Precision and practical applications | Year 10

01 / LEARN

Understand the idea.

Plan before calculating

Choose the model from the geometry: Pythagoras needs perpendicular sides; right-triangle ratios connect an acute angle with suitable sides; the sine rule needs opposite pairs; the cosine rule uses an included angle or three sides.

Before calculating, name the requested quantity and label the sketch. A horizontal distance and a cable length are different sides. Include eye height, use consistent units and confirm degree mode for degree angles.

Round at the end and check the result

Keep exact roots or the calculator's stored value through intermediate steps. Round only the final requested quantity; replacing a trigonometric ratio with a short decimal can change a later reported measurement.

Check the geometry as well as the arithmetic. A hypotenuse must be longest, triangle angles total $180^{\circ}$ , and sine or cosine must lie between $- 1$ and $1$ . State assumptions such as level ground and vertical objects.

02 / WORKED EXAMPLES

See the method in action.

EXAMPLE 01

Retain the calculator value

An observer eye is 1.5 m high and 20 m from a vertical pole on level ground. Elevation is 37°. Find pole height to 0.1 m.

Worked solution

1
Form the triangle above eye level.
$tan 37^{\circ} = \frac{h - 1.5}{20}$
On level ground, the rise from the observer's eyes is opposite the elevation angle.
2
Calculate the total height without early rounding.
$h = 20 tan 37^{\circ} + 1.5 \approx 16.571081 m$
The tangent value is retained before adding eye height.
3
Round the requested quantity.
$h \approx 16.6 m$
The hundredths digit rounds the tenths digit upward.

Answer

16.6 m

EXAMPLE 02

Check what the ratio measures

A right-triangle tower is 41 m high and the ground angle of a cable is 25°. Find the cable length to 0.1 m.

Worked solution

1
Identify the opposite side and hypotenuse.
$sin 25^{\circ} = \frac{41}{L}$
The tower height is opposite the ground angle; the cable is the hypotenuse.
2
Rearrange for cable length.
$L = \frac{41}{sin 25^{\circ}} \approx 97.0143 m$
Multiplying by $L$ and dividing by the sine isolates the required length.
3
Round and check against the height.
$L \approx 97.0 m > 41 m$
The cable must be longer than the perpendicular tower height.

Answer

97.0 m

03 / YOUR TURN

Try it yourself.

Check your understanding

Should a right triangle with height 18 m have hypotenuse 12 m?

Show answer & reasoning

No.

No. In a right triangle the hypotenuse is longer than either leg. A $12 m$ hypotenuse cannot accompany an $18 m$ height.

Keep practising

6 worksheet questions · 15 marks

Open worksheet Worked solutions

Read the questions online

A right triangle has hypotenuse 14.2 cm and leg 8.1 cm. Find its other leg to 0.1 cm.
3 marks
A student uses $tan 37^{\circ} \approx 0.75$ for a height with horizontal distance 20 m. Compare their height with $20 tan 37^{\circ}$ , both to 0.1 m.
3 marks
A calculation gives $sin A = 1.08$ for a triangle. What should you conclude?
2 marks
A map shows horizontal separation 80 m and a line of sight rises 25 m. Should 80 m be used as the hypotenuse? Explain.
2 marks
Sides 12 cm and 17 cm include $52^{\circ}$ . State an appropriate method for the third side and write its formula.
3 marks
A survey estimates a tower height assuming level ground. Why should the final report state that assumption?
2 marks

Read the worked solutions

Equivalent correct methods earn the corresponding marks. Include units where asked.

A right triangle has hypotenuse 14.2 cm and leg 8.1 cm. Find its other leg to 0.1 cm.
3 marks
$x = \sqrt{{14.2}^{2} - {8.1}^{2}}$ .
Retain the square-root result until rounding.
11.7 cm
A student uses $tan 37^{\circ} \approx 0.75$ for a height with horizontal distance 20 m. Compare their height with $20 tan 37^{\circ}$ , both to 0.1 m.
3 marks
Rounded-ratio height $= 20 (0.75) = 15.0$ m.
Full-ratio height $= 20 tan 37^{\circ} = 15.071 \dots$ m.
15.0 m versus 15.1 m.
A calculation gives $sin A = 1.08$ for a triangle. What should you conclude?
2 marks
A sine value must lie between $- 1$ and 1.
Check the supplied lengths and opposite-pair substitution.
The data, model or calculation is inconsistent; no real angle has that sine.
A map shows horizontal separation 80 m and a line of sight rises 25 m. Should 80 m be used as the hypotenuse? Explain.
2 marks
The sloping line of sight is the hypotenuse.
Its length is $\sqrt{80^{2} + 25^{2}}$ m.
No; it is the horizontal leg.
Sides 12 cm and 17 cm include $52^{\circ}$ . State an appropriate method for the third side and write its formula.
3 marks
$c = \sqrt{12^{2} + 17^{2} - 2 (12) (17) cos 52^{\circ}}$ .
The known angle is included between the known sides.
Cosine rule.
A survey estimates a tower height assuming level ground. Why should the final report state that assumption?
2 marks
The right-triangle model used a horizontal base.
A conclusion is valid only under the assumptions used to form it.
Sloping ground changes the vertical difference and can change the inferred height.

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Curriculum reference & lesson scope

Australian Curriculum v9 and South Australian school course scope

Right-triangle applications include AC9M10M03. The approved chapter also develops non-right-triangle methods, exact values and trigonometric functions within one normal Year 10 course.

Official curriculum source ↗

In this lesson

Understand the idea.

Plan before calculating

Round at the end and check the result

See the method in action.

Retain the calculator value

Form the triangle above eye level.

Calculate the total height without early rounding.

Round the requested quantity.

Check what the ratio measures

Identify the opposite side and hypotenuse.

Rearrange for cable length.

Round and check against the height.

Try it yourself.

Check your understanding

Keep practising