IMPORTANT FACTS AND FORMULAE
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1.We
already know that:
In a rt.angled DOAB, where ÐBOA = q,
i)sin q = Perpendicular/Hypotenuse
= AB/OB;
ii)cos q = Base/Hypotenuse = OA/OB;
iii)tan q = Perpendicular/Base
= AB/OA;
iv)cosec q = 1/sin
q = OB/AB;
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vi)cot q = 1/tan
q = OA/AB.
2. Trigonometrical identities:
i)sin2q + cos2q = 1.
ii)1+tan2q = sec2q
iii)1+cot2q = cosec2q
3. Values of T-ratios:-
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q
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0
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30°
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45°
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60°
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90°
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Sin q
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0
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½
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1/Ö2
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Ö3/2
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1
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Cos q
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1
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Ö3/2
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1/Ö2
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½
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0
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Tan q
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0
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1/Ö3
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1
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Ö3
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Not defined
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4. Angle of
Elevation: Suppose a man from a point O looks up an object P, placed above
the level of his eye. Then, the angle which the line of sight makes with the
horizontal through O, is called the angle of elevation of P as seen from O.
\
Angle of elevation of P from O = ÐAOP.
5. Angle of Depression: Suppose a man from a point O looks down at an
object P, placed below the level of his eye, then the angle which the line of sight
makes with the horizontal through O, is called the angle of depression of P as
seen from O.
SOLVED EXAMPLES
Ex.1.If the height of a pole is 2Ö3 metres and the length of its shadow is 2 metres,
find the angle of elevation of the sun.
2Ö3m
q
C 2m A
Sol. Let AB be the pole and AC
be its shadow.
Let angle of elevation, ÐACB=q.
Then, AB = 2Ö3 m AC = 2 m.
Tan q = AB/AC = 2Ö3/2 = Ö3 Þ q = 60°
So, the angle of elevation is 60°
Ex.2. A ladder leaning against a
wall makes an angle of 60° with the ground. If the length of the ladder is
19 m, find the distance of the foot of the ladder from the wall.
B
2)
19m
 |
60°
C
A
X
Sol. Let AB be the wall and BC
be the ladder.
Then, ÐACB = 60° and BC = 19
m.
Let AC = x metres
AC/BC = cos 60° Þ x/19 = ½ Þ x=19/2 = 9.5
\Distance of the foot of the ladder from the wall
= 9.5 m
Ex.3. The angle of elevation of the top of a
tower at a point on the ground is 30°. On walking 24 m towards the tower, the angle of
elevation becomes 60°. Find the height of the tower.
B
h
30° 60°
C 24m
D A
Sol. Let AB be the tower and C
and D be the points of observation. Then,
AB/AD = tan 60° = Ö3 => AD
= AB/Ö3 = h/Ö3
AB/AC = tan 30° = 1/Ö3 AC=AB
x Ö3 = hÖ3
CD = (AC-AD) = (hÖ3-h/Ö3)
hÖ3-h/Ö3 = 24 => h=12Ö3 = (12´1.73) = 20.76
Hence, the height of the tower is 20.76 m.
Ex.4. A man standing on the bank of a river
observes that the angle subtended by a tree on the opposite bank is 60°. When he retires 36 m from
the bank, he finds the angle to be 30°. Find the breadth of the river.
B
h
30° 60°
C 36m D x A
Sol. Let
AB be the tree and AC be the river. Let C and D be the two positions of the
man. Then,
ÐACB=60°, ÐADB=30° and CD=36
m.
Let
AB=h metres and AC=x metres.
Then,
AD=(36+x)metres.
AB/AD=tan
30°=1/Ö3 => h/(36+x)=1/Ö3
h=(36+x)/ Ö3 .....(1)
AB/AC=tan
60°=Ö3 => h/x=Ö3
h=Ö3x .....(2)
From
(i) and (ii), we get:
(36+x)/ Ö3= Ö3x => x=18 m.
So,
the breadth of the river = 18 m.
Ex.5. A
man on the top of a tower, standing on the seashore finds that a boat coming
towards him takes 10 minutes for the angle of depression to change from 30° to 60°. Find the time taken by the
boat to reach the shore from this position.
B
h
30° 60°
D A
x C y
Sol. Let
AB be the tower and C and D be the two positions of the boat.
Let
AB=h, CD=x and AD=y.
h/y=tan
60°=Ö3 => y=h/Ö3
h/(x+y)=tan
30° = 1/Ö3 => x+y=Ö3h
x=(x+y)-y
= (Ö3h-h/Ö3)=2h/Ö3
Now,
2h/Ö3 is covered in 10 min.
h/Ö3 will be covered in (10´(Ö3/2h)´(h/Ö3))=5 min
Hence,
required time = 5 minutes.
Ex 6.
There are two temples, one on each bank of a river, just opposite to each
other. One temple is 54 m high. From the top of this temple, the angles of
depression of the top and the foot of the other temple are 30° and 60° respectively. Find the width
of the river and the height of the other temple.
B
30°
D E
h
60°
C A
Sol. Let
AB and CD be the two temples and AC be the river.
Then,
AB = 54 m.
Let
AC = x metres and CD=h metres.
ÐACB=60°, ÐEDB=30°
AB/AC=tan
60°=Ö3
AC=AB/Ö3=54/Ö3=(54/Ö3´Ö3/Ö3)=18m
DE=AC=18Ö3
BE/DE=tan
30°=1/Ö3
BE=(18Ö3´1/Ö3)=18 m
CD=AE=AB-BE=(54-18)
m = 36 m.
So,
Width of the river = AC = 18Ö3 m=18´1.73 m=31.14m
Height
of the other temple = CD= 18 m.
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