5-5: Solving Right Triangles
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5-5: SOLVING RIGHT TRIANGLES Alexis Mathis Aaron Goode Jessica Anderson
description
Alexis Mathis Aaron Goode Jessica Anderson. 5-5: Solving Right Triangles. Unit Circle. Pythagorean theorem. a²+b²=c² Used to determine the sides of a right triangle. Unit chart. Soh - Cah -Toa. Right triangle basics. Internal angle=180 Right angle=90 - PowerPoint PPT Presentation
Transcript of 5-5: Solving Right Triangles
5-5: SOLVING RIGHT TRIANGLES
Alexis Mathis
Aaron Goode
Jessica Anderson
Unit Circle
Pythagorean theorem
a²+b²=c² Used to determine the sides of a right
triangle
Unit chart0 30 45 60 90 120 135 150 180
sin 0 1/2 2/2 3/2 1 3/2 2/2 1/2 0
cos 1 3/2 2/2 1/2 0 -1/2 -2/2 -3/2 -1
tan 0 3/3 1 3 und -3 -1 -3/3 0
210 225 240 270 300 315 330 360
Sin -1/2 -2/2 -3/2 -1 -3/2 -2/2 -1/2 0
Cos -3/2 -2/2 -1/2 0 1/2 2/2 3/2 1
tan 3/3 1 3 Und -3 -1 -3/3 0
Soh-Cah-Toa
Right triangle basics
Internal angle=180 Right angle=90 The other two angles must add up to 90 The hypotenuse is the side of the
triangle vertical the right angle
Solve triangle ABC given side c = 25 cm and side b = 24 cm.
Solution. To find the remaining side a, use the Pythagorean theorem:
a² + 24² = 25² a² = 625 − 576 = 49 a = = 7. Next, to find angle A, we have
cos A = 2425 = 96100 , on multiplying each term by 4. = .96
0 30 45 60 90 120 135 150 180
Sin 0 1/2 2/2 3/2 1 3/2 2/2 1/2 0
Cos 1 3/2 2/2 1/2 0 -1/2 -2/2 -3/2 -1
Tan 0 3/3 1 3 und -3 -1 -3/3 0
210 225 240 270 300 315 330 360
Sin -1/2 -2/2 -3/2 -1 -3/2 -2/2 -1/2 0
Cos -3/2 -2/2 -1/2 0 1/2 2/2 3/2 1
tan 3/3 1 3 Und -3 -1 -3/3 0
inverse
The arcsine, arccosine, and arctangent of relations
Notation Definition Domain of X
Arcsine Y=arcsinx X=siny −1 ≤ x ≤ 1
Arccosine Y=arccosx X=cosy −1 ≤ x ≤ 1
Arctangent Y=arctanx X=tany All real numbers
Examples: arcsine
Sinx=4/9 X=arcsine4/9
Sinx=8/24 X=arcsine8/24
Examples: arccosine
Cosx=3/4 X=arccosine3/4
Cosx=5/7 X=arccosine5/7
Examples: Arctangent
Tanx=9/16 X=arctan9/16
Tanx=5/11 X=arctan5/11
Using inverse to solve right triangles a=4 c=6
SinA=4/6 A=arcsin4/6
Using inverse..
b=3 c=6
CosA=3/6 A=arccos3/6
Word problems
A six-meter-long ladder leans against a building. If the ladder makes an angle of 60° with the ground, how far up the wall does the ladder reach? How far from the wall is the base of the ladder?
Word problems
A five-meter-long ladder leans against a wall, with the top of the ladder being four meters above the ground. What is the approximate angle that the ladder makes with the ground?
