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Pythagoras and trigonometry - Mathematics Year 10

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Introduction

Choose a relationship from the information given. Two known sides in a right triangle suggest Pythagoras; an acute angle and a suitable side suggest a trigonometric ratio.

Solve six Pythagoras and six trigonometry questions with individual diagrams, then check signs, units and rounding. Exact values and non-right-triangle methods form a separately labelled Higher-tier or specification-sensitive extension.

Solve six right-triangle side questions

The hypotenuse is opposite the right angle and is the longest side. Relative to the selected acute angle θ, one remaining side is opposite and the other adjacent. Changing θ changes these latter roles; it does not change which side is the hypotenuse.

For a right triangle with perpendicular legs a, o and hypotenuse h, $a^{2} + o^{2} = h^{2} .$ Find a missing hypotenuse by adding the two leg squares; find a missing leg by subtracting the known leg square from the hypotenuse square. Take the positive root for a physical length and retain consistent units.

Six complete Pythagoras questions. Each table row and its matching schematic supplies a separate triangle. The marked right angle is at Q, so PR is the hypotenuse. “Not to scale” means that the drawing’s apparent side lengths cannot replace its stated data.

Task	Given and requested	Squared-length working	Answer
P 1	h 13 cm, a 5 cm; find o	o²=13²−5²=144	12 cm
P 2	a 6 cm, o 8 cm; find h	h²=6²+8²=100	10 cm
P 3	h 17 cm, o 8 cm; find a	a²=17²−8²=225	15 cm
P 4	a 4.8 m, o 3.6 m; find h	h²=4.8²+3.6²=36	6 m
P 5	a 7 cm, o 7 cm; find h	h²=7²+7²=98	7√2 cm≈9.90 cm
P 6	h 25 cm, a 24 cm; find o	o²=25²−24²=49	7 cm

P1: right angle at Q; angle θ at P.
        R
       /|
      / |
     P--Q
PQ = adjacent a: 5cm
QR = opposite o: ?cm
PR = hypotenuse h: 13cm
θ = not required
Schematic; not to scale.

P2: right angle at Q; angle θ at P.
        R
       /|
      / |
     P--Q
PQ = adjacent a: 6cm
QR = opposite o: 8cm
PR = hypotenuse h: ?cm
θ = not required
Schematic; not to scale.

P3: right angle at Q; angle θ at P.
        R
       /|
      / |
     P--Q
PQ = adjacent a: ?cm
QR = opposite o: 8cm
PR = hypotenuse h: 17cm
θ = not required
Schematic; not to scale.

P4: right angle at Q; angle θ at P.
        R
       /|
      / |
     P--Q
PQ = adjacent a: 4.8m
QR = opposite o: 3.6m
PR = hypotenuse h: ?m
θ = not required
Schematic; not to scale.

P5: right angle at Q; angle θ at P.
        R
       /|
      / |
     P--Q
PQ = adjacent a: 7cm
QR = opposite o: 7cm
PR = hypotenuse h: ?cm
θ = not required
Schematic; not to scale.

P6: right angle at Q; angle θ at P.
        R
       /|
      / |
     P--Q
PQ = adjacent a: 24cm
QR = opposite o: ?cm
PR = hypotenuse h: 25cm
θ = not required
Schematic; not to scale.

P 1 is the source example: 13²−5²=169−25=144, so the missing leg is 12 cm. Squaring and adding back gives 5²+12²=169=13². Adding 169 and 25 would instead calculate a hypotenuse from two legs, which is the wrong structure for this task.

P 2 and P 4 need addition because their unknown is the hypotenuse. P 3 and P 6 need subtraction. In P 5, the exact root√98 simplifies to 7√2; the decimal 9.90 cm is a final approximation. Keeping the exact form prevents an intermediate rounding decision from changing later work.

For every task, check that the hypotenuse exceeds either leg. The rectangle interpretation in P 4 uses two perpendicular sides 4.8 m and 3.6 m, so its diagonal is 6 m. A diagonal is a length; the same rectangle’s area 17.28 m² answers a different question. These are synthetic mathematical dimensions, not evidence of an actual site survey.

Choose ratios for six trigonometry questions

For an acute angle in a right triangle, sine compares opposite to hypotenuse, cosine compares adjacent to hypotenuse, and tangent compares opposite to adjacent. In symbols, $\sin θ = \frac{o}{h}, \cos θ = \frac{a}{h}, \tan θ = \frac{o}{a} .$ First label the side roles; then choose the ratio containing the known and requested quantities.

Six complete trigonometry questions. Use degree mode. Give requested angles to one decimal place and non-exact lengths to two decimal places. Keep calculator precision until the final answer. T 6 can be solved with a calculator here; the exact-value derivation is separated in the next chapter.

Task	Given and requested	Method	Answer
T 1	a 8 cm, h 10 cm; find θ	cosθ=8/10; θ=arccos 0.8	36.9°
T 2	h 12 cm, θ40°; find o	o=12 sin 40°	7.71 cm
T 3	a 9 cm, θ35°; find o	o=9 tan 35°	6.30 cm
T 4	o 7 cm, θ50°; find h	h=7/sin 50°	9.14 cm
T 5	o 12 cm, a 5 cm; find θ	θ=arctan 12/5	67.4°
T 6	h 14 cm, θ30°; find o	o=14 sin 30°	7 cm

T1: right angle at Q; angle θ at P.
        R
       /|
      / |
     P--Q
PQ = adjacent a: 8cm
QR = opposite o: unknown
PR = hypotenuse h: 10cm
θ = ?°
Schematic; not to scale.

T2: right angle at Q; angle θ at P.
        R
       /|
      / |
     P--Q
PQ = adjacent a: unknown
QR = opposite o: ?cm
PR = hypotenuse h: 12cm
θ = 40°
Schematic; not to scale.

T3: right angle at Q; angle θ at P.
        R
       /|
      / |
     P--Q
PQ = adjacent a: 9cm
QR = opposite o: ?cm
PR = hypotenuse h: unknown
θ = 35°
Schematic; not to scale.

T4: right angle at Q; angle θ at P.
        R
       /|
      / |
     P--Q
PQ = adjacent a: unknown
QR = opposite o: 7cm
PR = hypotenuse h: ?cm
θ = 50°
Schematic; not to scale.

T5: right angle at Q; angle θ at P.
        R
       /|
      / |
     P--Q
PQ = adjacent a: 5cm
QR = opposite o: 12cm
PR = hypotenuse h: unknown
θ = ?°
Schematic; not to scale.

T6: right angle at Q; angle θ at P.
        R
       /|
      / |
     P--Q
PQ = adjacent a: unknown
QR = opposite o: ?cm
PR = hypotenuse h: 14cm
θ = 30°
Schematic; not to scale.

T 1 is the source cosine example. Adjacent 8 and hypotenuse 10 give cosθ=0.8, so θ≈36.8699°, reported 36.9°. An inverse cosine finds an angle from a ratio; it is not the reciprocal 1/0.8. The unrounded ratio can be checked by a 6-8-10 triangle whose opposite is 6.

In T 2 and T 3, the unknown numerator is found by multiplication. T 4 instead has 7/h=sin 50°, so h=7/sin 50°. Multiplying 7 by sin 50°would produce a length smaller than the known leg and could not be the hypotenuse. Its unrounded value is approximately 9.137851 cm, reported 9.14 cm.

T 5 uses tangent because the given sides are opposite and adjacent, with no hypotenuse required. T 6 has opposite 7 cm under the stated 30°model. Reversing the selected acute angle would swap opposite and adjacent, so do not copy a ratio solely because a number appears beside a convenient edge.

The actual Corbettmaths Trigonometry Introduction preview labels hypotenuse, opposite and adjacent and displays all three ratios. Compare its selected acute angle with the explicit θ at P in our twelve schematics. Its reference drawing does not supply our task measurements. Original: corbettmaths, Trigonometry Introduction - Corbettmaths. The written explanations and native tasks work without sound or the external player.

An independent check uses a different relationship. For T 1, Pythagoras gives opposite 6, so sinθ=6/10 as well as cosθ=8/10. For a computed hypotenuse, confirm that its square exceeds each leg square and that the chosen ratio reconstructs the given side. These checks support mathematical consistency; they do not certify practical measurement accuracy.

Derive exact values and distinguish extensions

An acute triangle with a perpendicular altitude meeting its base. — A perpendicular altitude divides a suitable triangle into right subtriangles. The pictured triangle is not asserted to be equilateral or to carry our numerical dimensions;the equilateral and non-right examples below state their own assumptions.

Higher-tier or specification-sensitive exact-value extension. A unit square cut along a diagonal produces a 45°-45°-90°triangle with legs 1 and hypotenuse√2. Therefore $\sin 4 5^{\circ} = \cos 4 5^{\circ} = \frac{1}{\sqrt{2}} = \frac{\sqrt{2}}{2}, \tan 4 5^{\circ} = 1.$ The exact expressions are equal; rounding them before further work is unnecessary.

Split an equilateral triangle of side 2 along its perpendicular altitude. The halves have hypotenuse 2, short leg 1 and long leg√3, because the height squared is 4−1=3. The 30°angle is opposite the short leg. Thus sin 30°=1/2 and cos 60°=1/2, as in the source extension. Also cos 30°=sin 60°=√3/2, tan 30°=√3/3 and tan 60°=√3. Their roles follow from the selected angle, not from a memorised picture orientation.

For T 6, this reasoning gives opposite 14×1/2=7 exactly. The corresponding adjacent length is 7√3, whose square 147 plus opposite square 49 equals hypotenuse square 196. For a new right triangle with hypotenuse 18 and angle 30°, opposite 9 and adjacent 9√3 satisfy 81+243=324. This changes the values while retaining a checked exact structure.

Separate non-right-triangle extension, only where selected. Pythagoras and the simple right-triangle ratios need a right angle. For a triangle with sides b 8, c 10 and included angle A 60°, the cosine rule gives $a^{2} = b^{2} + c^{2} - 2 b c \cos A = 84, a = 2 \sqrt{21} \approx 9.17.$ The area is $\frac{1}{2} b c \sin A = 20 \sqrt{3} \approx 34.64.$ The included angle condition is essential; this is not an unmarked right triangle.

Another non-right example gives A 30°, B 45°and opposite side a 5. The third angle is 105°, and the sine rule gives b=5 sin 45°/sin 30°=5√2, approximately 7.07. Two angles fix the angle pattern here; do not silently generalise that every side-side-angle problem has a unique solution. These extensions are separately labelled and do not replace the twelve core right-triangle solutions.

First-error repair. If a learner adds squares when a leg is missing, start with the correctly labelled hypotenuse. If they multiply while the unknown is a denominator, rearrange the equation. If they obtain an inverse-ratio angle in radians, check degree mode. If they apply a right-triangle formula to the 60°included-angle example, use the general rule instead.

Retain all twelve given/unknown diagrams, full working, units and final precision. Distinguish exact values from displayed approximations and basic calculator work from specification-sensitive extensions. A later independent task or practical instrument demonstration counts as evidence only after it is actually completed or observed.

Interactive Tasks

Quiz: Test Your Knowledge

Hypotenuse 13 cm and leg 5 cm: other leg? (12 cm) (!√194 cm) (!8 cm) (!18 cm)

Perpendicular legs 6 cm and 8 cm: hypotenuse? (10 cm) (!14 cm) (!2 cm) (!48 cm)

Hypotenuse 17 cm and opposite 8 cm: adjacent? (15 cm) (!√353 cm) (!9 cm) (!25 cm)

Square with side 7 cm: exact diagonal? (7√2 cm) (!14 cm) (!49 cm) (!√14 cm)

Adjacent 8, hypotenuse 10: acute angle to one decimal degree? (36.9°) (!53.1°) (!0.8°) (!1.25°)

Hypotenuse 12 cm and angle 40°: opposite to two decimals? (7.71 cm) (!9.19 cm) (!18.67 cm) (!4.63 cm)

Adjacent 9 cm and angle 35°: opposite to two decimals? (6.30 cm) (!12.85 cm) (!5.16 cm) (!7.37 cm)

Opposite 7 cm and angle 50°: hypotenuse to two decimals? (9.14 cm) (!5.36 cm) (!10.89 cm) (!4.50 cm)

Opposite 12, adjacent 5: acute angle to one decimal degree? (67.4°) (!22.6°) (!2.4°) (!0.4°)

In the labelled exact-value extension, sin 30°equals? (1/2) (!√3/2) (!1) (!√2/2)

Memory Game

Hypotenuse	Side opposite the right angle
Opposite side	Leg across from the selected acute angle
Adjacent side	Leg beside the selected angle excluding the hypotenuse
Inverse trigonometric function	Finds an angle from a ratio within its stated range
Exact form	Value retained without decimal approximation
Degree mode	Calculator setting matching angles measured in degrees

Drag and Drop

Two known right-triangle sides	Choose the appropriate Pythagorean sum or difference
Opposite and hypotenuse	Use sine for the selected acute angle
Adjacent and hypotenuse	Use cosine for the selected acute angle
Opposite and adjacent	Use tangent for the selected acute angle
Unknown in the ratio denominator	Rearrange before substituting and rounding

Crossword Puzzle

HYPOTENUSE	Longest right-triangle side
OPPOSITE	Leg across from the chosen angle
ADJACENT	Leg beside the chosen angle excluding hypotenuse
TANGENT	Opposite divided by adjacent for an acute right-triangle angle
COSINE	Adjacent divided by hypotenuse
INVERSE	Describes finding an angle from a trigonometric ratio

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 every side role: Locate θ and the right angle before choosing a ratio.
Identify sum or difference: Distinguish P 1 fromP 2.
Explain degree mode: State the requested unit for T 1.
Separate exact and approximate answers: Compare 7√2 with 9.90.

Standard

Solve all six Pythagoras questions: Show the squared relationship and positive root with units.
Solve all six trigonometry questions: Use the individual diagrams and final precision.
Check the source cosine angle differently: Use the 6-8-10 relationship and sine.
Repair the unknown-denominator error: Explain why T 4 requires division and a hypotenuse longer than 7.

Advanced

Derive the exact-value table: Use the square and equilateral split within the selected specification.
Transfer the 30°model: Use hypotenuse 18 and independently check both legs.
Compare right and non-right methods: Solve the included 60°extension and state why Pythagoras is insufficient.
Audit a sine-rule conclusion: Use the given 30°and 45°case without generalising unique solutions to every SSAproblem.

Learning Check

Six Pythagoras solutions: Keep given/unknown diagrams and independent square checks.
Six trigonometry solutions: Use degree mode and final rounding.
Source angle and inverse check: Distinguish an inverse function from a reciprocal.
Exact-value extension: State the separate route and derive values.
Non-right extension: Use the included angle or two supplied angles explicitly.
First errors and evidence: Repair the earliest method or interpretation error.

Worked checks and assessment guidance

P 1=12 cm, P 2=10 cm, P 3=15 cm, P 4=6 m, P 5=7√2 cm≈9.90 cm, P 6=7 cm. Their square relationships are 25+144=169, 36+64=100, 225+64=289, 23.04+12.96=36, 49+49=98, and 576+49=625.
T 1≈36.9°, T 2≈7.71 cm, T 3≈6.30 cm, T 4≈9.14 cm, T 5≈67.4°, T 6=7 cm. Equations are cosθ=8/10, o=12 sin 40°, o=9 tan 35°, h=7/sin 50°, θ=arctan 12/5, and o=14 sin 30°.
Cosθ=0.8 gives θ≈36.8699°rather than 1/0.8. Pythagoras supplies opposite 6 for adjacent 8 and hypotenuse 10, so sinθ=0.6 checks the same acute angle. Opposite and adjacent swap for its complementary angle.
The unit square gives sin 45=cos 45=√2/2 and tan 45=1. Splitting an equilateral side 2 gives sin 30=cos 60=1/2 and cos 30=sin 60=√3/2. Hypotenuse 18 at 30°gives legs 9 and 9√3 with squares 81+243=324.
Sides 8, 10 with included 60°give opposite side√84=2√21≈9.17 and area 20√3≈34.64. With A 30°, B 45°, a 5, the third angle 105°and b 5√2≈7.07 follow. This is separate from the core right-triangle tasks and does not imply all SSAdata are unique.
Check the marked right angle, hypotenuse, selected θ, degree mode, unknown position and precision. A smaller-than-leg result cannot be its hypotenuse. Retain twelve complete diagrams and working; do not claim unobserved measurement accuracy or later retrieval that has not occurred.

Learning evidence

Retain six Pythagoras and six trigonometry solutions with individual labelled diagrams, unit checks, inverse verification, final precision, and separately identified exact-value and non-right-triangle extensions.

OER and Wikipedia

Department for Education:GCSE Mathematics subject content and assessment objectives. Source area:GCSE Mathematics subject content > Scope of study > Geometry and measures - Mensuration and calculation items 20 to 21. Year10 allocation is editorial within the GCSE pathway;Higher-only extensions are explicitly labelled. Original examples,teaching and feedback;Commons images retain their actual attribution,and videos identify original publishers. Reviewed12November2026.

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.