Label the sides from the angle, and the ratio picks itself.
The problem
Standing 40 m from a building, the angle of elevation to the top is 32°. How tall is the building?
The 40 m runs along the ground, adjacent to the angle, and the height is opposite — so it is TOA: h = 40 tan 32° ≈ 25 m. B uses sine, which needs the hypotenuse, and the ground distance is not the hypotenuse.
Opposite and adjacent are named relative to the angle, not the page.
| Side | How to find it | Here |
|---|---|---|
| hypotenuse | opposite the right angle | the line of sight |
| opposite | across from the angle | the height, h |
| adjacent | touching the angle | the 40 m |
sin = opphyp
cos = adjhyp
tan = oppadj
We have the adjacent and we want the opposite — the only ratio holding both is tan.
Tick the side you know and the side you want. Then pick the ratio that contains exactly those two. There is never more than one.
tan 32° = h40
opposite over adjacent
h = 40 · tan 32°
× 40 both sides
h = 40 · 0.6249
from the calculator
h ≈ 25.0
metres
In radian mode, tan 32 returns 0.6610 and the building becomes 26.4 m. Look for DEG on the screen before you start.
Same ratio, one extra step — and this is where marks get lost.
tan 32° = 25d
now the distance is unknown
d · tan 32° = 25
× d both sides
d = 25tan 32° ≈ 40
÷ tan 32°
Unknown on top — multiply. Unknown underneath — divide by the ratio. Reaching for tan⁻¹ here is the usual wrong turn; that button is for finding angles.
Both are measured from the horizontal, never from the vertical.
The two are equal — they are alternate angles between parallel horizontals.
An angle of depression is measured down from the horizon line, not across from the cliff face. Drawing the horizontal dashed line first prevents the mistake.
In right △PQR with ∠Q = 90°, cos P = 2029. Find sin R and tan R. A ratio is two sides of the triangle — draw them, then Pythagoras gives the third.
cos P = adj/hyp = PQ/PR = 2029
take PQ = 20, PR = 29
QR = √29² − 20² = √441 = 21
the missing side
from R: opp = PQ = 20, adj = QR = 21
re-label from the new angle
sin R = 2029, tan R = 2021
exact fractions
The side opposite P is adjacent to R. So sin R = cos P always, and tan R is tan P turned upside down. Answering 2120 reads tan from the wrong corner.
A kite on a straight 50 m string flies 25 m above the ground. Find the angle θ between the string and the ground.
sin θ = 2550 = 12
opposite and hypotenuse known
θ = sin⁻¹(12) = 30°
a special angle
Different: both sides are known and the angle is not. This is the one time the inverse button (or a special-angle value) is the right move.
A balloon is 30 m above the ground, at an angle of elevation of 60° from an observer. Find the straight-line distance to it.
sin 60° = 30L
opposite over hypotenuse
L = 30 ÷ (√32) = 60√3
unknown underneath: divide
L = 20√3 m ≈ 34.6 m
× √3√3
Different: the hypotenuse is wanted, so the answer must come out longer than 30. Multiplying, 30 × √32 ≈ 26, fails that test.
In right △ABC with ∠C = 90°, tan A = 23. Find sin A.
opp = 2, adj = 3
build the triangle
hyp = √4 + 9 = √13
not a triple
sin A = 2√13 = 2√1313
rationalise the denominator
Different: the third side is a root. Keep it exact and rationalise; the choices are usually written that way.
A sine given as a multiple of the cosine. If sin A = 3 cos A, divide both sides by cos A: tan A = 3 = 31. Build the triangle with opp 3, adj 1, hyp √10, so cos A = 1√10 = √1010 and sin A = 3√1010. A sine or cosine above 1 means the division was skipped.
Right △ABC has ∠C = 90°, AC = 5, BC = 12, and CD is the altitude to AB. Find cos ∠BCD.
∠BCD = 90° − ∠B = ∠A
both are complements of ∠B
AB = 13
5-12-13
cos ∠BCD = cos A = 513
read it in the big triangle
From P the top of a tower has elevation 60°. From Q, 30 m farther away in line, it has elevation 30°. Find the height h. Let the tower’s foot be x from P.
h = x tan 60° = x√3
from P
h = (x + 30) tan 30° = x + 30√3
from Q
3x = x + 30, x = 15
× √3, then solve
h = 15√3 m
≈ 26 m
Two more moves with the altitude CD. When it splits the hypotenuse into AD and DB, the two small triangles are similar, so CD² = AD × DB. With AD = 2 and DB = 8, CD = 4, and tan A = CD/AD = 42 = 2. And when the legs are known, find the altitude from the area two ways: legs 9 and 12 give hypotenuse 15, so ½ × 9 × 12 = ½ × 15 × CD and CD = 108 ÷ 15 = 7.2.
UPCAT-type papers seldom allow a calculator, so the angles are 30°, 45° and 60° (tan 30° = 1√3, tan 45° = 1, tan 60° = √3) or a ratio is given as a fraction to build a triangle from. Expect a ratio from the other acute angle, an angle hidden by an altitude, or two observers sharing one height. Ten-second check: sine and cosine are never above 1, and the side facing the bigger angle is longer.
The problem
A 10 m ladder leans against a wall at 65° to the ground. How far up the wall does it reach?
The ladder is the hypotenuse and the wall height sits opposite the 65°, so it is SOH: 10 sin 65° ≈ 9.1 m. C uses cosine, which gives the ladder's distance from the wall instead. Choose the ratio from the two sides you actually have.
Hint: this time you know the hypotenuse, not the adjacent — so tan is the wrong choice. You want the ratio holding opposite and hypotenuse.
