Transcript Trigonometry

APPS 4 US
MATHS
Laia Erra, Irene Lama, Aina Gràcia 1BAT1
Theory: TRIGONOMETRY
-HOW TO MEASURE THE ANGLES
Degree (angle)
1º= 60’ 1’=60’’
Gradiant
1º= 100’ 1‘=100’’
Radians
(PI)radians= 180 degrees
What is a radian?
A radian is the angle wich its single arc is equal
to its radiant.
1 radian= 57’296º
Arc
Radian
Radius
Equivalences
0º = 0
30º = π/6
45º = π/4
60º = π/3
90º = π/2
120 = 2π/3
180 = π
270 = 3π/2
360º = 2π
Internal angles
Degrees of a triangle: 180º
Degrees: 360º
5 sides
360º/5 = 72º
180º - 72º= 108º 108/2 = 54º
54º x 10 external angles = 540º
Trigonometric reasons
Trigonometric reason
sen a
Definition
opposite ray to a/hypotenuse
cos a
contiguous ray to a/hypotenuse
tang a
opposite ray to a/contiguous ray
to a
ctg a
1/tang a
sec a
cosc a
1/cos a
1/sen a
Sen and Cos theorem
Signs of the trigonometric reasons
Trigonometric
reason
Sign
Complementary angles
Supplementary angles
Addition and subtraction of angles
Double angle
Half angle
REMEMBER
Pythagoras
sin2 a + cos2 a = 1
1 + tag2 a = sec2 a
1 + ctg2 a = cosc2 a
Sines theorem
A/sin a = B/sin b = C/sin c
Cosines theorem
A2 = B2 + C2 – 2BC x cos a
B2 = A2 + C2 – 2AC x cos b
C2 = A2 + B2 – 2AB x cos c
Activities
cos(t)+sin(t)
cos(2t)
------------- = ------------cos(t)-sin(t) 1- sin(2t)
sin(a)+sin(b)+sin(c)
--------------------- = tan(b)
cos(a)+cos(b)+cos(c)
In the figure we see a triangle and a hexagon inscribed in the same circle.
Calculate the ratio of the areas of the two figures.
Solution
cos(2t)
---------- =
1- sin(2t)
sin(b-v)+sin(b)+sin(b+v)
------------------------- = tan(b) <=>
cos(b-v)+cos(b)+cos(b+v)
cos2(t)-sin2(t)
--------------- =
1-2sin(t)cos(t)
2sin(b)cos(v)+sin(b)
--------------------- = tan(b) <=>
2cos(b)cos(v)+cos(b)
(cos(t)-sin(t))(cos(t)+sin(t))
------------------------------------------ =
cos(t)cos(t)-2sin(t)cos(t)+sin(t)sin(t)
(cos(t)-sin(t))(cos(t)+sin(t))
------------------------------- =
(cos(t)-sin(t))(cos(t)-sin(t))
cos(t)+sin(t)
-----------cos(t)-sin(t)
sin(b)(2cos(v)+1)
----------------- = tan(b) <=>
cos(b)(2cos(v)+1)
sin(b)
------- = tan(b)
cos(b)
1.
2.
First Method:
The ratio of the areas of the two figures is independent of the radius of the circle. We can therefore assume
that the radius is 1.
3. Area of the hexagon:
4. Connect the center of the circle with the vertices. There arise 6 equilateral triangles with side 1. The area of
the hexagon is 6.(1/2).1.1.sin(60°) = 3.sin(60°)
5. Area of the triangle:
6. Connect the center of the circle with the vertices. There arise 3 isosceles triangles. The area of the triangle is
3.(1/2).1.1.sin(120°) = (3/2) sin(60°)
7. We see that the ratio of the areas of the two figures is 2.
8. Second method :
9. The ratio of the areas of the two figures is independent of the mutual position of the polygons. We rely on this
property to put the triangle in a favorable position. We connect the center with the vertices of the triangle.
10. There are 6 congruent triangles. Then it is immediately clear that the ratio of the areas of the two figures is 2.
Activities
A person is 100 meters from the base of a tree, he
observes that the angle between the ground and the top
of the tree is 18 degrees. Estimate the height (h) of the
tree.
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SOLUTION: Use the tangent
tan(18o) = h / 100
Solve for h to obtain
h = 100 tan(18o) = 32.5 meters.
Activities
The angle of elevation of a hot air balloon, climbing vertically, changes from 25
degrees at 10:00 am to 60 degrees at 10:02 am. The point of observation of
the angle of elevation is situated 300 meters away from the take off point.
What is the upward speed, assumed constant, of the balloon? Give the answer
in meters per second and round to two decimal places.
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Use the tangent to write
tan(25o) = h1 / 300
and
tan(60o) = (h1 + h2) / 300
Solve for h1 and h2
h1 = 300 tan(tan(25o))
and
h1 + h2 = 300 tan(60o)
Use the last two equations to find h2
h2 = 300 [ tan(60o) - tan(25o) ]
If it takes the balloon 2 minutes (10:00 to 10:02) to climb h2, the the
upward speed S is given by
S = h2 / 2 minutes
= 300 [ tan(60o) - tan(25o) ] / (2 * 60) = 3.16 m/sec
Activities
BH is perpendicular to AC. Find x the length of
BC.
ABC is a right triangle with a right angle at A.
Find x the length of DC.
Solution
1.
BH perpendicular to AC means that triangles
ABH and HBC are right triangles. Hence
2.
tan(39o) = 11 / AH or AH = 11 / tan(39o)
3.
4.
1.
Since angle A is right, both triangles ABC and
ABD are right and therefore we can apply
Pythagora's theorem.
2.
142 = 102 + AD2 , 162 = 102 + AC2
3.
Also x = AC - AD
4.
= sqrt( 162 - 102 ) - sqrt( 142 - 102 ) = 2.69
(rounded to 3 significant digits)
HC = 19 - AH = 19 - 11 / tan(39o)
Pythagora's theorem applied to right triangle
HBC: 112 + HC2 = x2
5.
solve for x and substitute HC: x = sqrt [ 112 +
(19 - 11 / tan(39o) )2 ]
6.
= 12.3 (rounded to 3 significant digits)
Activities
From the top of a 200 meters high building, the angle of depression to the bottom of a second building is 20 degrees.
From the same point, the angle of elevation to the top of the second building is 10 degrees. Calculate the height of the
second building.
Solution
1.
2.
3.
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5.
6.
7.
The diagram below show the two buildings and the angles of depression and elevation.
tan(20o) = 200 / L
L = 200 / tan(20o)
tan(10o) = H2 / L
H2 = L * tan(10o)
= 200 * tan(10o) / tan(20o)
Height of second building = 200 + 200 * tan(10o) / tan(20o)
Activities
In a right triangle ABC with angle A equal to 90o, find angle B and C so that sin(B) = cos(B).
SOLUTION:
1.Let b be the length of the side opposite angle B and c the length of the side opposite angle C and h the length of the
hypotenuse.
2.sin(B) = b/h and cos(B) = c/h
3.sin(B) = cos(B) means b/h = c/h which gives c = b
4.The two sides are equal in length means that the triangle is isosceles and angles B and C are equal in size of 45 o.
Activities
Calculate the length of the side x, given that tan θ = 0.4
Calculate the length of the side x, given
that sin θ = 0.6
Solution:
Solution:
Using Pythagoras’
theorem:
Examples
The angle of elevation is always measured from the ground up. Think of it like an elevator
that only goes up. It is always INSIDE the triangle.
In the diagram at the left, x marks the angle of elevation of the top of the tree as seen from a
point on the ground.
You can think of the angle of elevation in relation to the movement of your eyes. You are
looking straight ahead and you must raise (elevate) your eyes to see the top of the tree.
Examples
The angle of depression is always OUTSIDE the triangle. It is never inside the
triangle.
In the diagram at the left, x marks the angle of depression of a boat at sea from the top of
a lighthouse.
You can think of the angle of depression in relation to the movement of your
eyes. You are standing at the top of the lighthouse and you are looking straight
ahead. You must lower (depress) your eyes to see the boat in the water.
Activities
You are walking up a 500. meter high hill. The trail has an incline of 12 degrees. How far will you walk
to get to the top?
Activities
Calculating radius of the outer core
The S wave shadow zone is caused by the outer core not transmitting S waves. It crosses an arc of
105 degrees on the Earth (see the diagram on the left). Estimate the radius of the outer core. The
radius of the entire Earth is 6370 km.
A= cos(52.5)*6370km
so
A= 3877 km
Activities
The angle of repose for sand is typically about 35°. What is the sine of this angle?
SOLUTION:
1.Type 35 into your calculator
2.press the sin button.
3.Your calculator should read 0.574.
When driving, a steep hill is typically only 12°. What is the cosine of this angle?
SOLUTION:
1.Type 12 into your calculator