Menu
Sign up Sign in

AI HL · 12.10

Geometry & Trigonometry

0 of 79 skills done

0
12.10.8

Applications of trigonometry: 3D, bearings, elevation

12.10.8

Applications of trigonometry: 3D, bearings, elevation

Target: I can find lengths and angles in a cuboid by drawing a right-angled triangle and using 3D Pythagoras and trigonometry. (+5 more)

By the end of this section

Key points

Working in 3D

  • Do not work on the 3D picture. Pick out a right-angled triangle, draw it on its own, and label the lengths you know.
  • The diagonal of a cuboid with edges a, b and c is \(\sqrt{a^2+b^2+c^2}\). Find a face diagonal first when you need an angle.
  • The angle between a line and a plane is the angle between the line and its shadow on the plane. For AG and the base ABCD, the shadow is AC.
  • The four space diagonals of a cuboid are equal and cross at the centre. To find the angle between two of them, use the cosine rule in the triangle they make with a face diagonal.
  • Keep full calculator values until the final line. Give answers to 3 significant figures, and angles to 1 decimal place or 3 significant figures.

Pyramids, cones and prisms

  • In a right pyramid the apex is directly above the centre of the base. The foot of the height is the midpoint of the base diagonal. For an equilateral triangle base, the centre is \(\frac{s}{\sqrt3}\) from each corner.
  • The height of a pyramid is not the slant height. The slant height is the height of a sloping face.
  • Volume of a pyramid or cone: \(\frac13\times\)base area\(\times\)height. Volume of a prism: cross-section area \(\times\) length.
  • If a smaller shape is similar to a larger one with every length multiplied by k, areas are multiplied by \(k^2\) and volumes by \(k^3\). In a prism with the same length, the volume is multiplied by \(k^2\).
  • When a question says "state your assumptions", say what you treated as ideal: a solid shape, nothing inside it, whole units rounded up.

Bearings

  • A bearing is measured clockwise from north and written with 3 digits: 042°.
  • Draw a north line at every point where the direction changes. North lines are parallel, so you can use alternate and co-interior angles.
  • The back bearing differs by 180°. Add 180° if the bearing is under 180°. Subtract 180° otherwise.
  • For a journey of two legs, the angle inside the triangle at the turn is 180° minus the change of bearing. Use the cosine rule for the distance back.
  • To find a bearing from a triangle, find the angle with the sine or cosine rule, then add or subtract it from a known bearing. Check the answer against your sketch.
  • If a bearing problem has a right angle (for example due east and due south), use right-angled trigonometry.

Elevation and depression

  • Angles of elevation and depression are measured from the horizontal. The angle of depression from A to B equals the angle of elevation from B to A.
  • If the observer's eye is above the ground, add the eye height to the height you find.
  • With two angles from two positions, let x be the unknown distance and write two equations. Subtract the two expressions for the distance to get an equation in the height.

On the calculator

Degree mode and inverse functions

  • Check the calculator is in degrees before any trigonometry.
  • Use the inverse tangent, sine or cosine to turn a ratio into an angle.
  • Do the whole calculation in one go, or store the full value. Do not retype a rounded value.

Screenshot and video to come

Solving two equations for the height

  • Type the two equations tan 28° = h ÷ x and tan 19° = h ÷ (x + 10) into the solver.
  • Solve for h and x together. Read h.
  • Check that x is positive and the height is sensible.

Screenshot and video to come

Explore: a bearing and its back-bearing

Drag A and B on the map. Each point has a north line. The bearing is the angle clockwise from north, written with three figures.

Bearing of B from A
Bearing of A from B

One square is 1 km. Click a point, then use the arrow keys to move it by half a kilometre.

Drag A or B to change the bearing.

  1. Drag B right round A. When is the bearing of B from A less than 180°? What does the back-bearing do then?
  2. Put B due east of A, then due south, then due west. Write down both bearings each time. What do you notice?
  3. Why do the two bearings always differ by exactly 180°? Look at the two north lines.
  4. Tick the box and put B south-west of A. What does the working do that it did not do for north-east? Why?

Explore: a right-angled triangle inside a cuboid

Choose the angle you want. The right-angled triangle that holds it is shaded in the cuboid and drawn flat beside it. Change the three edges and watch the working.

Space diagonal AG
Check with cosine

Move a slider to change the cuboid.

  1. Make BC longer. Which angle gets smaller, the one with the base or the one with AB? Use the flat triangle to explain.
  2. Make AB, BC and AE equal, so the cuboid is a cube. What is the angle between AG and the base? Change the size of the cube. Does the angle change?
  3. In each triangle, where is the right angle, and why is it a right angle? Name the perpendicular lines.
  4. The working uses tan. Which two sides does it use? Check that cos with the hypotenuse AG gives the same angle.

Worked examples

Problems like the ones in class. Pick one, try it, then open the working.
A diagonal of a cuboid and its angle with the base

The cuboid ABCDEFGH has AB = 12 cm, BC = 7 cm and AE = 8 cm. (a) Find the length of the diagonal AG. (b) Find the angle between AG and the base ABCD.

12 cm7 cm8 cmABCDEFGH
Working
  1. The base diagonal

    \(AC=\sqrt{12^2+7^2}=13.9\text{ cm}\)

    AC is the shadow of AG on the base. Right-angled triangle ABC.

  2. AG by 3D Pythagoras

    \(AG=\sqrt{12^2+7^2+8^2}=16.0\text{ cm}\)

    Or use triangle ACG: AG² = AC² + CG².

  3. Triangle ACG

    \(\tan\theta=\frac{CG}{AC}=\frac{8}{13.9}\)

    CG = AE is the side opposite the angle. Right angle at C. Use the full value of AC from the calculator.

  4. The angle

    \(\theta=29.9^\circ\)

    The angle between a line and a plane is the angle between the line and its shadow on the plane.

Video walkthrough coming
A square-based pyramid: height and apex angle

VABCD is a right pyramid. The base ABCD is a square of side 12 cm. Each sloping edge is 16 cm long. (a) Sketch the triangle ACV and label every length you can. (b) Find the height of the pyramid. (c) Find the angle AVC.

12 cm16 cmABCV
Working
  1. The sketch

    Triangle ACV: AV = CV = 16 cm, AC = 17.0 cm. The height VM meets AC at its midpoint M.

    Draw ACV on its own. It is isosceles. Mark the height from V to the midpoint M of AC, with a right angle at M.

  2. The base diagonal

    \(AC=\sqrt{12^2+12^2}=17.0\text{ cm}, \quad AM=8.49\text{ cm}\)

    M is the centre of the base, so AM is half of AC.

  3. The height

    \(VM=\sqrt{16^2-8.49^2}=13.6\text{ cm}\)

    Right-angled triangle AMV. The sloping edge AV is the hypotenuse.

  4. The apex angle

    \(\sin\left(\tfrac12 AVC\right)=\frac{8.49}{16}, \quad AVC=64.1^\circ\)

    Half the angle is in the right-angled triangle AMV. Double it. Or use the cosine rule in triangle ACV.

Video walkthrough coming
A two-leg journey and the way back

A ferry leaves port P and sails 8.2 km on a bearing of 040° to Q. It then sails 6.5 km on a bearing of 125° to R. (a) Find the distance from R back to P. (b) Find the bearing on which the ferry must sail from R to return to P. Give the bearing to the nearest degree.

8.2 km6.5 km?PQRNN
Working
  1. The angle at Q

    \(\angle PQR=180^\circ-(125^\circ-40^\circ)=95^\circ\)

    Draw a north line at Q. The two north lines are parallel, so the turn at Q is the difference of the bearings. The angle inside the triangle is what is left of 180°.

  2. Cosine rule

    \(PR^2=8.2^2+6.5^2-2(8.2)(6.5)\cos 95^\circ\)

    Two sides and the angle between them.

  3. The distance

    \(PR=10.9\text{ km}\)

    Part (a).

  4. The angle at P

    \(\cos P=\frac{8.2^2+10.9^2-6.5^2}{2(8.2)(10.9)}, \quad \angle QPR=36.5^\circ\)

    Use the cosine rule again, or the sine rule. Keep the full value of PR on the calculator.

  5. The bearing

    \(\text{bearing of R from P}=40^\circ+36.5^\circ=76.5^\circ\)

    R is clockwise of Q as seen from P, so add the angle to 040°.

  6. The way back

    \(76.5^\circ+180^\circ=256°\)

    The back bearing differs by 180°. Give 3 digits.

Video walkthrough coming
Sketching a bearing, then a right-angled triangle

A rock R is 2.8 km from a lighthouse L on a bearing of 040°. A buoy B is due east of the lighthouse and due south of the rock. (a) Sketch the situation. Mark the north direction at L and the right angle. (b) Find the distance from the buoy to the rock.

2.8 km?040°LRBN
Working
  1. The sketch

    L at the bottom left, north line up from L. R is up and to the right. B is level with L and directly below R.

    Draw the north line first, then R at 040° clockwise from it. Then drop a vertical line from R to meet the east line from L.

  2. The angle at R

    \(\angle LRB=40^\circ\)

    RB is a north-south line, parallel to the north line at L. Alternate angles are equal. The right angle is at B.

  3. Choose the ratio

    \(\cos 40^\circ=\frac{BR}{2.8}\)

    BR is adjacent to the angle at R. LR = 2.8 km is the hypotenuse.

  4. Solve

    \(BR=2.8\cos 40^\circ=2.14\text{ km}\)

    A check: BR is the northward part of the journey, so it is less than 2.8 km.

Video walkthrough coming
Two angles of elevation

A student stands at P, level with the foot B of a vertical tower, and measures the angle of elevation of the top T as 28°. She walks 10 m directly away from the tower to Q and measures the angle of elevation as 19°. Find the height of the tower.

hx10 m28°19°BTPQ
Working
  1. Two equations

    \(\tan 28^\circ=\frac{h}{x}, \quad \tan 19^\circ=\frac{h}{x+10}\)

    Let h be the height and x the distance BP. Each angle gives one right-angled triangle.

  2. Write x in terms of h

    \(x=\frac{h}{\tan 28^\circ}, \quad x+10=\frac{h}{\tan 19^\circ}\)

    Make the distance the subject of both.

  3. Subtract

    \(10=\frac{h}{\tan 19^\circ}-\frac{h}{\tan 28^\circ}=h\left(\frac{1}{\tan 19^\circ}-\frac{1}{\tan 28^\circ}\right)\)

    The difference of the two distances is the 10 m walked.

  4. Solve for h

    \(h=9.77\text{ m}\)

    Or put both equations into the calculator's solver. Check: x = h ÷ tan 28° = 18.4 m.

Video walkthrough coming
The roof space of a barn

The roof of a barn is a triangular prism. Its end is an isosceles triangle with a base of 12 m. Each side of the roof makes an angle of 25° with the horizontal. The barn is 20 m long. (a) Find the height of the ridge above the base of the roof. (b) Find the volume of the roof space.

12 m25°
Working
  1. Half the end triangle

    \(\tan 25^\circ=\frac{h}{6}\)

    The height cuts the isosceles triangle into two right-angled triangles. Each has a base of half the width.

  2. The height

    \(h=6\tan 25^\circ=2.80\text{ m}\)

    Part (a).

  3. Area of the end

    \(\tfrac12\times 12\times 2.80=16.8\text{ m}^2\)

    Use the full base of the triangle here.

  4. Volume

    \(V=16.8\times 20=336\text{ m}^3\)

    A prism: area of the cross-section times the length.

Video walkthrough coming