Calculations in a right-angled triangle

Geometry, grade 10 (4 units) Lesson 11 of 20Learning

Calculations in a right-angled triangle

Finding a side

When an acute angle and one side are known, choose the function that links the known side to the side you want – by where they are next to the angle.

Example: in the triangle $\displaystyle ABC$ the angle $\displaystyle C$ is right, $\displaystyle {\angle A}={30^{\circ}}$ , and the hypotenuse $\displaystyle {AB}={12}$ .

$\displaystyle BC$ is the side opposite $\displaystyle A$ , and $\displaystyle AB$ the hypotenuse – the sine: $\displaystyle {BC}={12\cdot\sin30^{\circ}}={12\cdot\frac{1}{2}}={6}$ ;
$\displaystyle AC$ is the side adjacent to $\displaystyle A$ – the cosine: $\displaystyle {AC}={12\cdot\cos30^{\circ}}={12\cdot\frac{\sqrt{3}}{2}}={6\sqrt{3}}$ ;
a check by Pythagoras: $\displaystyle {6^{2}+(6\sqrt{3})^{2}}={36+108}={144}={12^{2}}$ .

For $\displaystyle BC$ the property from grade 9 works too: the side opposite an angle of $\displaystyle 30^{\circ}$ is half the hypotenuse.

Finding an angle

When two sides are known, their ratio is the sine, cosine or tangent of the angle, and the calculator gives the angle.

Example: the legs are $\displaystyle 3$ and $\displaystyle 4$ . The angle $\displaystyle \alpha$ opposite the leg $\displaystyle 3$ :

\tan\alpha=\frac{3}{4}\quad\Rightarrow\quad\alpha=\tan^{-1}\frac{3}{4}\approx36.87^{\circ}

And the other angle: $\displaystyle {90^{\circ}-36.87^{\circ}}={53.13^{\circ}}$ .

When to round

With $\displaystyle 30^{\circ}$ , $\displaystyle 45^{\circ}$ , $\displaystyle 60^{\circ}$ the result is exact: $\displaystyle 6\sqrt{3}$ , not $\displaystyle 10.39$ .
With another angle, write the exact expression, $\displaystyle 12\sin40^{\circ}$ , and round only in the last line: $\displaystyle \approx7.71$ .
A rounded number that you go on calculating with adds an error at each step.
Each step gets a reason: which function, which theorem, which property of the triangle.

Exercise 1

In the triangle $\displaystyle ABC$ the angle $\displaystyle C$ is right, $\displaystyle {\angle B}={60^{\circ}}$ , and $\displaystyle {AB}={12}$ .
Calculate exactly.

a

\displaystyle AC

/ – fraction bar · Tab or the arrows – between numerator and denominator, then a note above or below · Enter – check and a new line · * – times · ^ – power, ↓ – end of the power

/ – fraction bar · Tab or the arrows – between numerator and denominator, then a note above or below · Enter – a new line · * – times · ^ – power, ↓ – end of the power

b

\displaystyle BC

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c

The triangle’s area

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/ – fraction bar · Tab or the arrows – between numerator and denominator, then a note above or below · Enter – a new line · * – times · ^ – power, ↓ – end of the power

Exercise 2

The legs of a right-angled triangle are $\displaystyle 5$ and $\displaystyle 12$ long.

a

The angle opposite the leg

\displaystyle 5

, rounded to two decimal places

/ – fraction bar · Tab or the arrows – between numerator and denominator, then a note above or below · Enter – check and a new line · * – times · ^ – power, ↓ – end of the power

/ – fraction bar · Tab or the arrows – between numerator and denominator, then a note above or below · Enter – a new line · * – times · ^ – power, ↓ – end of the power

b

The hypotenuse

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Exercise 3

One angle of a right-angled triangle is $\displaystyle 45^{\circ}$ , and the hypotenuse is $\displaystyle 8$ long.
Calculate exactly.

a

The length of a leg

/ – fraction bar · Tab or the arrows – between numerator and denominator, then a note above or below · Enter – check and a new line · * – times · ^ – power, ↓ – end of the power

/ – fraction bar · Tab or the arrows – between numerator and denominator, then a note above or below · Enter – a new line · * – times · ^ – power, ↓ – end of the power

b

The perimeter

/ – fraction bar · Tab or the arrows – between numerator and denominator, then a note above or below · Enter – check and a new line · * – times · ^ – power, ↓ – end of the power

/ – fraction bar · Tab or the arrows – between numerator and denominator, then a note above or below · Enter – a new line · * – times · ^ – power, ↓ – end of the power

c

The area

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Exercise 4

In the triangle $\displaystyle ABC$ the angle $\displaystyle C$ is right, and $\displaystyle CM$ is the median to the hypotenuse.
$\displaystyle {CM}={5}$ and $\displaystyle {\angle B}={30^{\circ}}$ .
Calculate exactly, and give a reason for each step.

a

\displaystyle AB

/ – fraction bar · Tab or the arrows – between numerator and denominator, then a note above or below · Enter – check and a new line · * – times · ^ – power, ↓ – end of the power

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b

\displaystyle BC

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/ – fraction bar · Tab or the arrows – between numerator and denominator, then a note above or below · Enter – a new line · * – times · ^ – power, ↓ – end of the power

c

The triangle’s area

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Exercise 5

A ladder $\displaystyle 6$ m long leans against a wall.
The angle between the ladder and the floor is $\displaystyle 70^{\circ}$ .
How high up the wall does the ladder reach?
Round to two decimal places.

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Formulas: geometry, grade 10 (4 units)

Analytic geometry

The distance between the points $\displaystyle (x_1,y_1)$ and $\displaystyle (x_2,y_2)$
$\displaystyle d$ $\displaystyle =$ $\displaystyle \sqrt{(x_2-x_1)^{2}+(y_2-y_1)^{2}}$
The midpoint of the segment
$\displaystyle M\left(\frac{x_1+x_2}{2},\ \frac{y_1+y_2}{2}\right)$
The slope of a line through two points
$\displaystyle m$ $\displaystyle =$ $\displaystyle \frac{y_2-y_1}{x_2-x_1}$ $\displaystyle x_1\neq x_2$
A line’s equation in explicit form – $\displaystyle m$ the slope, $\displaystyle b$ where it meets the $\displaystyle y$ -axis
$\displaystyle y=mx+b$
A line’s equation in implicit form
$\displaystyle Ax+By+C=0$
The line of slope $\displaystyle m$ through the point $\displaystyle (x_1,y_1)$
$\displaystyle y-y_1$ $\displaystyle =$ $\displaystyle m(x-x_1)$
Parallel lines
$\displaystyle m_1$ $\displaystyle =$ $\displaystyle m_2$
Perpendicular lines
$\displaystyle m_1\cdot m_2$ $\displaystyle =$ $\displaystyle -1$
The slope – $\displaystyle \alpha$ the angle between the line and the positive direction of the $\displaystyle x$ -axis
$\displaystyle m$ $\displaystyle =$ $\displaystyle \tan\alpha$

Similar triangles and midsegments

The similarity ratio – corresponding sides
$\displaystyle \frac{a_1}{a_2}=\frac{b_1}{b_2}=\frac{c_1}{c_2}=k$
The ratio of the perimeters
$\displaystyle \frac{P_1}{P_2}$ $\displaystyle =$ $\displaystyle k$
The ratio of the areas
$\displaystyle \frac{S_1}{S_2}$ $\displaystyle =$ $\displaystyle k^{2}$
The ratio of corresponding altitudes, medians and angle bisectors
$\displaystyle \frac{h_1}{h_2}=\frac{m_1}{m_2}=\frac{d_1}{d_2}=k$
A triangle’s midsegment – $\displaystyle DE$ parallel to $\displaystyle BC$
$\displaystyle DE$ $\displaystyle =$ $\displaystyle \frac{1}{2}BC$
A trapezoid’s midsegment – $\displaystyle a$ , $\displaystyle b$ the bases
$\displaystyle EF$ $\displaystyle =$ $\displaystyle \frac{a+b}{2}$

Trigonometry

Sine – $\displaystyle a$ the side opposite the angle, $\displaystyle c$ the hypotenuse
$\displaystyle \sin\alpha$ $\displaystyle =$ $\displaystyle \frac{a}{c}$
Cosine – $\displaystyle b$ the side adjacent to the angle
$\displaystyle \cos\alpha$ $\displaystyle =$ $\displaystyle \frac{b}{c}$
Tangent – the opposite side over the adjacent side
$\displaystyle \tan\alpha$ $\displaystyle =$ $\displaystyle \frac{a}{b}$
The tangent from the sine and the cosine
$\displaystyle \tan\alpha$ $\displaystyle =$ $\displaystyle \frac{\sin\alpha}{\cos\alpha}$
The identity from Pythagoras
$\displaystyle \sin^{2}\alpha+\cos^{2}\alpha$ $\displaystyle =$ $\displaystyle 1$
Angles whose sum is $\displaystyle 90^{\circ}$
$\displaystyle \sin(90^{\circ}-\alpha)$ $\displaystyle =$ $\displaystyle \cos\alpha$
$\displaystyle \cos(90^{\circ}-\alpha)$ $\displaystyle =$ $\displaystyle \sin\alpha$
The sine of an obtuse angle
$\displaystyle \sin(180^{\circ}-\alpha)$ $\displaystyle =$ $\displaystyle \sin\alpha$
A triangle’s area – $\displaystyle \gamma$ the angle between the sides $\displaystyle a$ and $\displaystyle b$
$\displaystyle S$ $\displaystyle =$ $\displaystyle \frac{1}{2}ab\sin\gamma$

Exact values

$\displaystyle \sin30^{\circ}=\cos60^{\circ}=\frac{1}{2}$
$\displaystyle \sin45^{\circ}=\cos45^{\circ}=\frac{\sqrt{2}}{2}$
$\displaystyle \sin60^{\circ}=\cos30^{\circ}=\frac{\sqrt{3}}{2}$
$\displaystyle \tan30^{\circ}$ $\displaystyle =$ $\displaystyle \frac{\sqrt{3}}{3}$
$\displaystyle \tan45^{\circ}$ $\displaystyle =$ $\displaystyle 1$
$\displaystyle \tan60^{\circ}$ $\displaystyle =$ $\displaystyle \sqrt{3}$

Special lines in a triangle

Where the medians meet, $\displaystyle G$ divides the median $\displaystyle AD$ in the ratio $\displaystyle 2:1$
$\displaystyle AG$ $\displaystyle =$ $\displaystyle 2GD$
Where the medians meet, in coordinates
$\displaystyle G\left(\frac{x_1+x_2+x_3}{3},\ \frac{y_1+y_2+y_3}{3}\right)$
The median $\displaystyle AD$ splits the triangle into two triangles of equal area
$\displaystyle S_{ABD}$ $\displaystyle =$ $\displaystyle S_{ACD}$
The angle bisector theorem – $\displaystyle AD$ bisects the angle $\displaystyle A$
$\displaystyle \frac{BD}{DC}$ $\displaystyle =$ $\displaystyle \frac{AB}{AC}$