Practice: trigonometry

Geometry, grade 10 (4 units) Lesson 15 of 20Practice

Practice: trigonometry

The vertices of the rectangle $\displaystyle ABCD$ are $\displaystyle A(-5,-5)$ , $\displaystyle B(-6,-1)$ , $\displaystyle C(2,1)$ and $\displaystyle D(3,-3)$ .
Find the angles, rounded to two decimal places.
(A hint: $\displaystyle {AB}={\sqrt{17}}$ and $\displaystyle {BC}={2\sqrt{17}}$ .)

The angle between the diagonal

\displaystyle AC

and the side

\displaystyle AB

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The angle between the diagonal

\displaystyle AC

and the side

\displaystyle BC

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The acute angle between the diagonals

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The vertices of the triangle $\displaystyle ABC$ are $\displaystyle A(-1,1)$ , $\displaystyle B(5,1)$ and $\displaystyle C(3,5)$ .

Write on the board the equations of the lines $\displaystyle AC$ and $\displaystyle BC$ .

\displaystyle A=(-1,1)

\displaystyle B=(5,1)

\displaystyle C=(3,5)

What is the equation of the side $\displaystyle AB$ ?
And the equation of the line containing the altitude from $\displaystyle C$ to the side $\displaystyle AB$ ?

Find the triangle’s angles, rounded to two decimal places.

\displaystyle \angle A

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\displaystyle \angle B

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\displaystyle \angle C

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Find the triangle’s perimeter exactly.

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Find the triangle’s area in two ways, exactly.
Did you get the same area?

A side times the altitude to it, over

\displaystyle 2

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\displaystyle \frac{1}{2}\cdot AB\cdot AC\cdot\sin A

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In an isosceles triangle the legs are $\displaystyle 10$ long, and the base angles $\displaystyle 70^{\circ}$ .
Calculate, rounded to two decimal places.

The length of the base

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The triangle’s area

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Two sides of a triangle are $\displaystyle 8$ and $\displaystyle 5$ long, and the angle between them is $\displaystyle 135^{\circ}$ .
Find the triangle’s area exactly.

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$\displaystyle \alpha$ is an acute angle, and $\displaystyle {\tan\alpha}={2}$ .
Calculate exactly.
(A hint: a right-angled triangle whose legs are $\displaystyle 1$ and $\displaystyle 2$ .)

\displaystyle \sin\alpha

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\displaystyle \cos\alpha

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The diagonals of a rhombus are $\displaystyle 6$ and $\displaystyle 8$ long.

The rhombus’s side

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The rhombus’s acute angle, rounded to two decimal places

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