98 Higher Engineering Mathematics
Now try the following exercise
Exercise 44 Further problems on the
theorem of Pythagoras
1. In a triangle CDE, D = 90 ◦ , C D = 14.83 mm
and C E = 28.31 mm. Determine the length of
D E.
[24.11 mm]
2. Triangle PQR is isosceles, Q being a right
angle. If the hypotenuse is 38.47 cm find (a)
the lengths of sides P Q and Q R, and (b) the
value of ∠QPR. [(a) 27.20 cm each (b) 45 ◦ ]
3. A man cycles 24 km due south and then 20 km
due east. Another man, starting at the same
time as the first man, cycles 32 km due east and
then 7 km due south. Find the distance between
the two men.
[20.81 km]
4. A ladder 3.5 m long is placed against a perpendicular wall with its foot 1.0 m from the wall.
How far up the wall (to the nearest centimetre)
does the ladder reach? If the foot of the ladder is now moved 30 cm further away from the
wall, how far does the top of the ladder fall?
[3.35 m, 10 cm]
5. Two ships leave a port at the same time. One
travels due west at 18.4 km/h and the other due
south at 27.6 km/h. Calculate how far apart the
two ships are after 4 hours.
[132.7 km]
6. Figure 11.4 shows a bolt rounded off at one
end. Determine the dimension h. [2.94 mm]
R 5 45 mm
h
r
516 mm
Figure 11.4
7. Figure 11.5 shows a cross-section of a
component that is to be made from a round bar.
If the diameter of the bar is 74 mm, calculate
the dimension x.
[24 mm]
72 mm
␾ 7 4 m m
x
Figure 11.5
11.3 Trigonometric ratios of acute
angles
(a) With reference to the right-angled triangle shown
in Fig. 11.6:
(i)
sine θ =
opposite side
hypotenuse
i.e.
sin θ =
b
c
(ii)
cosine θ =
adjacent side
hypotenuse
i.e.
cos θ =
a
c
(iii)
tangent θ =
opposite side
adjacent side
i.e.
tan θ =
b
a
(iv)
secant θ =
hypotenuse
adjacent side
i.e.
sec θ =
c
a
(v)
cosecant θ =
hypotenuse
opposite side
i.e. cosec θ =
c
b
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