The sine of an obtuse angle and a triangle's area

Geometry, grade 10 (4 units) Lesson 12 of 20Learning

The sine of an obtuse angle and a triangle's area

The sine of an obtuse angle

An obtuse angle is not one of the acute angles of a right-angled triangle, so the definition of the sine as a ratio of sides does not fit it. Its sine is defined by the angle adjacent to it, $\displaystyle 180^{\circ}-\alpha$ , which is acute:

\displaystyle \sin(180^{\circ}-\alpha)

\displaystyle =

\displaystyle \sin\alpha

$\displaystyle {\sin120^{\circ}}={\sin60^{\circ}}={\frac{\sqrt{3}}{2}}$ ;
$\displaystyle {\sin150^{\circ}}={\sin30^{\circ}}={\frac{1}{2}}$ ;
the calculator gives the same: $\displaystyle {\sin100^{\circ}}={\sin80^{\circ}\approx0.98}$ .

The definition fits a right angle too: $\displaystyle {\sin(180^{\circ}-90^{\circ})}={\sin90^{\circ}}={1}$ .

Exercise 1

Calculate exactly.

/ – 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

Two angles with the same sine

Every number between $\displaystyle 0$ and $\displaystyle 1$ is the sine of two angles between $\displaystyle 0^{\circ}$ and $\displaystyle 180^{\circ}$ : an acute angle $\displaystyle \alpha$ , and the obtuse angle $\displaystyle 180^{\circ}-\alpha$ .

$\displaystyle {\sin\gamma}={\frac{1}{2}}$ : $\displaystyle {\gamma}={30^{\circ}}$ or $\displaystyle {\gamma}={150^{\circ}}$ ;
with $\displaystyle \sin^{-1}$ the calculator gives only the acute one, and the obtuse one is calculated: $\displaystyle 180^{\circ}-30^{\circ}$ ;
in a triangle both are possible, until another given decides which.

A triangle’s area from two sides and an angle

A triangle’s area is half a side times the altitude to that side. When two sides $\displaystyle a$ , $\displaystyle b$ and the angle $\displaystyle \gamma$ between them are known, the sine gives the altitude:

\displaystyle S

\displaystyle =

\displaystyle \frac{1}{2}ab\sin\gamma

$\displaystyle \gamma$ acute: in the right-angled triangle the altitude makes, $\displaystyle h$ is the side opposite $\displaystyle \gamma$ and $\displaystyle b$ the hypotenuse, so $\displaystyle {h}={b\sin\gamma}$ ;
$\displaystyle {S}={\frac{1}{2}a\cdot h}={\frac{1}{2}ab\sin\gamma}$ .

$\displaystyle \gamma$ obtuse: the altitude falls outside the triangle, on the extension of the side $\displaystyle a$ , and in the right-angled triangle it makes, the angle opposite $\displaystyle h$ is $\displaystyle 180^{\circ}-\gamma$ ;
$\displaystyle {h}={b\sin(180^{\circ}-\gamma)}={b\sin\gamma}$ – exactly the same formula.

Example: an area with an obtuse angle

Two sides of a triangle are $\displaystyle 6$ and $\displaystyle 10$ long, and the angle between them is $\displaystyle 150^{\circ}$ :

S=\frac{1}{2}\cdot6\cdot10\cdot\sin150^{\circ}=30\cdot\frac{1}{2}=15

Exercise 2

Find the triangle’s area from the two sides and the angle between them.
An exact result where possible; otherwise – rounded to two decimal places.

\displaystyle 6

\displaystyle 8

\displaystyle 60^{\circ}

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\displaystyle 5

\displaystyle 4

\displaystyle 120^{\circ}

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\displaystyle 7

\displaystyle 9

\displaystyle 50^{\circ}

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Exercise 3

A triangle’s area is $\displaystyle 15$ , and two of its sides are $\displaystyle 6$ and $\displaystyle 10$ long.
Find the sine of the angle between the two sides.
What can the angle be? Find both possibilities.

/ – 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

The acute angle

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The obtuse angle

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Exercise 4

In the parallelogram $\displaystyle ABCD$ , $\displaystyle {AB}={6}$ , $\displaystyle {AD}={3.5}$ , and $\displaystyle {\angle DAB}={45^{\circ}}$ .
Find the parallelogram’s area exactly.
(A hint: the diagonal $\displaystyle BD$ splits it into two congruent triangles.)

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Formulas: geometry, grade 10 (4 units)

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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 $\displaystyle A$
$\displaystyle \frac{BD}{DC}$ $\displaystyle =$ $\displaystyle \frac{AB}{AC}$