PRODI PENDIDIKAN MATEMATIKA
FAKULTAS KEGURUAN DAN ILMU PENDIDIKAN
UNIVERSITAS MUHAMMADIYAH TANGERANG
Jl. Perintis Kemerdekaan 1/33 Cikokol-Tangerang-Banten
Nama
NIM
Semester/Kelas
Jenis
: Ufi Luthfiyah Saeruroh
: 10.84-202.099
: VI/B1 (malam)
: Tugas Individu
Jurusan/Prodi
Mata Kuliah
Dosen
Chapter
: Pendidikan Matematika
: Matematika Advance Level
: Drs. Hairul Saleh, M.Si.
: 10 (Trigonometry)
No.
Exercise
9.
Find the least positive value of the angle A for which
a) sin Ao = 0.2 and cos Ao is negative;
Answer :
Solving the equation sin 𝐴𝑜 = 0,2
Step 1
Use your calculator to find sin-1(0,2) = 11,54
This is one root in the interval −180 ≤ 𝐴 ≥ 180.
Step 2
Use the symmetry property sin (180 – A) = sin A to show that 180 – (11,54) = 168,5… is another
root. Unfortunately it is not in the required interval.
Step 3
Use the periodic property, sin (𝜃 ± 360) = 168,5 – 360 = – 191,5 . which is a root in the required
interval.
So, the least positive value of the angle A is 168,5.
b) tan Ao = –0.5 and sin Ao is negative;
Answer :
Solving the equation tan 𝐴𝑜 = – 0,5
Step 1
Use your calculator to find tan-1( – 0,5) = – 26,57
This is one root in the interval −180 ≤ 𝐴 ≥ 180.
Step 2
Use the symmetry property tan (180 + A) = tan A to show that 180 + (– 26,57) = 153,4… is
another root. Unfortunately it is not in the required interval.
Step 3
tan (180 + A) = tan A to show that 180 + (153,4) = 333,4
So, the least positive value of the angle A is 333,4.
c) cos Ao = sin Ao and both are negative;
d) sin Ao = –0.2275 and A > 360.
Answer :
Solving the equation sin 𝐴𝑜 = –0.2275
Step 1
Use your calculator to find sin-1(–0.2275) = –13,15
This is one root in the interval −180 ≤ 𝐴 ≥ 180.
Step 2
Use the symmetry property sin (180 – A) = sin A to show that 180 – (–13,15) = 193,15… is
another root. Unfortunately it is not in the required interval.
Step 3
Use the periodic property, sin (𝜃 ± 360) = 193,15 + 360 = 553,1 . which is a root in the required
interval.
So, the least positive value of the angle A is 553,1
19.
The road to an island close to the shore is sometimes covered by the tide. When the water rises to the
level of the road, the road is closed. On a particular day, the water at high tide is a height 4.6 metres
above mean sea level. The height, h metres, of the tide is modeled by using the equation h = 4.6 cos
kto, where t is the time in hours from high tide; it it also assumed that high tides occur every 12 hours.
a) Determine the value of k
Answer :
360𝑜
𝑘=
= 300
12
Show that :
h = 4.6 cos kto
h = 4,6 cos (30 x 12)o
h = 4,6 cos (360)o
h = 4,6 cos 1
h = 4,6
b) On the same day, a notice says that the road will be closed for 3 hours. Assuming that this notice
is correct, find the height of the road above sea level, giving your answer correct to two decimal
places.
Unknown : hwater = 4,6 ; twater = 12 ; troad = 3
Asked : hroad
Anwer :
ℎ𝑤𝑎𝑡𝑒𝑟 𝑡𝑤𝑎𝑡𝑒𝑟
=
ℎ𝑟𝑜𝑎𝑑
𝑡𝑟𝑜𝑎𝑑
4,6 12
=
𝑥
3
12x = 4,6 x 3
12x = 13,8
x = 1,15
so, height of the road above sea level is
hwater - hroad = 4,6 – 1,15 = 3,45 m
c) In face, a road repair has raised its level, and it is impassble for only 2 hours 40 minutes. By how
many centimetres has the road level been raised?
Answer :