122.
If 4 cos2 A – 3 = 0 then prove that
(i)
sin 3 A = 3 sin A – 4 sin3 A.
(ii)
cos 3 A = 4 cos3 A – 3 cos A
Wednesday, 31 July 2019
TRIGONOMETRICAL IDENTITIES - PART – 87
121.
(i) If 2 sin A – 1 = 0,
show that sin
3A = 3 sin A – 4 sin3 A.
(ii)
If 4 cos2 A – 3 = 0,
show that cos
3A = 4 cos3 A – 3 cos A.
TRIGONOMETRICAL IDENTITIES - PART – 86
120.
If tan A = n tan B and
sin A = m sin B,
prove that: cos2 A = (m2
– 1) / (m2 + 1).
Tuesday, 30 July 2019
TRIGONOMETRICAL IDENTITIES - PART – 84
118.
If x = a cosꝊ and
y = b cotꝊ,
show that:
(a2
/x2) – (b2 /y2).
TRIGONOMETRICAL IDENTITIES - PART – 83
117.
If cos A / cos B = m and
cos A / sin B = n
show that (m2 + n2)
cos2 B = n2
Monday, 29 July 2019
TRIGONOMETRICAL IDENTITIES - PART – 82
116.
If x = r cos A cos B,
y = cos A sin B
and then
z = r sin A, then prove that
x2 + y2 + z2
= r2
TRIGONOMETRICAL IDENTITIES - PART – 81
115.
If sin A + cos A = m and
sec
A + cosec A = n show that
n (m2 – 1) = 2 m
TRIGONOMETRICAL IDENTITIES - PART – 80
114.
If x = r sin A cos B,
y = sin A sin B
and then
z = r cos A, then prove that
x2 + y2 + z2
= r2
TRIGONOMETRICAL IDENTITIES - PART – 79
113.
If m = a sec A + b tan A and
n = a tan A + b sec A, then prove that
m2
– n2 = a2 – b2
TRIGONOMETRICAL IDENTITIES - PART – 78
112.
If p = sin A + cos A and
q = sec A + cosec A, then prove that
q (p2
– 1) = 2 p
Saturday, 27 July 2019
TRIGONOMETRICAL IDENTITIES - PART – 77
111.
If x cos A + y sin A = m and
x sin A – y cos A = n, then prove that:
x2
+ y2 = m2 + n2
TRIGONOMETRICAL IDENTITIES - PART – 76
110.
If x = a sec A cos B,
y = b sec A sin B and z
= c tan A
show that
x2/a2
+ y2/b2 + z2/c2 = 1
TRIGONOMETRICAL IDENTITIES - PART – 75
109.
If tan A + sin A = m and tan A – sin A = n;
prove that
m2 – n2
= 4 (mn)1/2
Friday, 26 July 2019
TRIGONOMETRICAL IDENTITIES - PART – 74
108.
[(cos
A) / (cosec A + 1)]
+
[(cos
A )/ (cosec A – 1)]
= 2 tan A
TRIGONOMETRICAL IDENTITIES - PART – 71
105.
(sin Ꝋ + 2sec Ꝋ)2 + (cos Ꝋ + cosec Ꝋ)2
=
(1
+ sec Ꝋ cosec Ꝋ)2
TRIGONOMETRICAL IDENTITIES - PART – 61
95.
(sinꝊ – cosꝊ + 1) / (sinꝊ + cosꝊ – 1)
=
(1/
(secꝊ – tanꝊ)
TRIGONOMETRICAL IDENTITIES - PART – 59
93. [(1 + cot A) / cos A] + [(1 + tan
A) / sin A]
=
2 (sec A + cosec A)
TRIGONOMETRICAL IDENTITIES - PART – 58
92. sin2 A tan A +
cos2 A cot A + 2 sin A cos A
=
tan A + cot A
Tuesday, 23 July 2019
TRIGONOMETRICAL IDENTITIES - PART – 54
88.
[1 / cos A + sin A – 1]
+
[1 / cos A + sin A + 1]
=
cosec A + sec A
TRIGONOMETRICAL IDENTITIES - PART – 53
87.
[tan A + 1 / cos A]2 + [tan A – 1 / cos A)]2
=
[1
+ sin A / 1 – sin A]
TRIGONOMETRICAL IDENTITIES - PART – 50
84.
[(cos3A + sin3A) / (cos A + sin A)]
+
[(cos3A – sin3A) / (cos
A – sin A)]
=
2
TRIGONOMETRICAL IDENTITIES - PART – 49
83.
[sin A – sin B / cos A + cos B]
+
[cos
A + cos B / sin A + sin B]
=
0
TRIGONOMETRICAL IDENTITIES - PART – 48
82.
[cot A / 1 – tan A] + [tan A / 1 – cot A]
=
sec
A cosec A + 1
TRIGONOMETRICAL IDENTITIES - PART – 47
81.
[Cot A / 1 – tan A] + [tan A / 1 – cot A]
=
1 + tan A + cot A
TRIGONOMETRICAL IDENTITIES - PART – 43
77.
(sin A + cos A / sin A – cos A)
+
(sin A –
cos A / sin A + cos A)
=
(2 /2sin2 A – 1)
TRIGONOMETRICAL IDENTITIES - PART – 36
70. (sin A – cos A)
(1 + tan A + cot A)
=
[(sec A / cosec2 A) – (cosec2 A /
sec2 A)]
APPLICATION OF TRIGONOMETRY – PART – 33
PROBLEM
– 38
For the top of a hill, the angles of
depression are found to be 30 degrees and 45 degrees respectively. Find the
distances of the two stones form the foot of the hill.
APPLICATION OF TRIGONOMETRY – PART – 32
PROBLEM
– 37
An aeroplane flying horizontally 1 km above
the ground and going away from the observer is observed at an elevation of 60
degrees. After 10 seconds, its elevation is observed to be 30 degrees; find the
uniform speed of the aeroplane in km per hour.
APPLICATION OF TRIGONOMETRY – PART – 31
PROBLEM
– 36
The length of the shadow of a tower standing
on level plane is found to be 2y metres longer when the sun’s altitude is 30
degrees then when it was 45 degrees. Prove that the height of the tower is y (√3
+ 1) meters.
APPLICATION OF TRIGONOMETRY – PART – 30
PROBLEM
– 35
The
horizontal distance between two towers is 75 m and the angular depression of
the top of the first tower as seen from the top of the second, which is 160 m
high, is 45 degrees. Find the height of the first tower.
APPLICATION OF TRIGONOMETRY – PART – 29
PROBLEM
– 34
A
person standing on the bank of a river observes that the angle of elevation of
the top of a tree standing on the opposite bank is 60 degrees. When he moves 40
m away from the bank, he find the angle
of elevation to be 30 degrees. Find
(i)
the height of the tree, correct to 2 decimal places.
(ii) the width of the
river.Wednesday, 17 July 2019
APPLICATION OF TRIGONOMETRY – PART – 28
PROBLEM
– 33
A man in a boat rowing away from a lighthouse
150 m high, takes 2 minutes to change the angle of elevation of the top of the
lighthouse from 60 degrees to 45 degrees. Find the speed of the boat.
APPLICATION OF TRIGONOMETRY – PART – 27
PROBLEM
– 32
A man on a cliff observes a boat, at an angle
of depression 30 degrees, which is sailing towards the shore to the point
immediately beneath him. Three minutes later, the angle of depression of the
boat is found to be 60 degrees. Assuming that the boat sails at a uniform
speed, determine:
(a) how much more time it will take to reach
the shore?
(b) the speed of the boat in metre per second,
if the height of the cliff is 500 m.
Tuesday, 16 July 2019
APPLICATION OF TRIGONOMETRY – PART – 26
PROBLEM
– 31
From
the top of the cliff, 60 m high, the angles of depression of the top and bottom
of a tower are observed to be 30 degrees
and 60 degrees. Find the height of the tower.
APPLICATION OF TRIGONOMETRY – PART – 25
PROBLEM
– 30
The angle of elevation of the top of a tower
is observed to be 60 degrees. At a point, 30 m vertically above the first point
of observation, the elevation is found to be 45 degrees. Find
(i) the height of the tower
(ii) its horizontal distance from the points
of observation.
APPLICATION OF TRIGONOMETRY – PART – 24
PROBLEM
– 29
From
the figure, given below, calculate the length of CD.
APPLICATION OF TRIGONOMETRY – PART – 23
PROBLEM
– 28
Two pillars of equal heights stand on either
side of a roadway, which is 150 m wide. At a point in the roadway between the
pillars the elevation of the tops of the pillars are 60 degrees and 30 degrees;
find the height of the pillars and position of the point.
Sunday, 14 July 2019
APPLICATION OF TRIGONOMETRY – PART – 22
PROBLEM
– 27
From
the top of a light house 100 m high, the angle of depression of two ships are
observed as 48 degrees and 36 degrees respectively. Find the distance between
the two ships if:
(a)
the ships are on the same side of the light house.
(b) the ships are on
the opposite sides of the light house.APPLICATION OF TRIGONOMETRY – PART – 21
PROBLEM
– 26
Find the height of a building, when it is
found that on walking towards it 40 m in a horizontal line through its base the
angular elevation of its top changes from 30 degrees to 45 degrees.
Saturday, 13 July 2019
APPLICATION OF TRIGONOMETRY – PART – 20
PROBLEM
– 25
Find the height of a tree when it is found
that on walking away from it 20 m, in a horizontal line through its base, the
elevation of its top changes from 60 degrees to 30 degrees.
APPLICATION OF TRIGONOMETRY – PART – 19
PROBLEM
– 24
In the fig, given below, it is given that AB
is perpendicular to BD and is of length X metres. DC = 30 m. Triangle ADB = 30
degrees and Triangle ACB = 45 degrees. Without using tables, find X.
APPLICATION OF TRIGONOMETRY – PART – 18
PROBLEM
– 23
A vertical pole and a vertical tower are on
the same level ground. From the top of the pole the angle of elevation of the
top of the tower is 60 degrees and angle of depression of the foot of the tower
is 30 degrees. Find the height of the tower if the height of the pole is 20 m.
APPLICATION OF TRIGONOMETRY – PART – 17
PROBLEM
– 22
Form
a point on the ground, the angle of elevation of the top of a vertical tower is
found to be such that its tangent is 3 / 5. On walking 50 m towards the tower,
the tangent of the new angle of elevation of the top of the tower is found to
be 4 / 5. Find the height of the tower.
APPLICATION OF TRIGONOMETRY – PART – 16
PROBLEM
– 21
The angle of elevation of a stationary cloud
from a point 25 m above a lake is 30 degrees and the angle of depression of its
reflection in the lake is 60 degrees. What is the height of the cloud above the
lake-level.
Wednesday, 10 July 2019
APPLICATION OF TRIGONOMETRY – PART – 15
PROBLEM
– 20
A man on the top of the vertical observation
tower observes a car moving at a uniform speed directly towards it. If it takes
12 minutes for the angle of depression to change from 30 degrees to 45 degrees,
how soon after this will the car reach the observation tower.
APPLICATION OF TRIGONOMETRY – PART – 14
PROBLEM
– 19
An observer on the top of the cliff; 200 m
above the sea-level, observes the angles of depression of the two ships to be
45 degrees and 30 degrees respectively. Find the distance between the ships, if
the ships are
(a) on the same side of the cliff
(b) on the opposite
sides of the cliff.Tuesday, 9 July 2019
APPLICATION OF TRIGONOMETRY – PART – 13
PROBLEM
– 18
From
the top of a cliff 92 m high, the angle of depression of a buoy is 20 degrees.
Calculate to the nearest metre, the distance of the buoy from the foot of the
cliff.
APPLICATION OF TRIGONOMETRY – PART – 12
PROBLEM
– 17
A man stands 9 m away
from a flag pole. He observes that angle of elevation of the top of the pole is
28 degrees and the angle of depression of the bottom of the pole is 13 degrees.
Calculate the height of the pole.
APPLICATION OF TRIGONOMETRY – PART – 11
PROBLEM
– 16
Two climbers are at points A and B on a
vertical cliff face. To an observer C, 40 m from the foot of the cliff, on the
level ground. A is at an elevation of 48 degrees and 57 degrees. What is the
distance between the climbers?
APPLICATION OF TRIGONOMETRY – PART – 10
PROBLEM
– 15
Two vertical poles
are on either side of a road. A 30 m long ladder is placed between the two
poles. When the ladder rests against one pole. When the ladder rests against
one pole, it makes angle 30 degree with pole and when it is tuned to rest
against another pole, it makes angle 30 degrees with the road. Calculate the
width of the road.APPLICATION OF TRIGONOMETRY – PART – 09
PROBLEM
– 14
At a particular time, when the sun’s altitude
is 30 degrees, the length of the shadow of a vertical tower is 45 m. calculate
(a) the height of the tower (b) the length of the shadow of the same tower,
when the sun’s altitude is (i) 45 degrees (ii) 60 degrees.
APPLICATION OF TRIGONOMETRY – PART – 08
PROBLEM
– 13
The angle of elevation of the top of an unfinished tower at a point distance 80 m from its base is 30 degrees. How much higher must the tower be raised so that its angle of elevation at the same point may be 60 degrees.
The angle of elevation of the top of an unfinished tower at a point distance 80 m from its base is 30 degrees. How much higher must the tower be raised so that its angle of elevation at the same point may be 60 degrees.
Thursday, 4 July 2019
APPLICATION OF TRIGONOMETRY – PART – 07
APPLICATION
OF TRIGONOMETRY – PART – 07
PROBLEM
– 12
The upper part of a palm tree, broken over by
the wind makes an angle of 45 degrees with the ground; and the distance from the root to the point where the
top of the tree touches the ground is 15 m. What was the height of the tree
before it was broken.
APPLICATION OF TRIGONOMETRY – PART – 06
APPLICATION
OF TRIGONOMETRY – PART – 06
PROBLEM
– 11
The length of the shadow of a vertical tower
on level ground increases by 10 m, when the altitude of the sun changes from 45
degrees to 30 degrees. Calculate the height of the tower.
APPLICATION OF TRIGONOMETRY – PART – 05
APPLICATION OF TRIGONOMETRY – PART – 05
PROBLEM – 09
PROBLEM – 09
Two people standing on the same side of a tower in a straight line with it, measure the angle of elevation of the top of the tower as 25 degrees and 50 degrees respectively. If the height of tower is 70 m, find the distance between the two people.
PROBLEM – 10
A boy 1.5 m tall , is 25 m away from a tower and observes the angle of elevation of the top of the tower to be (i) 45 degrees and (ii) 60 degrees. Find the height of the tower in each case.
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