Answers[Trigonometry]
1.
+
=
(
)
(
)
=
=
=
= 2 cosec Q
Ans
=
=
2
2.
+ tan Q
×
=
×
Ans
= 3 sec2 Q – 4 tan2 Q + tan2 Q
OR
= 3 sec2 Q – 3 tan2 Q
= tan Q = = =
= 3 (sec2Q – tan2Q)
=3×1=3
Ans
=x√
3. sin2A cot2A + cos2A tan2A
Now, =
×
2
= sin A ×
+
So, h = √
2
=
+ cos A
= sin2A + cos2A = 1
+
×
–
×
×
×
=
Ans
6. Simply put Q = 45°
4. sin4Q + sin2 Q cos2 Q
= sin2 Q (sin2Q + cos2Q)
2
= sin Q
Ans
7. Go through option
2 × 1 – 3 × 1 = -1
So, Q = 90°
5.
–
8. 2 sin2 Q + 4 cos2 Q = 3
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Answers[Trigonometry]
= 2sin2Q + 2cos2Q + 2cos2Q = 3
= 2 + 2cos2Q = 3
= cotQ =
We know that tanQ × cotQ =1
2
= 2 (1 + cos Q) = 3
= 1 + cos2Q =
= h2 = xy
= cos2Q = - 1 =
= cosQ =
= × =1
=h =
so, Q = 45°
√
12.
9. sin2A cos2B – cos2A sin2B
X
= (1 – cos2A) cos2B – cos2A (1 – cos2B)
Q
= cos2B – cos2A cos2B – cos2A + cos2
cos2B
2
2
√3 x
= tanQ =
= cos B – cos A
= 1 – sin2B – (1- sin2A)
= 1 – sin2B – 1 + sin2A
√
=
√
= tanQ = tan 30
=Q
= 30°
= sin2A – sin2B
13. AB = BC
10. Between and
Now, tanQ =
A Q
=1
11. tanQ =
= tanQ = tan 45° B
=
C
Q = 45°
= tan (90 – Q) =
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Answers[Trigonometry]
then go through option for easy solve
otherwise you can apply quadratic
formula that is
14.
x=
±√
h
60
30
x
x
√2 x
Q
20
x
= h = tan 30 × (x + 20)
= tan 30 =
so, h : (P or b) = √2 x : x
= tan 60 =
= √2 : 1
= √3
=
= h
= √3 x
And, h =
= √3 x =
= 3x
√
× (x + 20)
√
= x + 20
x = 10
So, h = √3 x = √3 ×10 = 10√3
2000
B
60 45
C
= ABC = 60°
= DCB = 45°
In △ ABC, tan60 =
OR,
h2 = x × (x + 20)
h=
16. A
(x + 20)
= 2 x = 20
=
Ans
× ( + 20)
= √3 =
= BC =
√
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Answers[Trigonometry]
In △ DBC, tan45° =
=
= BC =
= 200 m/sec
=
BC = DB
=
DB =
√
Ans
18. A
So, AD = AB – BD
E
= 2000 -
√
√
= 2000 (
√
h
)
Ans
2h
90Q
17. tan60 =
= √3 =
=
√
= BC = 1500 mt
Again, tan30 =
=
=
√
=
B
√
Q
C
D
60
Now, tan(90 – Q) =
= cotQ =
Again, tanQ =
√
CE = 1500 × 3
Now, tanQ × cotQ = 1
=
×
=1
= 4500
= 2h2 = 900
Now, BE = 4500 – 1500
= h2 = 450
= 3000
So, it takes 15sec to travel 3000 mt
So, speed of aeroplane =
= h
= 15√2 m
So, 2h = 30 √2 m
Ans
19. now tan45 =
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Answers[Trigonometry]
= √3 RS ------- 1
= 1
=
= QR
= AB
= BC
Again, tan 60 =
= √3 =
A
= QR =
45
B
60
D
√
= √3 RS =
= RS =
Again, tan30 =
=
--------2
From eqn 1 & 2 we can write
30
C
√
√
=
√ ×√
21. tan 45 =
= CD = √3 h – h
= AB
= 60 = h (√3 - 1)
Now, tan 30 =
= h =
=
=
Ans
=
= h + CD = √3 h
=
m
√
(√
√
)
=
√
AD
= 1500 m
=
= 1500 √3
√
(√
The distance covered = 1500
)
= 30 (√3 + 1)
√3 – 1500
Ans
= 1500 (√3 – 1)
= 1500 (√3 – 1)
Ans
20. tan 30 =
=
√
=
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Answers[Trigonometry]
22.
=
h =
√
Now, x √3 =
√
h
=3x
60
= x + 10
30
=
X
x =5
10
So, h = x√3
= 5√3
Using formula , h =
h=
=
(√
√
–
23. tan30 =
)
×√
=
= 5√3
Ans
Ans
=
√
=
y =
√
m
OR
Now,
h
x
30
10
= cot 60 =
=
=
√
h
= x
+ 100 = x
=
60
+ 10
√
=
= x √3
= 100
(
)
= 10
=
×
√
=x
=x
=x
Now, x + y =
√
+
×
√
Again, cot 30 =
=
√
= √3 =
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Answers[Trigonometry]
=
= 10√3
√
Ans
= 2a = c
25.
Ans
A
24.
60
x
Q1
B
Q2
a
c
=
= tanQ2 =
=
As, Q1 + Q2 = 90°
C
Now, cos 60° = =
= tanQ1 = = tan 60° =
= tan 30° =
6.5
=
AC
.
= 6.5 × 2
= 13mt
Ans
So, tanQ1 = cot (90 – Q2)
Or, tanQ2 = cot (90 – Q2)
26. In △ PMS , tan 30 =
= tan 60° =
In △ PQR , tan 60 =
=
√3 =
=
√3 a = x
= tan 60 =
=
√3 =
=
QR =
Again, tan 30° =
=
=
√
=
x =
= 18√3
√
Again, tan 30 =
√
Now, √3 a =
=
√
=
√
=
√
PM = 18
= 3a = a + c
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Answers[Trigonometry]
So, RS = PQ – PM = MQ
=
= 54 – 18
= 36 mt
=
Ans
×√
√
√ ×(√
(√
√
(
=
27.
)
)
√ )
= 30 (√3 + 3)
= 30 (√3 + 3)
h
29.
60
30
1km
12m
h=
=
Ans
√
=
Q
4√3
√
=
Now, tanQ = =
√
Ans
√
= tanQ =
√
√
= tanQ = √3
28.
= tanQ = tan 60, Q = 60°
Ans
h
60
45
60
Using formula, h =
=
30. The top of two buildings of height
24m and 36m are connected by a
cable wire. If the cable wire makes an
angle of 60° with the horizontal then
the length of the wire is
√
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Answers[Trigonometry]
Now, sin 60 =
= cosQ = =
=
x
=
= cosQ =
=
x
=
=
=
√
√
= cosQ = cos 45
√
=
×
Q = 45°
Ans
√
= 8√3 m
Ans
33.
31.
A
x
√2 x
Q
x
B
Q
C
= sinQ =
=
=sinQ =
= sinQ = sin 45 = Q = 45°
√3 X
= tanQ =
√
=
√
√
√
= tanQ = tan 30°
=
32.
34.
Q = 30°
A
A
Q
√2 x
B
Now, tanQ =
Q
B
x
C
C
= tanQ = 1
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Answers[Trigonometry]
= tanQ = tan 45°, Q = 45°
= 2 sinQ – sinQ = cosQ + cosQ
= sinQ = 2 cosQ
35. = tan (90 – Q) = cotQ
=
So, tan (20 + 45°) = cot 3Q
=
=1
=
= 90 – (2Q - 45°) = 3Q
= 90 – 45 – 2Q
= cotQ =
= 3Q
Ans
= 45° = 3Q + 2Q
= Q
= 15°
39. cosQ + secQ = √3
Ans
36. = sec4Q – tan4Q = (sec2Q + tan2Q)
(sec2Q – tan2Q)
= (sec2Q + tan2Q) × 1
= sec4Q – tan4Q =
= cos3Q + sec3Q = (cosQ + secQ)3 –
3(cosQ + secQ)
= (√3 )3 – 3 × √3
= 3√3 - 3√3 = 0
Ans
Ans
40. sinQ + cosecQ = 2
37. = tan 10 . tan 15 . tan 75 . tan 80
If x + = 2 , then xn +
= cot 80 . cot 75 . tan 75 . tan 80
(tan (90 – Q) = cotQ)
So, sin28Q + cosec28Q = 2
= cot 80 × tan 80 × cot 75 × tan 75
= 1 × 1= 1
41. (
)(
Ans
-
)
Ans
=(
38. =
+
=2
=1
= 2 sinQ – cosQ = sinQ + cosQ
=
+
)(
-
)
×
=
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Answers[Trigonometry]
=
= 1
A
Ans
42. tan7Q tan2Q = 1
2x
B
So, (7Q + 2Q) = 90°
= 9Q
= 90°
=
= 10°
Q
x
= AC = (2 ) + ( )
+
= √4
So, tan 3Q = tan 30°
= √5 x
=
So, sinA + cotC
Ans
√
=
=
(
=
=
)(
)
(
(
)
(
)
√
Ans
√
45. tanQ – cotQ = 0
= tanQ = cotQ
= tan45 = cot45
=
So, =
+
√
=
)
+
√
=
43. =
C
Ans
=
Q = 45°
So, sinQ + cosQ
= sin45 + cos45
44.
=
√
+
√
=
√
= √2
Ans
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Answers[Trigonometry]
46. cos 90° = 0
49. Always 1
So, the total value = 0
50. a sinQ + b cosQ
47. sinQ =
Greatest value = √
= sinQ = sin30
= √3 + 4
=
= √25
Q = 30°
= 5
+ Q = 90° ( )
So,
+
Ans
= 60°
Now, sin
= sin 60° =
√
Ans
48. sin4Q + cos4Q = (sin2Q + cos2Q)
(sin2Q – cos2Q)
= sin4Q + cos2Q = sin2Q – cos2Q
=
√
So, value will remain same i.e.,
√
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