Ashoka Institution, Malkapur, Hyderabad Prof.S.Rajendiran, Ashoka Institution Basic and fundamental to be recollected while joining Engineering Stream
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Ashoka Institution,Malkapur,
Hyderabad
Prof.S.Rajendiran,Ashoka Institution
Basicand
fundamentalto be
recollected while joiningEngineering
Stream
A
B C
900
Right angled triangle ABC
Sin = BC
AC
cos = AB
AC
tan = BC
AB
BC = opposite side with respect to angle
AB = adjacent side with respect to angle
AC = hypotenuses side the side opposite to right angle is 900
Step by Step Engineering mechanics
= Theta
= alpha
Angle can be represented By and and so on
1
+ = 900
A
B C
900
Right angled triangle ABC
Sin = AB
AC
cos = BC
AC
tan = AB
BC
AB = opposite side with respect to angle
BC = adjacent side with respect to angle
AC = hypotenuses side the side opposite to right angle is 900
Step by Step Engineering mechanics
2
A
B C
900
opposite side
adjacent side
A
B C
900
opposite side
adjacent side
Pl note opposite side and adjacent are referred with respect to angle under consideration
Step by Step Engineering mechanics
3
A
B C
900
Right angled triangle ABC
Sin = BC
AC
cos = AB
AC
tan = BC
AB
Step by Step Engineering mechanics
tan = Sin
cos =
BCACAB
AC
= BC
AC
AC
AB=
BC
AB
4
A
B C
900
Right angled triangle ABC
Sin = AB
AC
cos = BC
AC
tan = AB
BC
Step by Step Engineering mechanics
tan = Sin
cos =
ABACBC
AC
= AB
AC
AC
BC=
AB
BC
5
A
B C
900
Right angled triangle ABC
Step by Step Engineering mechanics
As per Pythagoras theorem
AC2 = AB2 + BC2
Let us see the Proof of sin2 + cos2 =1
sin2 + cos2 = AB2
AC2+
BC2
AC2=
AB2 + BC2
AC2
=AC2
AC2
= 1
sin2 + cos2 = 1
6
A
B C
900
Right angled triangle ABC
Sin = BC
AC
tan = BC
AB
Step by Step Engineering mechanics
(Sin) = BCAC
BC = (Sin) AC
(cos) = ABACACAB (cos) =
Pl note 0pposite side = (sin)x hypotenuses side----1
Pl note adjacent side = (cos)x hypotenuses side---2
Equation 1 and 2 very much important for Engineering Mechanics 7
opposite side
adjacent side
hypotenuses side
1800
3600
180-
180-
180-
180-
Please NoteWhen two parallel line cut by a line the angle created by the line and angle similarities
Straight line will have 1800
Circle consist of 3600
8
A
B Ca
bc
A
B C
a
Sin A=
b
Sin B=
c
Sin C
Lame's Theorem
Triangle Sides and its angle relation Angles, A+B+C=180
Side BC > AB+AC
9
A
B CD
Let us draw a line AD perpendicular to BC
Proof of Lame's Theorem
Sin B =
a
bcADc
AD = (Sin B) x c 1
Sin C = ADb
AD = (Sin c) x b 2
Comparing equation 1 and 2
AD = (Sin c) x b (Sin B) x c=
(Sin c) x b (Sin B) x c= Re arranging this equation we getb
Sin B=
c
Sin CSimilarly we can prove a
Sin A=
b
Sin B=
c
Sin C
CB
A
10
A
BC
cb
a
a
b
c
a=
b=
c
Sin A Sin B Sin C
a=
b=
c
Sin(180- A) Sin(180- B) Sin(180- C)
Both are parallel
Both are parallel
Both are parallel
Fig 1 TriangleFig 2 force diagram
Extension of Lame's theorem force diagram( Proof)
a=
b=
c
Sin A Sin B Sin C
C
A
11
A
Ba
bc
C
Triangle Equation related to angle and sides most useful for Engineering mechanics
CB
A
a2 = b2 + c2 - (b x c x cosA)
12
Area formula for regular shape
13
r
Circle Perimeter 2πr or πd
d (diameter)= 2r
r
L = arc length whose radius is r = r
l
14
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