7/18/2019 Trigonometrical Formulae

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Hyp.

r

·

(iii) sec x = = -

Base

p

Base b

iv)

cos y = =-

Hyp.

n

t

Base b

v)

.co y = -

Perp. m

Hyp. n

vi) cosec y = -

Perp.

m

vii)

. Perp.

u

Stn Z =

= -

Hyp. n

Base

k

viii)

cos

z

=

= -

Hyp. n

Perp. u

tan z = =-

A

s

Base k

0

•

xi)

Q. 2.

In the given figure, LB

90°,

AB

= 4

units and BC = 3 units. Find :

A

i) sin A (ii) cos A

i i i) cot A

v) sec C

iv)

sin C

vi) tan C

4 units

c

B

Sol.

In

MBC,

LB

=

90°

AB

=

4

units and BC =

3

units

But AC

2

=

AB

2

+ BC2

•

. .

Now

Pythagoras Theorem)

=

4)2 + 3)2

= 16

+9

=

25

= 5)

2

AC= 5 units

i) sin A

=

Perp.

=

BC

= .

Hyp. AC 5

(ii) cos A = Base = AB = i

Hyp. AC 5

cot A = Base = AB = i

Perp. BC

3

iii)

sin C = Perp.

=

AB

=

i

Hyp. AC 5

iv)

v)

sec C =

Hyp.

= AC =

Base BC 3

Perp. AB 4

vi) tan C = = = - Ans

Base BC 3 ·

Q. 3.

From the given figure, write down the

values

of:

i)

sin B (ii) tan B

(iii) cos c

iv)

cot c

v)

sin B cos C

+cos

B sin

C)

vi) sec

2

C - tan

2

C)

A

L ~ ~ ~ ~ ~ ~ ~ ~ ~ ~ ~ ~ ~ c

17 units

Sol. In MBC,

LA

90°

BC = 17 units, AB = 15 units

But BC

2

= AB

2

+

AC2

Pythagoras Theorem)

>

17)

2

=

15)

2

+ AC

2

>

289 = 225

AC

2

> AC

2

= 289

- 225

=

64

=

8)

2

,

:. AC= 8

units

r·)

sin B = Perp. =AC =

_. _

1

Hyp. BC 17

ii) tan B = Perp. = AC =

_ _

Base AB

15

(iii) cos

C

=

Base =

AC

= _ _

Hyp.

BC

17