1. Ashoka Institution,
Malkapur,
Hyderabad
Prof.S.Rajendiran,
Ashoka Institution
Basic
and
fundamental
to be
recollected
while joining
Engineering
Stream

2. 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

3. 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

4. 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

5. 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 =
BC
AC
AB
AC
= BC
AC
AC
AB
=
BC
AB
4

6. 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 =
AB
AC
BC
AC
= AB
AC
AC
BC
=
AB
BC
5

7. 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

8. 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) = ABAC
ACAB (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

9. 1800
3600


180-
180-

180-
180-

Please Note
When 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

10. A
B C
a
b
c
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

11. A
B C
D
Let us draw a line AD perpendicular to BC
Proof of Lame's Theorem
Sin B =
a
bc
AD
c
AD = (Sin B) x c 1
Sin C = AD
b
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 get
b
Sin B
=
c
Sin C
Similarly we can prove a
Sin A
=
b
Sin B
=
c
Sin C
CB
A
10

12. A
B
C
c
b
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 Triangle
Fig 2 force diagram
Extension of Lame's theorem force diagram( Proof)
a
=
b
=
c
Sin A Sin B Sin C
C
A
11

13. A
B
a
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

14. Area formula for regular shape
13

15. r
Circle Perimeter 2πr or πd
d (diameter)= 2r
r
L = arc length whose radius is r = r
l

14