Trigonometry Assignment: Solving Engineering Measurement Problems

This assignment solution covers a range of trigonometry concepts relevant to engineering applications. Task 1 addresses circular and triangular measurement problems using sine, cosine, and tangent functions, including calculations of ladder lengths and angles, and conversions between ...Read More

Trigonometry Assignment
Student’s Name
Institution Affiliation

Task 1
Part a
I. How far up does the ladder reach
Using the sine rule , 1.4
Sinθ = 5
sin 90
Sinθ=1.4 sin 90
5 =0.28
θ=sin−1 0.28=16.26 °
Wall length=5 cos 16.26 °=4.80 m
We obtain the same answer using Pythagoras theorem as follows.
x= √52−1.42=4.8 m
II. Minimum length of the ladder. Given that the value of height =4.5 and the
θ=90 °−74 °=16°
cos ( 16 ° ) = 4.5
Ladder Length

Ladder Length= 4.5
cos ( 16 ° ) =4.68 m
III. The angle at which the ladder meets the horizontal ground, given that the ration for
ladder horizontal to vertical is 1:4
Tanθ= Opposite
Adjacent = 4
1 =4
θ=tan−1 (4)=75.96° ≅ 76.0 °
Part b
I. 150 ° to radian
180 °=π radian,
Therefore, 150 °will be given by
150°
180° π =5
6 π radian
II. Area of the region covered
Area=rad
2 π ∗π r2
¿ 5 π
6∗2 π ∗π∗122

¿ 188.50 km2
III. Perimeter of the region covered
Perimeter= rad
2 π ∗2 πr +12+12
¿ 5 π
6∗2 π ∗2∗π∗12+24
¿( 31.42+24 ) km
¿ 55.42 km
Task 2
Trigonometric graphs
Angle in Radians(x) 0 π
3
2 π
3
π 4 π
3
5 π
3
2 π
y=sinx 0 1 0.8660 0 -0.866 -
0.866
0
y=cosx 1 0.5 -0.5 -1 -0.5 0.5 1
y=tanx 0 1.732 -1.732 0 1.732 -
1.732
0
A graph of y=sin x

A graph of y=cos x
A graph of y=tan x

Task 3
a. Task

Vertical Component of first force=50 sin 45 °=35.3553 kN
Horizontal Component of first force=50 cos 45°=35.3553 kN
The second force has horizontal component only that is equal to 50 kN .
Total Vertical Component of the resultant force R=35.3553 kN + 0=35.3553 kN
Total Horizontal Component of the resultant force R=35.3553 kN +50 kN =85.3553 kN
Total Resultant force Using PythagorasTheorem , R= √ 35.35532 +85.35532
¿ √ 8535.5244762
¿ 92.3879 kN
tanr °= 35.3553
85.3553 =0.4141136
r °=tan−1 0.4141136=22.5 °
b. Task

Cosine rule: c2=a2+b2−2 abcosC
In the above case, the length of the tie will be given by
( length ) 2=5.12 ×62 −2× 5.1× 6 ×cos 28=7.9736
length= √7.9736=2.8238 m
The value of θ, will be given by
6
sinα = 2.8238
sin 28
sinα= 6 sin 28
2.8238 =0. 99753
α =sin−1 0. 99753=85.97 °
The θ will be given by
θ=180° −85.97 °=94.03 °
Task 4
I. Open cylinder:
Volume of the open cylinder

V =π r2 l
¿ π∗422∗150
¿ 831 , 265.416 mm3
Surface area of an open cylinder:
S . A=π r2 +2∗2 πrl
¿ π∗422+2∗π∗42∗150
¿ 5 , 541.77+39 ,584.06=45 ,125.84 mm2
II. Triangular prism
Volume of a triangular prism is given by:
V = 1
2 ( bh )∗L
¿ 1
2 ( 16∗72 )∗38
¿ 21 , 888 mm3
Surface area of the triangular prism
S . A=bh+lb+2 ls
s= √722 +82 =72.443 mm
S . A=16 ×72+38 ×16 +2× 38 ×72.443=¿ 7265.668 mm2
III. Sphere

Volume of sphere is given by:
V = 4
3 π r3= 4
3 ∗π∗253=65 , 449.85 mm3
Surface area of sphere is given by:
S . A=4 π r 2=4∗π∗252=7 , 853.98 mm2