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Recap D — Trigonometry

SOH CAH TOA, exact values, Pythagoras, angles of elevation and depression, and two-triangle problems — 15 worked examples, each with a diagram.

The Big Idea

Trigonometry connects the angles of a right-angled triangle with the lengths of its sides. First, name the sides from the angle you are using : Hypotenuse (H) — the longest side, opposite the right angle Opposite (O) — the side across from the angle Adjacent (A) — the side next to the angle (not the hypotenuse)

Four right-angled triangles in different positions, each with the hypotenuse, opposite and adjacent sides labelled for the marked angle. — $The labels move when the triangle turns — always start from the marked angle.$

1. SOH CAH TOA

Exact values — learn this table (3 = 1.732, 2 = 1.414): 30 ° 45 ° 60 ° sin \frac{1}{2} = 0.5 \frac{1}{2} \approx 0.707 \frac{3}{2} \approx 0.866 cos \frac{3}{2} \approx 0.866 \frac{1}{2} \approx 0.707 \frac{1}{2} = 0.5 tan \frac{1}{3} \approx 0.577 1 3 \approx 1.732

	$30$ °	$45$ °	$60$ °
sin	$\frac{1}{2} = 0.5$	$\frac{1}{2} \approx 0.707$	$\frac{3}{2} \approx 0.866$
cos	$\frac{3}{2} \approx 0.866$	$\frac{1}{2} \approx 0.707$	$\frac{1}{2} = 0.5$
tan	$\frac{1}{3} \approx 0.577$	$1$	$3 \approx 1.732$

Right-angled triangle with a 60 degree angle, adjacent side 10 m and the opposite side unknown. — $Example 1$

Example 1 — find the opposite (tan)

Question: In the triangle, the angle is 60 ° and the adjacent side is 10 m . Find the opposite side. Answer: Know A, want O \to tan . tan 60° = \frac{O}{10}, so O = 10 \times tan 60° = 10 \times 1.732 = 17.32 m

Right-angled triangle with a 30 degree angle, opposite side 8 m and the hypotenuse unknown. — $Example 2$

Example 1 — find the hypotenuse (sin)

Question: The angle is 30 ° and the opposite side is 8 m . Find the hypotenuse. Answer: Know O, want H \to sin . sin 30° = \frac{8}{H}, so H = \frac{8}{0.5} = 16 m

Right-angled triangle with a 60 degree angle, hypotenuse 20 m and the adjacent side unknown. — $Example 3$

Example 1 — find the adjacent (cos)

Question: The angle is 60 ° and the hypotenuse is 20 m . Find the adjacent side. Answer: Know H, want A \to cos . cos 60° = \frac{A}{20}, so A = 20 \times 0.5 = 10 m

Right-angled triangle with a 45 degree angle, adjacent side 7 cm, opposite and hypotenuse unknown. — $Example 4$

Example 1 — a 45° triangle

Question: The angle is 45 ° and the adjacent side is 7 cm. Find the opposite side and the hypotenuse. Answer: tan 45° = 1, so O = 7 \times 1 = 7 cm. cos 45° = \frac{7}{H}, so H = 7 \times 2 = 7 \times 1.414 = 9.90 cm. (Dividing by \frac{1}{2} is the same as multiplying by 2 .)

Right-angled triangle with opposite 5 m, adjacent 5 m and the angle unknown. — $Example 5$

Example 1 — find the angle

Question: The opposite side is 5 m and the adjacent side is 5 m . Find the angle θ. Answer: Know O and A \to tan . tan θ = \frac{5}{5} = 1 . From the table, tan 45° = 1, so θ = 45° .

2. Pythagoras' Theorem

Right-angled triangle with shorter sides 6 cm and 8 cm and the hypotenuse unknown. — $Example 6$

Example 1 — find the hypotenuse

Question: The two shorter sides are 6 cm and 8 cm. Find the hypotenuse. Answer: h^{2} = 6^{2} + 8^{2} = 36 + 64 = 100, so h = 100 = 10 cm

Right-angled triangle with hypotenuse 13 cm, one side 5 cm and the other side unknown. — $Example 7$

Example 1 — find a shorter side

Question: The hypotenuse is 13 cm and one side is 5 cm. Find the other side. Answer: a^{2} = 1 3^{2} - 5^{2} = 169 - 25 = 144, so a = 12 cm

A ladder 2.5 m long leaning against a wall, with its foot 0.7 m from the wall; the height is unknown. — $Example 8$

Example 1 — a ladder

Question: A 2.5 m ladder leans against a wall. Its foot is 0.7 m from the wall. How high up the wall does it reach? Answer: The ladder is the hypotenuse. h^{2} = 2. 5^{2} - 0. 7^{2} = 6.25 - 0.49 = 5.76, so h = 5.76 = 2.4 m

3. Angles of Elevation and Depression

A building 40 m away seen at an angle of elevation of 30 degrees; the height is unknown. — $Example 9$

Example 1 — elevation, no eye height

Question: From a point 40 m from a building, the angle of elevation of the top is 30 °. How tall is the building? Answer: tan 30° = \frac{h}{40}, so h = 40 \times \frac{1}{3} = \frac{40}{1.732} = 23.09 m

A person with eyes 1.6 m above the ground, 30 m from a tree, looking up at 45 degrees. — $Example 10$

Example 1 — elevation with eye height

Question: Kumari stands 30 m from a tree. The angle of elevation of the top is 45 °, and her eyes are 1.6 m above the ground. How tall is the tree? Answer: 1 Height above her eyes: h = 30 tan 45° = 30 \times 1 = 30 m 2 Add her eye height: 30 + 1.6 = 31.6 m

A person with eyes 1.5 m above the ground, 25 m from a tower, looking up at 60 degrees. — $Example 11$

Example 1 — elevation 60° with eye height

Question: A student 25 m from a tower sees the top at 60 °. Eye level is 1.5 m . Find the height of the tower. Answer: 1 h = 25 tan 60° = 25 \times 1.732 = 43.30 m 2 Tower = 43.30 + 1.5 = 44.80 m

A 50 m cliff with a boat below at an angle of depression of 30 degrees; the distance is unknown. — $Example 12$

Example 1 — angle of depression

Question: From the top of a 50 m cliff, the angle of depression of a boat is 30 °. How far is the boat from the foot of the cliff? Answer: The angle of depression at the top equals the angle of elevation from the boat (alternate angles), so the angle at the boat is 30 °. tan 30° = \frac{50}{d}, so d = 50 \times 3 = 50 \times 1.732 = 86.60 m

A kite on a 60 m string at 30 degrees to the horizontal; the height is unknown. — $Example 13$

Example 1 — a kite

Question: A kite string is 60 m long and makes an angle of 30 ° with the horizontal. How high is the kite above the person's hand? Answer: The string is the hypotenuse; the height is opposite the angle \to sin . h = 60 sin 30° = 60 \times 0.5 = 30 m

4. Two Triangles Sharing a Side

Right-angled triangle ABD with the right angle at B. Point C lies on DB. Angle ADB is 30 degrees, angle ACB is 60 degrees, DC is 24 m and CB is 12 m. — $Used in Examples 14 and 15 .$

Example 1 — find AB

Question: In the diagram, find the height AB. Answer: Use triangle ABC (you know CB = 12 m and the 60 ° angle). From 60 °, CB is adjacent and AB is opposite \to tan : tan 60° = \frac{A B}{12}, so A B = 12 \times 1.732 = 20.78 m

Example 2 — find AD and AC

Question: Now find AD and AC. Answer: AD — use triangle ABD . From 30 °, AB is opposite and AD is the hypotenuse \to sin : sin 30° = \frac{A B}{A D}, so A D = \frac{20.78}{0.5} = 41.57 m AC — in triangle ABC, cos 60° = \frac{C B}{A C}, so A C = \frac{12}{0.5} = 24 m

Your turn — try these on paper, then check the answers below. 1 . An angle of 30 ° has an opposite side of 9 m . Find the hypotenuse. 2 . From 50 m away, the angle of elevation of a tower is 60 °. Eye level is 1.5 m . Find the height of the tower. 3 . The shorter sides of a right-angled triangle are 9 cm and 12 cm. Find the hypotenuse.

Example 1 — Answers to Your Turn

1 . H = 9 \div 0.5 = 18 m 2 . 50 \times 1.732 = 86.6 m, plus 1.5 \to 88.1 m 3 . 81 + 144 = 225, so h = 15 cm

Next: Recap E — Descriptive Statistics

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