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Pythagoras and trigonometry in right angled triangles - Mathematics Year 9

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Introduction

Which information determines a missing side or angle? Choose Pythagoras when two right-triangle sides are known, and a trigonometric ratio when an acute angle and suitable sides are involved.

Solve four Pythagoras and four trigonometry tasks with labelled schematics. Keep exact values until final rounding, check degree mode, and explain why opposite and adjacent change when the selected acute angle changes.

Label right triangles and solve four Pythagoras tasks

A right triangle has one 90° angle. Its hypotenuse is opposite that angle and is its longest side. The other two sides are legs. Relative to one selected acute angle θ, the opposite leg is across from θ and the adjacent leg touches θ but is not the hypotenuse. Rotating the picture does not change these roles; choosing the other acute angle exchanges the two leg roles.

For legs a, o and hypotenuse h, Pythagoras gives $h^{2} = a^{2} + o^{2}$ . This is a relation between squared lengths, so h is the positive square root of their sum. A missing leg instead requires subtracting the known leg square from h². Adding squares for every missing-side question ignores which side is the hypotenuse.

Four Pythagoras tasks. Each schematic states a right angle at Q, selected acute angle at P, and the given side roles. Units are centimetres. They are not scale drawings; derive the unknown rather than measuring the screen.

Task	Adjacent leg a	Opposite leg o	Hypotenuse h	Squared relation	Positive length
P 1	5 cm	12 cm	?	h²=25+144=169	h=13 cm
P 2	8 cm	6 cm	?	h²=64+36=100	h=10 cm
P 3	?	15 cm	17 cm	a²=289−225=64	a=8 cm
P 4	7 cm	?	25 cm	o²=625−49=576	o=24 cm

P1:
          R
         /|
        / |
       /  |
      /   |
     P----Q
Right angle at Q; θ at P.
adj=5cm; opp=12cm; hyp=?
θ: selected acute angle
Schematic; not to scale.

P2:
          R
         /|
        / |
       /  |
      /   |
     P----Q
Right angle at Q; θ at P.
adj=8cm; opp=6cm; hyp=?
θ: selected acute angle
Schematic; not to scale.

P3:
          R
         /|
        / |
       /  |
      /   |
     P----Q
Right angle at Q; θ at P.
adj=?; opp=15cm; hyp=17cm
θ: selected acute angle
Schematic; not to scale.

P4:
          R
         /|
        / |
       /  |
      /   |
     P----Q
Right angle at Q; θ at P.
adj=7cm; opp=?; hyp=25cm
θ: selected acute angle
Schematic; not to scale.

In source P 1, the legs 5 and 12 give h²=25+144=169, h=13 cm. Independently 13²−12²=25 rebuilds the other leg square. The numerical root −13 also squares to 169, but a physical length is positive. P 2 similarly gives 10 cm from 6 and 8.

For P 3, the hypotenuse 17 cm is already known, so a²=17²−15²=64 and a=8 cm. Check 8²+15²=17². For P 4, the missing leg is√(25²−7²)=24 cm, with 49+576=625 as an independent original-relation check. Each result is smaller than its given hypotenuse.

The right-angle condition is essential. Sides 5 and 12 in an arbitrary triangle do not force the third to be 13; the included angle can vary. No unmarked diagram is silently promoted to a right triangle because its outline looks familiar. Retain the side identification, squared relation, positive root and unit for all four tasks.

Connect similarity, sine, cosine and tangent

An acute triangle with a perpendicular height from its top vertex to the base. — The height marker creates right-angled subtriangles. It does not make the entire outside triangle right-angled. Choose a marked right subtriangle and identify its own hypotenuse before applying these methods.

Similar right triangles with the same acute angle have equal corresponding side ratios. This makes the following ratios depend on the selected acute angle rather than on a triangle's overall size:

$\sin θ = \frac{o}{h}, \cos θ = \frac{a}{h}, \tan θ = \frac{o}{a}$

Here o is the opposite leg, a the adjacent leg and h the hypotenuse. The letters name roles, not fixed positions on every drawing. Sine and cosine of an acute angle lie strictly between 0 and 1 because a leg is shorter than the hypotenuse. Tangent can exceed 1. These ratios are dimensionless when both lengths use the same units.

For a right triangle with opposite 6 cm, adjacent 8 cm, hypotenuse 10 cm, the ratios are 6/10=.6, 8/10=.8 and 6/8=.75. The squared-side check 36+64=100 confirms the given right-triangle data. Its similarity-scaled 12/16/20 version has the same three ratios. This is why the angle need not change when every length doubles.

To find an angle from a ratio, use the inverse function, not its reciprocal. The source $θ = \sin^{- 1} (0.6) \approx 36. 9^{\circ}$ uses degree mode and rounds only at the end to one decimal place. The calculator's inverse-sine notation here means an angle whose sine is.6; it does not mean 1/.6. The other acute angle is approximately 53.1° because the pair totals 90°.

Choosing that other acute angle exchanges opposite and adjacent. Its sine is.8 and cosine.6, while its hypotenuse remains 10 cm. Label θ before selecting SOH, CAH or TOA; memorising a mnemonic cannot repair a mismatched side role. If a proposed acute-angle sine is 1.6, recheck the side identification or input rather than accepting an impossible acute ratio.

The original Corbettmaths introduction explicitly labels opposite, adjacent and hypotenuse and discusses degree mode. Its reference angle is at a different drawn location from our Greek-labelled Commons figure; match roles relative to θ instead of copying screen positions. All ratios and task data are also written in English. Original: corbettmaths, Trigonometry Introduction - Corbettmaths. The written explanations and native tasks work without sound or the external player.

Use a simple diagnostic: with two sides and a missing third side, Pythagoras may avoid angle rounding. With a known acute angle and one side, choose the ratio containing that known side and the required side. With two side lengths and an unknown acute angle, form an appropriate ratio and use its inverse function. Always retain the given right-angle premise.

Solve four trigonometry tasks with diagrams and precision

Four trigonometry tasks. Each task below has its own labelled schematic and a stated right angle. An answer's displayed decimal precision is part of its meaning; trailing zeros in 8.40 and 14.00 retain the requested two decimal places rather than claiming exact equality.

Task	Given quantities	Find	Chosen equation	Rounded result	Precision
T 1	o=6 cm, h=10 cm	θ	sinθ=6/10	36.9°	1 decimal place
T 2	h=10 cm, θ=50°	a	a=10 cos 50°	6.43 cm	2 decimal places
T 3	a=12 cm, θ=35°	o	o=12 tan 35°	8.40 cm	2 decimal places
T 4	o=9 cm, θ=40°	h	h=9/sin 40°	14.00 cm	2 decimal places

T1:
          R
         /|
        / |
       /  |
      /   |
     P----Q
Right angle at Q; θ at P.
adj=not given; opp=6cm; hyp=10cm
θ: unknown
Schematic; not to scale.

T2:
          R
         /|
        / |
       /  |
      /   |
     P----Q
Right angle at Q; θ at P.
adj=unknown; opp=not given; hyp=10cm
θ: 50°
Schematic; not to scale.

T3:
          R
         /|
        / |
       /  |
      /   |
     P----Q
Right angle at Q; θ at P.
adj=12cm; opp=unknown; hyp=not given
θ: 35°
Schematic; not to scale.

T4:
          R
         /|
        / |
       /  |
      /   |
     P----Q
Right angle at Q; θ at P.
adj=not given; opp=9cm; hyp=unknown
θ: 40°
Schematic; not to scale.

T 1 uses inverse sine in degree mode: arcsin(.6)=36.869897…°, so θ≈36.9°. Its complementary angle and the corresponding 8 cm adjacent side offer independent checks through cosine or Pythagoras. Do not round.6 to a different ratio before finding the angle.

T 2 uses cosine because adjacent and hypotenuse are the selected sides: a/10=cos 50°, so a=10 cos 50°≈6.427876…cm, or 6.43 cm. It is shorter than 10 cm as required. Multiplying by sine 50° would find the other leg, not the requested adjacent one.

T 3 uses tangent: o/12=tan 35°, so o≈8.402490…cm, or 8.40 cm. Check the unrounded quotient o/12 against tan 35°. T 4 has the unknown in the denominator: 9/h=sin 40°. Multiply both sides by h and divide by sin 40° to obtain h≈14.001514…cm, or 14.00 cm. Multiplying 9 by sin 40° would produce a value below 9 and cannot be the hypotenuse here.

Unfamiliar transfer. In a right triangle with h 14 cm and θ30°, the opposite leg is 14 sin 30°=7 cm and adjacent leg 7√3 cm, approximately 12.12 cm. The squared relation 49+147=196 independently confirms h². A tangent ratio 7/(7√3)=1/√3 agrees with tan 30°. State exact radicals separately from final rounded lengths.

First-error and evidence check. A calculator set to radians interprets 35 as a radian angle, not the stated 35°. Check the angle unit before evaluation. A correct numerical button sequence with swapped opposite/adjacent labels still solves the wrong problem. Retain all four Pythagoras and four trig methods, schematics, units, precision and an original-relation or reasonableness check.

These are mathematical model tasks. A displayed length or angle is not a real survey measurement or evidence of practical instrument competence. Keep actual later retrieval and any actual drawing or measurement assessment separately; the numerical answer key does not fabricate them.

Interactive Tasks

Quiz: Test Your Knowledge

Legs 5 cm and 12 cm in a right triangle give hypotenuse? (13 cm) (!17 cm) (!7 cm) (!169 cm)

Hypotenuse 17 cm and leg 15 cm give the other leg? (8 cm) (!32 cm) (!2 cm) (!22.67 cm)

The hypotenuse lies opposite which angle? (The right angle) (!The smallest acute angle) (!Any angle selected at random) (!An exterior angle)

Relative to θ, sine uses? (Opposite divided by hypotenuse) (!Adjacent divided by hypotenuse) (!Opposite divided by adjacent) (!Hypotenuse divided by opposite)

Opposite 6, hypotenuse 10: θ to one decimal place in degrees? (36.9°) (!0.6°) (!53.1°) (!60.0°)

Hypotenuse 10, θ50°: adjacent to two decimal places? (6.43 cm) (!7.66 cm) (!15.56 cm) (!5.00 cm)

Adjacent 12, θ35°: opposite to two decimal places? (8.40 cm) (!17.14 cm) (!9.83 cm) (!6.88 cm)

Opposite 9, θ40°: hypotenuse to two decimal places? (14.00 cm) (!5.79 cm) (!11.75 cm) (!7.55 cm)

Choosing the other acute angle does what? (Exchanges opposite and adjacent roles) (!Changes which side is the hypotenuse) (!Makes every length negative) (!Makes both angles 90°)

Inverse sine of.6 means? (An angle whose sine is.6) (!The reciprocal 1/.6) (!A length.6 cm) (!A proof that the triangle is not right-angled)

Memory Game

Hypotenuse	Side opposite the right angle
Opposite	Leg across from the selected acute angle
Adjacent	Leg touching the selected angle other than the hypotenuse
Sine	Opposite divided by hypotenuse
Cosine	Adjacent divided by hypotenuse
Tangent	Opposite divided by adjacent

Drag and Drop

13 cm	Source 5-and 12 hypotenuse
8 cm	Missing leg with hypotenuse 17 and leg 15
36.9°	Source inverse-sine angle to one decimal place
6.43 cm	T 2 adjacent length to two decimal places
14.00 cm	T 4 hypotenuse to two decimal places

Crossword Puzzle

HYPOTENUSE	Side opposite the right angle
OPPOSITE	Leg across from the selected acute angle
ADJACENT	Non-hypotenuse leg beside the selected angle
SINE	Ratio of opposite to hypotenuse
COSINE	Ratio of adjacent to hypotenuse
TANGENT	Ratio of opposite to adjacent

LearningApps

Optional extra practice. Choose an appropriate activity with your teacher. If the external collection is unavailable, use the native activities and independent tasks in this course.

Cloze Text

Open Tasks

Easy

Label side roles: Use the marked right angle and selected θ on each schematic.
Solve the source 5/12 triangle: Show squared sum 169 and positive root 13.
Form source trigonometric ratios: Use 6/8/10 to get sine.6, cosine.8, tangent.75.
Explain degree mode: Distinguish an inverse-sine angle from a reciprocal.

Standard

Solve four Pythagoras tasks: Retain each diagram, side roles, add/subtract choice, positive length and unit.
Solve four trig tasks: Use all four labelled T diagrams and the specified final precision.
Check an answer independently: Use Pythagoras, complementarity or a rebuilt ratio.
Repair an unknown-denominator error: Explain why T 4 uses 9/sin 40° rather than 9 sin 40°.

Advanced

Transfer the 30° triangle: Find exact 7 and 7√3 from hypotenuse 14 and verify both squares and ratios.
Connect ratios to similarity: Double 6/8/10 and show all three ratios unchanged.
Compare efficient methods: Choose two-side Pythagoras versus angle-and-side trig without early rounding.
Retain actual later evidence: Keep eight diagrams/methods, error correction and a real delayed response without fabricated measurement.

Learning Check

Four Pythagoras solutions: Use the correct side role and positive root.
Four trigonometry answers: State ratio, degree mode and final precision.
Similarity and angle roles: Explain the ratios for 6/8/10 and a doubled triangle.
Independent checks and first error: Rebuild a relationship and reject an impossible result.
Unfamiliar exact transfer: Use the 30° right triangle with hypotenuse 14.
Portfolio and evidence limits: Retain all required diagrams, methods and actual work.

Worked checks and assessment guidance

P 1: √(5²+12²)=13 cm; P 2: √(8²+6²)=10 cm; P 3: √(17²−15²)=8 cm; P 4: √(25²−7²)=24 cm. Original square checks 169, 100, 289, 625 all agree; the right angle is a given in every schematic.
T 1: arcsin(6/10)≈36.9° to 1 dp. T 2: 10 cos 50°≈6.43 cm to 2 dp. T 3: 12 tan 35°≈8.40 cm to 2 dp. T 4: 9/sin 40°≈14.00 cm to 2 dp. Use unrounded intermediates and the stated selected angle.
Sine.6, cosine.8, tangent.75 are unchanged for 12/16/20. Choosing the complementary acute angle exchanges leg roles, sine.8/cosine.6, while the 10 cm hypotenuse remains opposite 90°.
13²−12²=25; 8²+15²=289. In T 4, h must exceed 9, so 9 sin 40° is impossible as its hypotenuse. Inverse sine is not 1/.6, and degree tasks require degree mode. Rounded values are approximate, not exact equations.
Opposite 14 sin 30°=7 cm; adjacent 7√3 cm≈12.12 cm. Squares 49+147=196 verify the hypotenuse, and tangent 1/√3 agrees with 30°. Keep radical exactness separate from rounded measurements.
Keep four Pythagoras and four trig solutions with labelled right-angle schematics, side roles, units, precision, independent checks and first-error repair. Online answers do not certify unobserved instrument competence or future mastery; retain actual later evidence when performed.

Learning evidence

Retain four Pythagoras and four trigonometry solutions with their own labelled diagrams, method choices, units, final precision, independent checks, first-error correction and exact unfamiliar transfer.

OER and Wikipedia

Department for Education: England mathematics programme of study, Key Stage 3, number, ratio, fluency, reasoning and problem solving. Year 9 allocation is editorial within KS3. Numerical models, tasks and feedback are original; images retain actual Commons attribution and videos name their original publisher. Reviewed 12 November 2026.

Right triangle trig.svg, Dimitris131, Public domain. Unchanged original; caption and learning activities are original.

Triangle (acute) bh.PNG, H Padleckas at English Wikibooks, Public domain. Unchanged original; caption and learning activities are original.