_1=2\), \(\theta_1=63.4^\circ\). \(L_2\): gradient \(-\frac13\), \(\theta_2=18.4^\circ\)

    Angles with the x-axis, from the size of each gradient.

  3. Combine

    \(L_1\) slopes up and \(L_2\) slopes down: 63.4° + 18.4° = 81.9°

    Lines sloping opposite ways: add the angles. If the total is over 90°, the acute angle is 180° minus it.

Video walkthrough coming
Angle first, then the area

In triangle ABC, AB = 7 cm, BC = 9 cm and CA = 12 cm. Find the area of the triangle.

7 cm9 cm12 cmθBCA
Working
  1. No angle is given

    The area formula needs an angle between two known sides. Find angle B first.

    B is between AB and BC, opposite the longest side.

  2. Cosine rule for the angle

    \(\cos B=\frac{7^2+9^2-12^2}{2\times7\times9}=-0.1111\), \(B=96.4^\circ\)

    A negative cosine means an obtuse angle. That is fine.

  3. The area

    \(\text{Area}=\frac12\times7\times9\times\sin 96.38^\circ=31.3\) cm²

    Use the unrounded B.

Video walkthrough coming
Two angles of elevation

A mast BT stands on top of a hill. The foot of the hill is O, with OB = h metres and BT = d metres. From a point P on level ground the angle of elevation of B is 25° and of T is 31°. (a) Show that \(h=\dfrac{d\tan 25^\circ}{\tan 31^\circ-\tan 25^\circ}\). (b) The mast is 8 m tall. Find h.

hd25°, 31°PBT
Working
  1. Two expressions for PO

    \(PO=\frac{h}{\tan 25^\circ}\) and \(PO=\frac{h+d}{\tan 31^\circ}\)

    Both triangles are right-angled with the same base PO.

  2. Set them equal

    \(h\tan 31^\circ=(h+d)\tan 25^\circ\)

    Multiply both sides by tan 31° and tan 25°.

  3. Collect h

    \(h(\tan 31^\circ-\tan 25^\circ)=d\tan 25^\circ\), so \(h=\frac{d\tan 25^\circ}{\tan 31^\circ-\tan 25^\circ}\)

    This is the result to show.

  4. Put in d = 8

    \(h=27.7\) m

    From the calculator, in degree mode.

Video walkthrough coming
Lengths in terms of h

A vertical post OT of height h metres stands between two points P and Q on level ground. P, O and Q are in a straight line and PQ = 60 m. The angle of elevation of T from P is 24° and from Q is 41°. Write PO and OQ in terms of h, then find h.

60 mh24°41°PQT
Working
  1. PO

    \(\tan 24^\circ=\frac{h}{PO}\), so \(PO=\frac{h}{\tan 24^\circ}\)

    The base is in the denominator.

  2. OQ

    \(OQ=\frac{h}{\tan 41^\circ}\)

    The same idea in the second triangle.

  3. One equation

    \(\frac{h}{\tan 24^\circ}+\frac{h}{\tan 41^\circ}=60\)

    PO + OQ = PQ.

  4. Solve for h

    Solve for h: h = 17.7

    Use the calculator's solver. Do not rearrange by hand unless you want to check.

Video walkthrough coming