Download trigonometric ratios

TRIGONOMETRIC RATIOS
1) If cos A =
, calculate sin A and cot A.
SOLUTION:
In ∆ ABC,
cos A =
If AB = 12k, AC = 13k
By Pythagoras Theorem,
(AC) 2 = (BC) 2 + (AB) 2
(13k) 2 = (BC) 2 + (12k) 2
169k2 = (BC) 2 + 144k2
(BC) 2 = 169k2 144k2
(BC) 2 = 25k2
BC = 5k
sin A =
=
=
cot A =
=
=
C
B
12k
A
2) In ABC, right-angled at B, if AC = 5 cm and BC = 3 cm, find all the six
trigonometric ratios of angle A.
SOLUTION:
In ∆ ABC,
By Pythagoras Theorem,
(AC) 2 = (BC) 2 + (AB) 2
(AB) 2 = (AC) 2 (BC) 2
(AB) 2 = (5)2 (3)2
(AB) 2 = 25 9
(AB) 2 = 16
C
3cm
B
A
TRIGONOMETRIC RATIOS
AB = 4cm
sin A =
=
cos A =
=
tan A =
=
cosec A =
=
sec A =
=
cot A =
=
3) In Δ ABC, right-angled at B, AC + BC = 25 cm and AB = 5 cm. Determine
the values of sin A, cos A and tan A.
SOLUTION:
AC + BC = 25; AB= 5 cm
Let BC = x cm
AC = (25-x) cm
In ∆ ABC,
By Pythagoras Theorem,
(AC) 2 = (BC) 2 + (AB) 2
(25 x) 2 = ( ) 2 + (5)2
625 50 + 2 = 2 + 25
625 25 = 50
600 = 50
=
= 12 cm
[Given]
C
B
5cm
A
TRIGONOMETRIC RATIOS
BC = 12 cm
AC = (25 12) = 13 cm
sin A =
=
cos A =
=
tan A =
=
4) If tan θ = , evaluate :
SOLUTION:
In ∆ ABC,
BAC =Ө.
tan Ө =
=
=
= k (say)
[where k is a constant]
AB= 3k and
By Pythagoras Theorem,
(AC) 2 = (BC) 2 + (AB) 2
(AC) 2 = (4k) 2 + (3k) 2
(AC) 2 = 16k 2 + 9k 2
(AC) 2 = 25k 2
AC = 5k
sin A= sin
=
C
4k
Ө
=
B
cos A= cos
=
=
=
3k
A
TRIGONOMETRIC RATIOS
=
=
=
=
Or
If you know Trigonometric identities, use this method.
=
=
5) If Sin A =
=
=( ) =
[
]
, calculate cos A and tan A.
SOLUTION:
In ∆ ABC,
Sin A =
If BC = 5k, AC= 13k
By Pythagoras Theorem,
(AC) 2 = (BC) 2 + (AB) 2
(13k) 2 = (5k) 2 + (AB) 2
169k 2 = 25k2 + (AB) 2
(AB) 2 = 169k2 25k2
(AB) 2 = 144k2
AB = 12k
cos A =
=
=
tan A =
=
=
C
5k
B
A
TRIGONOMETRIC RATIOS
6) In ABC, right-angled at B, if AB = 4 cm and BC = 3 cm, find all the six
trigonometric ratios of angle A.
SOLUTION:
In ∆ ABC,
By Pythagoras Theorem,
(AC) 2 = (BC) 2 + (AB) 2
(AC) 2 = (4) 2 + (3) 2
(AC) 2 = 16 + 9
(AC) 2 = 25
AC = 5cm
sin A =
=
cos A =
=
tan A =
=
C
3cm
B
cosec A =
4cm
=
sec A =
=
cot A =
=
7) In Δ ABC, right-angled at B, AC + BC = 49 cm and AB = 7 cm. Determine
the values of sin A, cos A and tan A.
SOLUTION:
AC + BC = 49; AB = 7 cm
Let BC = cm
AC = (49 x) cm
(given)
A
TRIGONOMETRIC RATIOS
In ∆ ABC,
By Pythagoras Theorem,
(AC) 2 = (BC) 2 + (AB) 2
(49
) 2 = ( ) 2 + (7) 2
2401 98 + 2 = 2 + 49
2401 49 = 98
2352 = 98
C
=
= 24 cm
BC = 24 cm
AC = (49 24) = 25 cm
sin A =
=
cos A =
=
tan A =
=
B
8) If cot θ = , evaluate :
SOLUTION:
In ∆ ABC,
BAC = Ө.
cot Ө =
=
=
= k (say), where k is a positive number.
AB = 3k and BC = 4k
7cm
A
TRIGONOMETRIC RATIOS
By Pythagoras Theorem,
(AC) 2 = (BC) 2 + (AB) 2
(AC) 2 = (4k) 2 + (3k) 2
(AC) 2 =16k2 +9k2
(AC) 2 = 25k2
AC = 5k
sin Ө =
=
cos Ө =
=
C
4k
=
Ө
B
=
=
=
=
Or
If you know Trigonometric identities, use this method.
=
=
=(
2
=
3k
A
TRIGONOMETRIC RATIOS OF SOME SPECIFIC ANGLES
1) Evaluate:
SOLUTION:
=
=
=
=
(
)(
)
(
)(
)
=
=
=
2) In ABC, right-angled at B, if A = 60o and AB = 12 cm, determine BC
and AC.
SOLUTION:
In ∆ ABC,
C
cos 60o =
AC =12 x 2
AC =
sin 60o =
60o
B
12 cm
A
TRIGONOMETRIC RATIOS OF SOME SPECIFIC ANGLES
BC = 24
BC =
3) If cos (15o + ) = sin 45o, find the value of .
SOLUTION:
cos (15o + ) = sin 45o
cos (15o + ) =
cos (15o + ) = cos 45o
15o +
= 45o
= 45o
= 30o
4) If cosec (A + B) =
[ cos 45o =
]
15o
and sec (A
B) =
, find A and B.
SOLUTION:
cosec (A + B) =
and sec (A
cosec (A + B) =
= cosec 60o
B) =
(given)
(A + B) = 60o………………………………………………….(1)
TRIGONOMETRIC RATIOS OF SOME SPECIFIC ANGLES
sec (A
B) =
= sec 30o
B) = 30o………………………………………………….(2)
(A
Solving (1) and (2), we get
A = 45o
B = 15o
5) If A = 30o and B = 60o, prove that sin (A+B)= sin A cos B + cos A sin B
SOLUTION:
A = 30o and B = 60o
(given)
LHS = sin (A + B)
= sin (30 + 60 )
= sin 90
=1
RHS= sin A cos B + cos A sin B
= sin 30 cos 60 + cos 30 sin 60
=(
)+
= +
=
=1
LHS = RHS
sin (A+B) = sin A cos B + cos A sin B
TRIGONOMETRIC RATIOS OF SOME SPECIFIC ANGLES
6) Evaluate:
SOLUTION:
=
=
=
=
(
)(
)
(
)(
)
=
=
=
7) In ABC, right-angled at B, if A = 30oand AB = 12 cm, determine BC and
AC.
SOLUTION:
In ∆ ABC,
cos 30o =
C
AC = 12
30o
AC =
B
AC = 8√3 cm
sin 30o =
12 cm
A
TRIGONOMETRIC RATIOS OF SOME SPECIFIC ANGLES
BC =8√3
BC =
BC=4√3 cm
8) If sin (15o + ) = cos 45o, find the value of x.
SOLUTION:
sin (15o + ) = cos 45o
sin (15o + ) =
sin (15o + ) = sin 45o
15o +
= 45o
= 45o
= 30o
9) If sin (A + B) =
[ sin 45o =
]
15o
and cos (A
B) =
, find A and B.
SOLUTION:
sin (A + B) =
and cos (A
sin (A + B) =
= sin 60o
B) =
(given)
(A + B) = 60o………………………………………………….(1)
TRIGONOMETRIC RATIOS OF SOME SPECIFIC ANGLES
cos (A
B) =
= cos 30o
(A B) = 30o………………………………………………….(2)
Solving (1) and (2), we get
A =45o
B = 15o
10) If A = 60o and B = 30o, prove that sin (A + B)= sin A cos B + cos A sin B
SOLUTION:
A = 60o and B = 30o
(given)
LHS= sin (A + B)
= sin (60o + 30o)
= sin 90o
=1
RHS= sin A cos B + cos A sin B
= sin 60o cos 30o + cos 60o sin 30o
=
=
+(
)
+
=
=1
LHS = RHS
sin (A + B) = sin A cos B + cos A sin B
TRIGONOMETRIC RATIOS OF SOME COMPLEMENTARY ANGLES
1) Evaluate:
SOLUTION:
=
=
=
=1
2) Show that cos 26o cos 64o
sin 26o sin 64o = 0
SOLUTION:
LHS = cos 26o cos 64o sin 26o sin 64o
= cos (90o 64o) cos (90o 26o) sin 26o sin 64o
= sin 64o sin 26o sin 26o sin 64o
=0
= RHS
LHS=RHS
3) If cos 3A = sin (A
42o), where 3A is an acute angle, find the value of A.
SOLUTION:
cos 3A = sin (A 42o)
sin (90o 3A) = sin (A
(90o 3A) = (A 42o)
4A = 132o
A = 33o
42o)
[
o
[ (90
cos Ө = sin (90o
3A) and (A
o
Ө)]
42 ) are both acute angles]
TRIGONOMETRIC RATIOS OF SOME COMPLEMENTARY ANGLES
4) If tan A = cot B, prove that A + B= 90o
SOLUTION:
tan A = cot B
tan A = tan (90o
A = (90o – B)
A + B = 90o
B)
[
[ A and (90o
tan (90o Ө)]
B) are both acute angles]
5) Express cos67o+sin80o in terms of trigonometric ratios of angles between 0o
and 45o.
SOLUTION:
cos67o + sin 80o
= cos (90o 23o) + sin (90o 10o)
= sin 23o + cos 10o
[ sin (90o Ө) = cos Ө and cos (90o Ө) = sin Ө]
6) Evaluate:
SOLUTION:
=
=
=
=1
TRIGONOMETRIC RATIOS OF SOME COMPLEMENTARY ANGLES
7) Show that cos 23o cos 67o
sin 23o sin 67o=0
SOLUTION:
LHS= cos 23o cos 67o sin 23o sin 67o
= cos (90o 67o) cos (90o 23o) sin 23o sin 67o
= sin 23o sin 67o sin 23o sin 67o
=0
= RHS
LHS = RHS
8) If sin 2A = cos (A
42o), where 2A is an acute angle, find the value of A.
SOLUTION:
sin 2A= cos (A - 42o)
cos (90o 2A)= cos (A
(90o 2A) = (A 42o)
3A= 132o
A = 44o
42o)
[ cos (90o
o
[ (90
2A) and (A
Ө) = sin Ө]
42o) are both acute angles]
9) If sec A = cosec B, prove that A + B = 90o
SOLUTION:
Sec A = cosec B
Sec A = sec (90o
A = (90o – B)
A + B = 90o
B)
[ sec (90o Ө) = cosec Ө]
[ A and (90o – B) are both acute angles]
TRIGONOMETRIC RATIOS OF SOME COMPLEMENTARY ANGLES
10) Express sin 53o + cos 86o in terms of trigonometric ratios of angles between
0o and 45o.
SOLUTION:
sin 53o + cos 86o
= sin (90o 37o) + cos (90o 4o)
= cos 37o + sin 4o
[ sin (90o Ө) = cos Ө and cos (90o Ө) = sin Ө]