Ship travels 12 km due north then 9 km due east — find PB and bearing.
The path north then east forms a right angle at A.
PB = √(12² + 9²) = √(144 + 81) = √225 = 15 km
Bearing of B from P: the angle east of north at P.
tan θ = 9/12 = 0.75
θ = tan−1(0.75) ≈ 36.87°
Bearing = 037° (to nearest degree, measured clockwise from north)
Angle of elevation of 40 m building from 60 m away.
The situation forms a right triangle: height = 40 m, horizontal distance = 60 m.
tan θ = opposite/adjacent = 40/60 = 2/3
θ = tan−1(2/3) ≈ tan−1(0.6667)
θ ≈ 33.7°
Space diagonal of rectangular box: l = 8 m, w = 6 m, h = 3 m.
Step 1: Diagonal of base = √(l² + w²) = √(64 + 36) = √100 = 10 m.
Step 2: Space diagonal using this as base of vertical right triangle:
d = √(10² + 3²) = √(100 + 9) = √109
d ≈ 10.4 m
Alternatively: d = √(8² + 6² + 3²) = √(64 + 36 + 9) = √109 ≈ 10.4 m
Two angles of elevation from points 20 m apart — find height.
Let d = horizontal distance from A to base of building. Height = h.
From A: tan 50° = h/d, so h = d tan 50°
From B (20 m further): tan 35° = h/(d + 20), so h = (d + 20) tan 35°
Setting equal: d tan 50° = (d + 20) tan 35°
d × 1.1918 = (d + 20) × 0.7002
1.1918d = 0.7002d + 14.004
0.4916d = 14.004
d = 28.49 m
Height: h = 28.49 × tan 50° = 28.49 × 1.1918 ≈ 33.9 m
Yacht: 15 km on bearing 040° to A, then 10 km on bearing 130° to B. Find PB and bearing of B from P.
Interior angle at A:
Back-bearing of leg PA (from A looking back to P) = 040 + 180 = 220°
Bearing of AB = 130°
Interior angle at A = 220 − 130 = 90° (right angle at A)
Bearing of B from P:
sin(∠APB) = AB/PB = 10/18.03 = 0.5547
∠APB ≈ 33.7°
Bearing = bearing of PA + ∠APB = 040 + 33.7 ≈ 074°
Right pyramid: square base side 10 m, vertical height 12 m. Find slant height and angle of slant edge with base.
(a) Slant height (apex to midpoint of base edge):
The midpoint of a base edge is 5 m from the centre of the base (half the side length).
Slant height = √(5² + 12²) = √(25 + 144) = √169 = 13 m
(b) Angle of slant edge with base:
The slant edge goes from apex to a base vertex.
Distance from centre of base to a vertex = √(5² + 5²) = 5√2 ≈ 7.071 m
Slant edge length = √(7.071² + 12²) = √(50 + 144) = √194 ≈ 13.93 m
Angle with base: tan θ = 12/7.071
θ = tan−1(12/7.071) = tan−1(1.697) ≈ 59.5°
Cliff 80 m high; angles of depression of two boats 25° and 40°. Find distance between boats.
Let d1 = horizontal distance to nearer boat (depression 40°).
Let d2 = horizontal distance to farther boat (depression 25°).
For nearer boat: tan 40° = 80/d1
d1 = 80/tan 40° = 80/0.8391 ≈ 95.3 m
For farther boat: tan 25° = 80/d2
d2 = 80/tan 25° = 80/0.4663 ≈ 171.6 m
Distance between boats = d2 − d1 = 171.6 − 95.3 = 76.3 m
Towns A and B are 50 km apart. C is on bearing 070° from A and bearing 310° from B. Find AC and BC.
Assume A is due west of B (AB is due east, so bearing of B from A = 090°).
Angle at A (between AB and AC):
Bearing of AB from A = 090°; bearing of AC = 070°
Angle CAB = 90 − 70 = 20°
Angle at B (between BA and BC):
Bearing of BA from B = 270°; bearing of BC = 310°
Angle ABC = 310 − 270 = 40°
Angle at C:
Angle ACB = 180 − 20 − 40 = 120°
Sine rule:
AB/sin C = AC/sin B = BC/sin A
50/sin 120° = AC/sin 40° = BC/sin 20°
AC = 50 × sin 40°/sin 120° = 50 × 0.6428/0.8660 ≈ 37.1 km
BC = 50 × sin 20°/sin 120° = 50 × 0.3420/0.8660 ≈ 19.7 km
Rectangular field ABCD: AB = 80 m, BC = 60 m. Vertical pole TP at corner A. Angle of elevation of T from C is 20°. Find height, CT, and elevation from midpoint M of BC.
Diagonal AC (horizontal distance from A to C):
AC = √(80² + 60²) = √(6400 + 3600) = √10000 = 100 m
Height TA:
tan 20° = TA/AC
TA = 100 × tan 20° = 100 × 0.3640 = 36.4 m
Distance CT (slant from C to top of pole):
CT = √(AC² + TA²) = √(100² + 36.4²) = √(10000 + 1324.96) = √11324.96 ≈ 106.4 m
Elevation of T from midpoint M of BC:
M is the midpoint of BC, so BM = 30 m.
Horizontal distance AM: A is at one corner, B is adjacent. AB = 80 m, BM = 30 m.
AM = √(AB² + BM²) = √(80² + 30²) = √(6400 + 900) = √7300 ≈ 85.44 m
Angle of elevation = tan−1(TA/AM) = tan−1(36.4/85.44) = tan−1(0.4260) ≈ 23.1°
Three points A, B, C in horizontal plane: AB = 120 m, AC = 150 m, angle BAC = 55°. Vertical tower of 60 m at C. Find area of triangle ABC and elevation of tower top from B.
Area of triangle ABC:
Area = ½ × AB × AC × sin(BAC)
= ½ × 120 × 150 × sin 55°
= 9000 × 0.8192
= 7373 m²
Distance BC (horizontal) using cosine rule:
BC² = AB² + AC² − 2 × AB × AC × cos(BAC)
= 120² + 150² − 2 × 120 × 150 × cos 55°
= 14400 + 22500 − 36000 × 0.5736
= 36900 − 20650
= 16250
BC = √16250 ≈ 127.5 m
Angle of elevation of tower top from B:
The tower is vertical at C with height 60 m. Horizontal distance from B to base of tower = BC ≈ 127.5 m.
tan θ = 60/127.5 = 0.4706
θ = tan−1(0.4706) ≈ 25.2°