Sponsor Area
Let AB = 24 cm
BC = 7cm Using Pythagoras theorem, we have
AC2 = AB2 + BC2
= (24 cm)2+ (7 cm)2
= 576 cm2 + 49 cm2
= 625 cm2
So, AC = 25 cm
Now,
Let
PQ = 12K
and
PR = 13K
Using Pythagoras theorem, we have
PR2 = PQ2 + QR2
⇒(13K)2 = (12K)2 + QR2
⇒169K2 = 144K2 + QR2
⇒QR2 = 169K2 - 144K2
⇒QR2 = 25K2
So, QR = 5K
Now,
Therefore,
If calculate cos A and tan A.
Let
AB = 8K, BC = 15K
Using Pythagoras theorem, we have
AC2 = AB2 + BC2 = (8K)2 + (15K)2
= 64K2 + 225K2
= 289K2
So,
AC = 17K
Now,
AC = 17K
Given calculate all other trigonometric ratios.
Using Pythagoras theorem, we have
AC2 = AB2 + BC2
⇒(13 K)2 = (12 K)2 + BC2
⇒169 K2 = 144 K2 + BC2
⇒BC2 = 169 K2-144 K2
BC2 = 25 K2 BC = 5 K.
Now,
and
If cot θ = evaluate:
Using Pythagoras theorem, we have
AC2 = AB2 + BC2
= (4K)2 + (3K)2
= 16K2 + 9K2 = 25 K2
So, AC = 5K
Now,
and
L.H.S. =
Now,
R.H.S. =
Hence, L.H.S. = R.H.S.
We have,
Let
Using Pythagoras theorem, we have
AC = 2K
So,
AC = 2K
and ,
Now,
(i) sin A cos C + cos A sin C
(ii) cos A. cos C - sin A .sin C
We have,
Let
PR + QR = 25 cm ..(i)
PR = x cm
QR = (25 - x) cm.
Using Pythagoras theorem, we have
PR2 = PQ2 + RQ2
⇒x2 = (5Y + (25 - x)2
x2 = 25 + 625 + x2 - 50x
⇒50x = 650
⇒x = 13 cm
Now,
PR = x = 13 cm QR = (25 - x) = 12 cm
and
PQ = 5 cm
Now,
PQ = 5 cm
and
(i) False, since and perpendicular may be longer than base.
(ii) True, since and hypotenuse being the longest side may be times the base.
(iii) False, since cos A is the abbreviation used for the cosine of angle A.
(iv) False, since cot A is used as an abbreviation for 'the cotangent of the angle A'.
(v) False, since the hypotenuse is the longest side in a right triangle. As such the value of sin A is always less than 1 (or, in particular equal to 1).
Choose the correct option and justify your choice:
sin 2A = 2 sin A is true when A =
(A) 0° (B) 30° (C) 45° (D) 60°
(iii) If A = 0
Then, L.H.S. = sin 2A = sin 2(0)
= sin 0° = 0
R.H.S. = 2 sin A = 2 sin 0°
= 2 x 0 = 0
Ans. (A).
Sponsor Area
If
Find A and B.
Take
Then
and
So, sin θ ≤θ cos θ
Hence, this statement is false.
Show that:
tan 48° tan 23° tan 42° tan 67° = 1.
LHS
= tan 48° . tan 23°. tan 42° . tan 67°
= tan 48° . tan 42° . tan 23° . tan 67°
= tan (90° - 42°). tan 42°. tan (90° - 67°). tan 67°
= cot 42° . tan 42° . cot 67° . tan 67°
= 1 x 1 = 1 = RHS Proved.
Show that:
cos 38° cos 52° - sin 38° sin 52° = 0.
LHS
= cos 38° . cos 52° - sin 38° . sin 52°
= cos (90°-52°). cos 52°-sin (90° - 52°). sin 52°
= sin 52° . cos 52° - cos 52° . sin 52°
= 0 = RHS Proved.
tan 2A = cot (A - 18°) = cot (90° - 2A) = cot (A-18°)
⇒90° - 2A = A - 18°
⇒-3A = -108°
⇒A = 36°.
We have,
sec 4A = cosec (A - 20°)
cosec (90° - 4A) = cosec (A - 20°)
⇒90° - 4A = A - 20°
⇒-5 A = -110°
⇒A = 22°.
sin 67° + cos 75°
= sin (90° - 23°) + cos (90° - 15°) = cos 23° + sin 15°.
sin (90° - 65°). cos 65° + cos (90° - 65°). sin 65°
= cos65°. cos 65° + sin 65° sin 65°
= cos2 65° + sin2 65° [Using sin2 A + cos2 A = 1]
= 1.
Sponsor Area
(sin A + cosec A)2 + (cos A + sec A)2 = 7 + tan2A + cot2 A
L.H.S.
= (sin A + cosec A)2 + (cos A + sec A)2
= (sin2 A + cosec2 A + 2 sin A. cosec A + cos2 A + sec2 A + 2 cos A. sec A
= (sin2 A + cos2 A) + 2 sin A. cosec A + 2 cos A. sec A + cosec2 A + sec2 A
= 1 + 2 + 2 + (cot2 A + 1) + (tan2 A + 1)
= 5 + 1 + 1 + cot2 A + tan2 A
= 7 + tan2 A + cot2 A = R.H.S.
Hence, L.H.S. = R.H.S.
sin2 69° + sin2 21° = sin2 (90 - 21°) + sin2 21°
= cos2 21° +sin2 21° = 1. 1
sin (65° + θ ) - cos (25° - θ)
= cos [90° - (65° + θ)] - cos (25° - θ)
= cos (90° - 65° - θ) - cos (25° - θ)
= cos (25° - θ) - cos (25° - θ) = 0.
sin 84° + tan 59°
= sin (90° -6°) + tan (90° -31°) = cos 6° + cot 31°.
sin θ = cos θ
⇒ cos (90-θ) = cos θ
⇒90-θ = θ
⇒2θ = 90 ⇒ θ = 45°.
= 3 cos (90° - 38°). cos 38°
=3 sin (90°-38°). sin 38°
= 3 sin 38° . cos 38° - 3 cos 38° . sin 38°
= 0.
sin 38° - cos 52°.
= sin (90° - 52°) - cos 52°
= cos 52° - cos 52° = 0
sin 25° - cos 65°.
= sin (90° - 65°) - cos 65°
= cos 65° - cos 65° = 0.
cot 85° + cos 75°
= cot (90° - 5°) + cos (90° - 15°)
= tan 5° + sin 15°.
Sponsor Area
If and
, then what is the value of cot B?
It is given that: ...(i)
and A + B = 90°
⇒ A = 90° - B ...(ii)
Putting the value of (ii) in (i), we get
Ans.
we have, sin θ = cos θ
cos (90° -θ) = cos θ
⇒ 90° - θ = θ
⇒ 2θ = 90°
⇒ θ = 45°
we have, tan θ = cot (30° + θ)
cot (90°-θ) = cot (30°+ θ)
⇒90° - θ = 30° + θ
2θ = 60°
⇒ θ = 30°
If what is the value of
If find the value of (sinA + cosA) secA.
If find the value of (tanA + cotA).
Solution not provided.
Ans. 24°
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Ans. 33°
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Ans. 12°
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Ans. 15°
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Ans. 20°
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Ans. 671/2°
Solution not provided.
Ans. 28°
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Ans. 10°
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Ans. 10°
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Ans. sec 36° + cos 18°
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Ans. cos 1° + cot 1°
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Ans. sec 21° + tan 21°
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Ans. sin 15° + tan 15°
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Ans. sin 20° + tan 20°.
Consider an equilateral ΔABC.
Since each angle in a equilateral A is 60°.
∴ ∠A = ∠B = ∠C = 60°
Draw a perpendicular from A on side BC.
Now, ΔABD= ΔACD => BD = DC
and ΔBAD = ΔCAD [C.P.C.T.]
Let AB = 2a. Then,
In right ΔADB, by Pythagoras theorem.
Now, in right ΔADB,
Sponsor Area
sec A (1 - sin A) (sec A + tan A) = 1
L.H.S. = sec A (1 - sin A) (sec A + tan A)
Hence, L.H.S. = R.H.S.
(1 + cot θ- cosec θ) (1 + tan θ+ sec θ) = 2.
Hence, L.H.S. = R.H.S. Proved.
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Ans. (i) 1, (ii)
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Evaluate the following:
3 cos2 30° + sec2 30° + 2 cos 0° + 3 sin 90° - tan2 60°
Solution not provided.
Ans.
Evaluate the following:
sin2 60° + sec2 30° + sin 90° + 3 cos 0° - tan2 60°
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Ans. 180 cm
If Find the value of cos (θ + β), if θ and β are both acute angles.
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Ans. θ = 24°
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Ans. k = 1
C.
20°Tips: -
cos (40 + x) = sin 30° = sin (90° – 60°) ⇒ cos (40° + x) = cos 60° ⇒ x = 20°.L.H.S. = (1+ cot A + tan A) (sin A - cos A)
L.H.S. = R.H.S.
Hence, (1 + cot A + tan A) (sin A - cos A) = sin A . tan A - cot A . cos A Proved
If then show that
Prove the following identity:
sec A (1 - sin A) (sec A + tan A) = 1
sec A (1 - sin A) (sec A + tan A) = 1
L.H.S. = sec A (1 - sin A) (sec A + tan A)
Hence, L.H.S. = R.H.S.
The ratio of the height of a tower and the length of its shadow on the ground is √3:1. What is the angle of elevation of the sun?
The angle of elevation of the sun from the ground is θ.
We have, x:y =√3:1
Now, In ∆ABC
A moving boat is observed from the top of a 150 m high cliff moving away from the cliff. The angle of depression of the boat changes from 60° to 45° in 2 minutes. Find the speed of the boat in m/h.
The angle of elevation of a cloud from a point 60 m above the surface of the water of a lake is 30° and the angle of depression of its shadow in water of lake is 60°. Find the height of the cloud from the surface of the water.
Let AB be the surface of the lake and P be the point of observation such that AP = 60 m. Let C be the position of the cloud and C be its reflection in the lake.
Then CB =
Draw PM⊥CB
Let CM = h
∴ CB = h + 60 m
In ∆CPM
⇒ 3h = h + 120 m
⇒ 2h = 120 m
⇒ h = 60 m
CB = h + 60m = 60m + 60m = 120m
Thus, the height of the cloud from the surface of the lake is 120 m.
As observed from the top of a 100 m high light house from the sea-level, the angles of depression of two ships are 30° and 45°. If one ship is exactly behind the other on the same side of the light house, find the distance between the two ships. [Use √3= 1.732]
Let ships are at distance x from each other,
The angle of depression of a car, standing on the ground, from the top of a 75 m high tower, is 30o. The distance of the car from the base of the tower (in m.) is
150
C.
Let AB be the tower of height 75 m and C be position of the car.
In
The angle of elevation of the top of a tower at a distance of 120 m from a point A on the ground is 45°. If the angle of elevation of the top of a flagstaff fixed at the top of the tower, at A is 60°, then find the height of the flagstaff. [Use ]
Let AB is the tower of height h meter and AC is flagstaff of height x meter.
The angle of depression of a car parked on the road from the top of a 150 m high tower is 30°. The distance of the car from the tower (in metres) is
75
A.
Let AB be the tower and BC be the distance between tower and car. Let be the angle of depression of the car.
According to the given information,
If a tower 30 m high, casts a shadow 10 3m long on the ground, then what is the angle of elevation of the sun?
Let AB be the tower and BC be its shadow.
On a straight line passing through the foot of a tower, two points C and D are at distances of 4 m and 16 m from the foot respectively. If the angles of elevation from C and D of the top of the tower are complementary, then find the height of the tower.
Let AB be the tower with height h.
Let x be the angle of elevation from c.
So, the angle of elevation from D is (90-x).
..........( Since the angle of elevation from C and D are complementary)
From the top of a tower, 100 m high, a man observe two cars on the opposite sides of the tower and in same straight line with its base, with its base, with angles of depression 30° and 45°. Find the distance between the cars.
[Take = 1.732]
The man is at the top of the tower AB.
In right angled triangle ,
Hence, the distance between X and Y is 273.20 m approximately.
The length of shadow of a tower on the plane ground is 3 times the height of the tower. The angle of elevation of sun is
450
300
600
900
B.
300
Let AB be the tower and BC be its shadow. Let be the angle of elevation of the sun.
According to the given information,
Hence, the angle of elevation of the sun is 300.
A kite is flying at a height of 45 m above the ground. The string attached to the kite is temporarily tied to a point on the ground. The inclination of the string with the ground is 60o. Find the length of the string assuming
that there is slack in the string.
Given: Position of kite is B.
Height of kite above ground is = 45 m.
Angle of inclination is = 600
Required length of string is = AB.
In right angled triangle AOB,
The angle of elevation of the top of a hill at the foot of a tower is 600 and the angle of depression from the top of the tower of the foot of the hill is 300. If the tower is 50 m high, find the height of the hill.
Let the height of hill is h.
In right triangle ABC,
Hence the height of hill is 150 m.
Sponsor Area
Sponsor Area