Maths In Focus - Exercise 6.5
Jessica leaves home and walks for 4.7 km on a bearing of 075 . She then turns and walks for 2.9 km on
a bearing of 115 and she is then due east of her home.
(a)
How far north does Jessica walk?
(b)
How far is she from home?
Solution:
The diagram required is displayed below.
(a)
1 Zoe
We are required to find how far north Jessica walked. This refers to past tense, and furthermore, the northerly distance at point T 1 . This is equivalent to finding the value of x. Using
W T H we refer to SOHCAHTOA. That is,
cos(75 )
=
x
4.9
=) x
=
4.9 ⇥ cos(75 ).
=) x
⇡
1.2 km (2 d.p.).
was correct in class. In fact, looking at the diagram it is the only conclusion we can reach.
1
(b)
We are required to find the distance, D, that is, the length of the interval HJ. The question
has stated that J is due east of her initial position, H. This implies that J is horizontally
joint to H 2 . We find D under the strategy
D = HZ + ZJ.
Since \W HT = 75
have
=) \T HZ = 90
75 = 15 . Apply SOHCAHTOA on
cos(15 )
=
HZ
4.7
=) HZ
=
4.7 ⇥ cos(15 ).
We leave our answer in exact form. Next, since \N 0 T J = 115
65 . Applying SOHCAHTOA on T ZJ we see that
sin(65 )
=
ZJ
2.9
=) ZJ
=
2.9 ⇥ sin(65 ).
=) \ZT J = 180
T HZ, we
115 =
We leave our answer in exact form.
Therefore,
D
=) D
=
HZ + ZJ
=
4.7 ⇥ cos(15 ) + 2.9 ⇥ sin(65 )
⇡
7.2 km (2 d.p.).
2 This is actually not mathematically true. However, when a question states due east, this is the assumption we must
always take.
2