# Homework 5.6 Right Triangle Trig by happo7

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```									Name:                                                                             Homework 5.6.L Right Triangle Trig
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Homework 5.6 Right Triangle Trig
MAT 125: Precalculus
Fall 2009: Sections 6, 9, and 13
Mr. Jonckheere

Due Homework

CP 5.6.1. Fill in the blanks to complete each theorem

a) If α is an accute angle of a right triangle, opp is the length of the side opposite α, adj is the length of
the side adjacent to α, hyp is the length of the hypotenuse, and x = m(α) then

sin ( x ) =          cos ( x ) =            tan ( x ) =

b) If a right triangle has legs of length a, and b, and a hypotenuse of length c then c2 = a2 + b2.
c) The sum of the primary measures of the angles in any triangle is π = 180°.
d) If x is in [0, π/2] then

arcsin(sin(x)) = x
arcos(cos(x)) = x

CP 5.6.2. Let α be an accute angle in a right triangle, and x = m(α). Let the side opposite α have length 6,
and the side adjecent to α have length 3. Evaluate all six trig functions at x.

h2 = 62 + 32 = 36 + 9 = 45
h = 45 = 3 5
opp   6         2                       5
sin ( x ) =     =   =                csc ( x ) =
hyp 3 5          5                     2
cos ( x ) =     =   =                sec ( x ) = 5
hyp 3 5          5
opp 6                                  3 1
tan ( x ) =     = =2                 cot ( x ) =    =
Name:                                                                          Homework 5.6.L Right Triangle Trig
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CP 5.6.3. A right triangle has an angle with measure 68°, and a hypotenuse of length 10. Solve the triangle.

x = 68°                                                                                       b
c = 10

y = 90° – 68° = 22°                                                                       y
c
sin ( x ) =
a                                                                                 a
c
a = c sin ( x )
= 10 sin ( 68° )                                                              x
≈ 9.3                                                                 a
b

b
cos ( x ) =
c
b = c cos ( x )
= 10 cos ( 68° )
≈ 3.7
Name:                                                                           Homework 5.6.L Right Triangle Trig
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CP 5.6.4. A right triangle has legs of lengths 20 and 31. Solve the triangle.
b
a = 20
b = 31
y
2      2       2
c =a +b                                                                           c
2       2   2
c = a + b = 20 + 31 = 1361
2                                                                  a

a
tan ( x ) =                                                                     x
b
a                                                          a
x = arctan                                                                         b
b
 20 
= arctan  
 31 
≈ 32.8°

y = 90° − x
≈ 90° − 32.8°
= 57.2°

CP 5.6.5. A 40 meter guy wire is attached to the top of a 35 meter atenna and to a point on the ground. What
is the primary measure (in degrees) of the angle made by the guy wire and the ground?

a
sin ( x ) =
c
a           35 
x = arcsin   = arcsin   = 61°
c           40 

CP 5.6.6. Two country roads meet at a right angle. A biker travelling on the North-South road and
approaching the intersection decides to take a shortcut across the field. When he reaches a large oak, he turns
40° and heads across the field for the East-West road. If he travelled a straight 0.25 miles across the field to
get to the East-West road, how far is the intersection from the oak? Round to the nearest 100th of a mile.

b
cos( x) =
c
b = c cos( x) = 0.25cos ( 40° ) ≈ 0.19 mi
Name:                                                                           Homework 5.6.L Right Triangle Trig
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CP 5.6.7. The forest ranger at the top of Kendrick mountain is watching a forst fire spread is her direction. In
10 minutes the angle of depression of the leading edge of the fire changed from 11° to 13°. At what speed is
the fire spreading in the direction of the ranger. Assume the ranger is 3430 feet above the fire.

a
tan ( x1 ) =
b1 + b2
a
tan ( x2 ) =
b2
b1 + b2 b2 b1
cot ( x1 ) − cot ( x2 ) =          − =
a      a a
b1 = a ( cot ( x1 ) − cot ( x2 ) )
b1
v=
t
a ( cot ( x1 ) − cot ( x2 ) )
=
t
3430 ( cot (11° ) − cot (13° ) )
=
10
≈ 279 ft/min
Name:                                                                          Homework 5.6.L Right Triangle Trig
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Extra Practice

CP 5.6.8. A triangle has a hypotenuse of length 100 and a leg of length 10. Solve the triangle.

c = 100                                                                                       b
a = 10

c2 = a2 + b2                                                                              y
b2 = c2 – a2                                                                     c
b = c −a
2         2                                                                           a
2       2
= 100 − 10
= 9900                                                                        x
= 90 11                                                                a
b

a
sin ( x ) =
c
a
x = arcsin  
c
 10 
= arcsin      
 100 
≈ 5.7°

y = 90° – x
≈ 90° – 5.7°
= 84.3°
Name:                                                                          Homework 5.6.L Right Triangle Trig
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CP 5.6.9. One angle of a right triangle measures 39° and the leg adjacent to it is 15 units long. Solve the
triangle.
b
x = 39°
b = 15
y
y = 90° – x                                                                      c
= 90° – 39°                                                                                a
= 51°

b                                                                  x
cos ( x ) =
c
b                                                                   a
c=                                                                                   b
cos ( x )
15
=
cos ( 39° )
≈ 19.3

a
tan ( x ) =
b
a = b tan ( x )
= 15 tan ( 39° )
≈ 12.1

CP 5.6.10. A tree casts a shadow which is 6.4 feet long. The primary measue of
the angle of elevation from the shadow of the tree’s top to the actual tree’s top is
71°. How tall is the tree to the nearest 100th of a foot?

a
tan ( x ) =
b
a = b tan ( x ) = 6.4 tan ( 71° ) ≈ 18.59
Name:                                                                       Homework 5.6.L Right Triangle Trig
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CP 5.6.11. A hot-air baloonist spots two mile-posts on the straight road over which he hovers. He measures
the angle of depression to mile post 184 to be 17° and the angle of depression to mile post 183 to be 21°.
How high is he?

a
tan ( x1 ) =
1+ b
a
tan ( x2 ) =
b
1+ b b 1
cot ( x1 ) − cot ( x2 ) =       − =
a   a a
1
a=
cot ( x1 ) − cot ( x2 )
1
=
cot (17° ) − cot ( 21° )
≈ 1.5 mi

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