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(a) From the figure (1) given below, find the values of:

(i) sin B

(i) cos C

(iii) sin B + sin C

(iv) sin B cos C + sin C cos B

(b) From the figure (2) given below, find the values of……..

(1) sin B

(ii) cos C

Answer:

From the right-angled triangle ABC,

By Pythagoras theorem, we get

\begin{array}{l} \mathrm{BC}^{2}=\mathrm{AC}^{2}+\mathrm{AB}^{2} \\ \mathrm{AC}^{2}=\mathrm{BC}^{2}-\mathrm{AB}^{2} \\ \mathrm{AC}^{2}=(10)^{2}-(6)^{2} \\ \mathrm{AC}^{2}=100-36 \\ \mathrm{AC}^{2}=64 \\ \mathrm{AC}^{2}=8^{2} \\ \mathrm{AC}=8 \end{array}

(i) sin B = perpendicular/ hypotenuse

= AC/BC

= 8/10

= 4/5

(ii) cos C = Base/hypotenuse

= AC/BC

= 8/10

= 4/5

(iii) sin B = Perpendicular/hypotenuse

= AC/BC

= 8/10

= 4/5

Sin C = perpendicular/hypotenuse

= AB/BC

= 6/10

= 3/5

Now,

Sin B + sin C = (4/5) + (3/5)

= (4 + 3)/5

= 7/5

(iv) sin B = 4/5

Cos C = 4/5

Sin C = perpendicular/ hypotenuse

= AB/BC

= 6/10

= 3/5

Cos B = Base/Hypotenuse

= AB/BC

= 6/10

= 3/5

sin B cos C + sin C cos B

= (4/5) × (4/5) × (3/5) × (3/5)

= (26/25) × (9/25)

= (16+9)/25

= 25/25

= 1

From Figure

AC = 13, CD = 5, BC =21,

BD = BC – CD

= 21 – 5

= 16

From right angled ∆ACD,

By Pythagoras theorem we get

\begin{array}{l} A C=A D^{2}+C D^{2} \\ A D^{2}=A C^{2}-C D^{2} \end{array}

\begin{array}{l} A D^{2}=(13)^{2}-(5)^{2} \\ A D^{2}=169-25 \\ \left.A D^{2}=144\right) \\ A D^{2}=(12)^{2} \\ A D=12 \end{array}

From right angled ∆ABD,

By Pythagoras angled ∆ABD

By Pythagoras theorem we get

\begin{array}{l} A B^{2}=A D^{2}+B D^{2} \\ A B^{2}=400 \\ A B^{2}=(20)^{2} \\ A B=20 \end{array}

(i) tan x = perpendicular/Base (in right angled ∆ACD)

=CD/AD

= 5/12

(ii) cos y = Base/Hypotenuse (in right angled ∆ABD)

= BD/AB

= (20)/12 – (5/3)

Cot y = Base/Perpendicular (in right ∆ABD)

=BD/AB

= 16/20 = 4/5

(iii) ) cos y = Hypotenuse/ perpendicular (in right angled ∆ABD)

BD/AB

= 20/12

= 5/3

Cot y = Base/Perpendicular (in right ∆ABD)

AB/AD

= 16/12

= 4/3

\begin{array}{l} \operatorname{cosec}^{2} y-\cot ^{2} y=(5 / 3)^{2}-(4 / 3)^{2} \\ =(25 / 9)-(16 / 9) \\ =(25-16) / 9 \\ =9 / 9 \\ =1 \end{array}

\text { Hence, } \operatorname{cosec}^{2} y-\cot ^{2} y=1

(iv) sin x = Perpendicular/Hypotenuse (in right angled ∆ACD)

= AD/AB

= 12/20

= 3/5

Cot y = Base/Perpendicular (in right angled ∆ABD)

= BD/AD

= 16/12

= 4/3

(5/sin x) + (3/sin y) – 3cot y

= 5/(5/13) + 3/(3/5) – 3 × 4/3

= 5 × 13/5 + 3 × 5/3 – 3 × 4/3

= 1 × 13/1 + 1 × 5/1 – 1 × 4/1

= 13 + 5 – 4 = 18 – 4

= 14

Hence , **5/sin x + 3/sin y – 3cot y = 14**

"hey guys welcome to lido q a video
and we neet your lido tutor bringing you
this question on your
screen question is
from the figure given below find the
values of
sine b cos c sine b
plus sine c sine b cos c
plus sine c cos b right
now we have two angles here
so we need the opposite and hypotenuse
with respect to both the angles
opposite adjacent and hypotenuse so
let us write it down right this is angle
b
and this is angle c
so
with respect
to angle c
opposite is
a b right
adjacent is
ac and
hypotenuse
is bc with respect to both the angles
that is
now
we are the sine and cos so just let let
us just write it down
sign b is equal to sohcahtoa
remember sohcahtoa always
so sine is opposite by hypotenuse
so a b by
b c right
and sine c sorry
cos b is adjacent by hypotenuse
right so we have ac by
bc that's all for angle b
now with
respect
sorry this was angle c
so let me fix that real quick
so this is angle c now with respect to
angle
b
so this is the opposite ac
is the opposite
and a b is the
adjacent right and bc is the hypotenuse
therefore sine b will be equal to
opposite
by hypotenuse
and sine
okay cos b will be equal to
a b by bc
okay now let us find the values
now in this figure this is 6 this is 10
so by pythagoras theorem
ac square is equal to a b square
minus sorry
b c square minus a b
square this is equal to 10 square minus
6 square
this is equal to 64. therefore ac is
equal to
8. now we know all the sides
and we know this
this is one two
three and four
right so let us see
number one
sine b is equal to
sine b is equal to ac by bc
from 1 so this is equal to
8 by 10 or 4 by
5
again
cos c
is equal to so cos c
from 2 is equal to ac by bc
from 2 so this is equal to
4 by 5 again right
number 3
sine b plus sine c is equal to
ac by bc plus
sine c is equal to a b by bc
so this is equal to
6 by 10 plus 4 by 10
sorry 8 by 10 so this is equal to 14 by
10
is equal to 7 by 5.
and the last one
sine a cos c
plus sine c cos
b so sine a is
sorry sine b is 4 by 5
into 4 by 5 plus sine
c is
6 by 10 that is 3 by 5 into cos b
is also 3 by 5 so this is equal to
16 plus 9 by 25
this is equal to 1 isn't that easy guys
if you still have a doubt please leave a
comment below
do like the video and subscribe to our
channel right
now let us look at the part 2 don't go
away yet
okay part 2 says from the figure given
below
find the values of sine v and cos c
so sine b is this
so from the figure if we are taking
angle b
right so opposite is 80
opposite is 80
so let us do number one
sine b is equal to a d by
b d right
now this is 21
and this is 5 so 21 minus 5 will be 16
80 is 16
80 is 16 now let us uh sorry b d is 16
let us find e d right
by
pythagoras theorem
v d sorry a d square
is equal to a c square minus
d c square this implies
a d square is equal to 169 minus
25 okay
so 169 minus 25 is 144
so a d is equal to root over 144
that is equal to 2n
therefore sine b will be equal to
12 by
16 so this is equal to
3 by 4. now let us look at number 2
cos c
cos c this is angle c so opposite is 80
but cos is car so adjacent
adjacent is 5 by hypotenuse is
13. okay
yeah okay sign b is not uh
yeah sign b is not a d by b d
but a d by is a b sorry yeah
a d by a b
because hypotenuse so this is a d by a b
right so a d is
16. and 80 is 12
right so 16
16 square plus 12 square
is equal to 400 so this will be 20.
right so let me fix that real quick
so this is 20 so this will be equal to 3
by 5
right so i'll see you in our next video
until then bye guys keep practicing
keep flourishing
"

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