This joke pokes fun at stereotypes about different professions. When a physicist and engineer call for help from a hot air balloon, they receive a response stating the obvious fact that they are in a hot air balloon. The physicist identifies the responder as a mathematician because the answer was technically correct but unhelpful in solving their problem. This highlights a stereotype about mathematicians providing precise but impractical solutions. The joke plays with perceptions of different approaches across fields.

1.  "A hundred years from now, it will
not matter what kind of car I drove,
what kind of house I lived in, how
much money I had in the bank...but
the world may be a better place
because I made a difference in the
life of a child." -- Forest Witcraft

2.  "Education would be much more
effective if its purpose was to ensure
that by the time they leave school
every boy and girl should know how
much they do not know and be
imbued with a lifelong desire to
know it." -- William Haley

3.  "One looks back with appreciation to
the brilliant teachers, but with
gratitude to those who touched our
human feelings. The curriculum is so
much necessary material, but
warmth is the vital element for the
growing plant and for the soul of the
child." -- Carl Jung

4.  There are two good
reasons to be a
teacher – June and
July.

5.  "We spend the first twelve
months of our children's lives
teaching them to walk and talk,
and the next twelve years telling
them to sit down and shut up."

6.  "A statistician can have his head
in an oven and his feet in ice, and
he will say that on the average
he feels fine."

7. • I have heard that parallel
lines do meet, but they are
very discrete

8. BY:
Mrs.ROOHI JILANI

9. Measuring Angles : In Degrees or Radians
θ
The angle, θ, can be
measured in degrees. This
represents the turn required
to move from one line to the
other in the direction shown.
This turn is measured in
degrees. Degrees are a unit
measuring turning where 360o
is a full turn.
360o

10. If we imagine a circle of
radius 1 unit, then a full
turn would be a full
circle and the point A
moves would be the
same as the
circumference of the
circle
Radians is another
measure for angles.
This time you represent
the angle as the
distance point A moves
around the
circumference of an
imaginary circle.
A
⇒360o
= 2π radians (or 2π c
)
⇒1o
= 2π c
360o
⇒1 c
= 360o
2π

11. θ
r
r
Length of arc, L
L = (2π r) θ
360o
Area of sector, A
A = (πr 2
) θ
360o
L = (2π r) θ = r θ
2π
A = (πr2
) θ = ½ r2
θ
2π
In degrees …In radians …
Here we have a sector
draw with angle θ. This
sector has an arc
length of L and an area
of A.
L
Area,A
Uses of radians

12. 1. Convert from degrees to radians
1. 30o
2. 145o
3. 500o
4. -60o
2. Convert from radians to degrees
1. 2
/3 π rads
2. 7
/5 π rads
3. -5
/8 π rads
4. 0.5 rads

13. Calculate length of the arc and areas for these sectors,
a)
b)
c)
Radius = 4cm
θ = 2
/9 π
Radius = 6.3cm
θ = 3
/7 π
Radius = 14cm
θ = 4.1
Note : angles in radians

14. Adjacent
Opposite
Hypotenuse
sine θ = opposite ÷ hypotenuse
cosine θ = adjacent ÷ hypotenuse
tangent θ = opposite ÷ adjacent

15. 0o
90o
45o
60o
60o
60o
30o
Hypotenuse = Adjacent
Opposite = 0
sin 0o
= 0 cos 0o
= 1 tan 0o
= 0
Hypotenuse = Opposite
Adjacent = 0
sin 90o
= 1 cos 90o
= 0 tan 90o
=
undefined
Adjacent = Opposite = x
Hypotenuse = x√2
sin 45o
= 1/√2 cos 45o
= 1/√2 tan 45o
= 1
x
x
x
x
x
For 60o
Hypotenuse = x
Adjacent = x Opposite = x √3
2 2
sin 60o
= √3 cos 60o
= 1 tan 60o
= √3
2 2
For 30o
Hypotenuse = x
Opposite = x Adjacent = x √3
2 2
sin 30o
= 1 cos 30o
= √3 tan 30o
= 1
2 2 √3
Some Standard Solutions …

16. θ
(deg)
0 30 45 60 90
θ
(rads)
0 π
6
π
4
π
3
π
2
sin θ 0 ½ 1
√2
√3
2
1
cos θ 1 √3
2
1
√2
½ 0
tan θ 0 1
√3
1 √3 -

17. 0o
180o
90o
270o
0o180o
90o
270o
0o
180o
90o
270o
0o180o
90o
270o
Opposite = +
Adjacent = +
Opposite = +
Adjacent = -
Opposite = -
Adjacent = -
Opposite = -
Adjacent = +

18. +
+
+
+
+
+
- -
-
-
-
-
Images from BBC AS Guru

19. Sin + All +
Tan + Cos +
CAST Diagram

20. Sin + All +
Tan + Cos +
Solve : sin x = 0.5 for the range 0 ≤ x ≤ 360o
∴ x = arcsin 0.5 = 30o
but sin is positive in two quadrants so
x = 30o
or (180 – 30)=150o

21. Solve : sin x = 0.5 for the range 0 ≤ x ≤ 360o
∴ x = arcsin 0.5 = 30o
but sin is positive in two quadrants so
x = 30o
or (180 – 30)=150o

22. Find all the angles (in degrees) in the given range
1. Sin x = - ½ for the range 0 ≤ x ≤ 360o
2. Cos 2x = √3
/2 for the range -360o
≤ x ≤ 360o
3. Tan (2x+40o
) = √3 for the range -180o
≤ x ≤ 180o
Find all the angles (in radian) in the given range
1. Sin x = √3
/2 for the range 0 ≤ x ≤ 2π
2. Cos 2x = -1
/2 for the range -2π ≤ x ≤ 2π
3. Tan (2x+ ½π) = 1 for the range -π ≤ x ≤ π

23.  A physicist and an engineer are in a hot-air
balloon. Soon, they find themselves lost in a
canyon somewhere. They yell out for help:
"Helllloooooo! Where are we?"
15 minutes later, they hear an echoing voice:
"Helllloooooo! You're in a hot-air balloon!!"
The physicist says, "That must have been a
mathematician."
The engineer asks, "Why do you say that?"
The physicist replied: "The answer was
absolutely correct, and it was utterly useless."