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Maths Quest A Year 11 for Queensland
WorkSHEET 5.2
Chapter 5 Right-angled triangles and trigonometry WorkSHEET 5.2
1
Right-angled triangles and trigonometry
Name: ___________________________
1
With respect to the labelled angle, state:
(a) hypotenuse
(b) adjacent side
(c) opposite side
2
Express the following angles in degrees,
minutes and seconds.
(a) 15.2

(b)
15 12
(c)
(d)
  tan 1 6.25
  sin 1 54

(a) hypotenuse = SP
(b) adjacent side = ST
(c) opposite side = PT
Use (DMS) or (o ’ ”) button on calculator.
(a)
15.2   15 12
(b)
15 12  15.5 

 15  30
  tan 1 6.25
(c)
 80.9097 
 80  5435
  sin 1 54 
(d)
 sin 1 0.8
 53.13
 53 7 48
3
Calculate the following correct to two decimal
places.
1
(a)
tan 30 
(b)
(c)
(d)
23  cos 4  5
(a)
(b)
(c)
75
sin 45
35 
(d)
1
sin 67  5437
1
tan 30 
 1.73
23  cos 4 5  22.94
75
sin 45
35 
 106.07
1
sin 67 5437

 37.77
Maths Quest A Year 11 for Queensland
Chapter 5 Right-angled triangles and trigonometry WorkSHEET 5.2
2
(a) Using Pythagoras’ Theorem:
4
hypot. 2  base 2  ht 2
x 2  52  52
 25  25
 50
Determine the length of the unknown side in
the triangle above using:
(a) Pythagoras’ Theorem
(b) trigonometry
x  50
 7.07 cm
(b)
opp.
hypot.
5

x

x sin 45  5
5
x
sin 45 
 7.07 cm
sin 45  
5
Find the values of unknown marked sides
correct to two decimal places.
6
2
Find the value of angle  correct to one decimal
cos  =
place.
5
cos  = 0.4
  = 66.4°
x
8
 x = 8  cos 27°
x = 7.13
y
sin 27° =
8
 y = 8  sin 27°
y = 3.63
cos 27° =
4
2
Maths Quest A Year 11 for Queensland
7
Chapter 5 Right-angled triangles and trigonometry WorkSHEET 5.2
A 6 m-long ladder rests against a vertical wall
and forms an angle of 40° to the horizontal
ground. How high up the wall does the ladder
reach, correct to two decimal places?
3
2
h
6
 h = 6  sin 40°
h = 3.86 m (correct to two decimal
places)
sin 40° =
8
A large heavy drum is pushed 3.5 m up an
inclined plane. If the inclined plane rises 2 m
vertically, find the angle the inclined plane
makes with the horizontal.
2
2
3 .5
sin  = 0.5714
  = 34.8°
sin  =
9
A 1.08 m-tall child flies a kite with 100 m of
released string which makes an angle of 70°
with the horizontal. How high is the kite
flying?
3
h
100
 h = 100  sin 70°
h = 9397 m
So, height of kite = 9397  1.08 m = 95.05 m
sin 70° =
Maths Quest A Year 11 for Queensland
10
Chapter 5 Right-angled triangles and trigonometry WorkSHEET 5.2
The angle of depression of a boat from a cliff
60 m high is 10°. How far is the boat from the
base of the cliff?
4
2
60
d
60
d=
tan 10
d = 340 m
tan 10° =