MathematicsExam code: 0580

Right Angled Triangles

Pythagoras and Trigonometry

PractiseFlashcards

Elevation and depression

  • An angle of elevation is measured up from the horizontal to the line of sight, and an angle of depression is measured down from it
  • Both are measured from the horizontal, never from the vertical, which is the usual misreading
An observer's eye with a dashed horizontal line drawn from it: the line of sight up to a bird makes the angle of elevation above that horizontal, and the line of sight down to a boat makes the angle of depression below it
Source: Elevation & depression by Save My Exams
  • Sketch the situation with the horizontal drawn in and the right angle marked, then apply SOHCAHTOA
The elevation problem turned into a right-angled triangle BFT: the dashed horizontal at the top of the tower makes the 35 degree angle of depression, which equals the 35 degree angle of elevation at B, with the 24 m height marked O and the sloping line of sight marked H
Source: Elevation & depression by Save My Exams
  • Tangent appears most often, because these questions usually pair a height with a horizontal distance
A vertical tower of height 24 m seen from a point B on the ground: the angle of depression of B from the top T is 35 degrees, measured down from the dashed horizontal at T, and the angle of elevation of a point M part way up is 18 degrees at B, with the height x m of M still to be found
Source: Elevation & depression by Save My Exams

Worked example

An angle of elevation

From a point 40 m from the base of a tower, the angle of elevation of the top is 32°. Find the height of the tower, correct to 1 decimal place.

A right-angled triangle standing on horizontal ground. The ground from the observation point to the foot of the tower is labelled 40 m, the tower rises vertically with a right angle marked at its foot, and the angle of elevation at the observation point is marked 32 degrees. The height of the tower is unlabelled.

Solution:

  • The horizontal distance is adjacent to the 32° angle and the height is opposite it
  • O and A means tangent
  • tan 32° = height ÷ 40
  • height = 40 × tan 32°
  • height = 24.9947…
  • The tower is 25.0 m tall to 1 decimal place

Worked example

An angle of depression between two masts

Two vertical masts stand on horizontal ground. Mast AB has its foot at B and is 37.5 m tall, and mast CD has its foot at D and is 9.5 m tall. B and D are 48 m apart. Find the angle of depression of C, the top of the shorter mast, from A, correct to 1 decimal place.

Two vertical masts on horizontal ground: the taller mast AB on the left, 37.5 m, with its foot B on the ground, and the shorter mast CD on the right, 9.5 m, with its foot D on the ground. B and D are 48 m apart along the ground, and right angles are marked at B and D. No line is drawn between the tops A and C, and no angle is marked. The figure is marked NOT TO SCALE.

Solution:

  • Draw a horizontal line from C across to AB, meeting it at E, so angle AEC = 90°
  • EC is the gap between the masts, 48 m, and E is 9.5 m above the ground
  • So AE is the difference in heights: 37.5 − 9.5 = 28 m
  • The angle of depression of C from A equals angle ACE, the angle of elevation of A from C, because the horizontals through A and C are parallel
  • From angle ACE, AE is opposite and EC is adjacent, so use tangent
  • tan(ACE) = 28 ÷ 48
  • Angle ACE = tan⁻¹(28 ÷ 48) = 30.256…
  • The angle of depression is 30.3° to 1 decimal place
  • Using the full height, tan⁻¹(37.5 ÷ 48) = 38.0°, gives the angle of depression of D, the foot of the shorter mast, instead

Shortest distance to a line

  • The shortest distance from a point to a line is the perpendicular distance, measured at a right angle to the line
  • The syllabus names this directly, and these papers ask for a "shortest distance" in 6 of the 21
  • Drawing that perpendicular creates a right-angled triangle, which Pythagoras or SOHCAHTOA then solves
  • The syllabus adds that knowledge of bearings may be required in two-dimensional problems like these, so a north line can appear on the same diagram

Exam tip

Draw the perpendicular before you calculate

Draw the perpendicular from the point to the line, and mark its right angle, before any calculation. The schemes give a method mark just for recognising that the shortest distance is this perpendicular, so it can score even if the working after it goes wrong. Where that recognition is shown only on the diagram, some schemes count it only if the right angle is marked.

Worked example

Shortest distance from a point to a line

In triangle FGH, FG = 14 cm, FH = 20 cm and angle GFH = 41°. Find the shortest distance from G to the line FH, correct to 3 significant figures.

Triangle FGH with FH along the bottom, F on the left and H on the right, and G above. FG is labelled 14 cm, FH is labelled 20 cm and the angle at F is marked 41 degrees; GH is not labelled. No perpendicular is drawn. The figure is marked NOT TO SCALE.

Solution:

  • The shortest distance is the perpendicular from G to FH: call its foot N, so angle GNF = 90°
  • In the right-angled triangle GNF, FG is the hypotenuse and GN is opposite the 41° angle
  • O and H means sine: sin 41° = GN ÷ 14
  • GN = 14 × sin 41° = 9.1848…
  • The shortest distance is 9.18 cm to 3 significant figures
  • FH = 20 cm is not needed for this route, but it gives a check through the area: ½ × 14 × 20 × sin 41° = 91.848… cm², and that area is also ½ × 20 × GN, so GN = 2 × 91.848… ÷ 20 = 9.18 cm
  • When the area is what you know, rearranging area = ½ × base × distance in the same way finds the distance directly

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