Question

If SinA =3/5 and CosB =12/13 Find the value of 1. Sin (A+B) 2. Cos (A-B)

Answer 1

Solution:-

Given:-

$\sf \to \: sinA = \dfrac{3}{5} \: \: and \: \: cosB = \dfrac{12}{13}$

To find :-

$\sf \to \: sin(A + B) \\ \sf \to \: cos(A + B)$

First of All we have to Find

$\sf \to \: sinB \: and \: \: cosA$

We know That

$\sf \to \: sinA = \dfrac{3}{5} = \dfrac{p}{h} \: \: and \: \: cosB = \dfrac{12}{13} = \dfrac{b}{h}$

Now Take

$\sf \to \: sinA = \dfrac{3}{5} = \dfrac{p}{h}$

We Have

$\sf \to \: p = 3,h = 5 \: \: and \: b \: = x$

Now using Pythagoras theorem

$\sf \to \: {h}^{2} = {p}^{2} + {b}^{2}$

$\sf \to \: {5}^{2} = {3}^{2} + {b}^{2}$

$\sf \to \: 25 = 9 + {b}^{2}$

$\sf \to \: {b}^{2} = 16$

$\sf \to \: b = 4$

So we get

$\sf \to \: p = 3,h = 5 \: \: and \: b \: = 4$

so

$\sf \to \: cos \: A = \dfrac{4}{5}$

Now find SinB

$\sf \to cosB = \dfrac{12}{13} = \dfrac{b}{h}$

$\sf \to \: p = x,h = 13 \: \: and \: b \: = 12$

Now using Pythagoras theorem

$\sf \to \: {h}^{2} = {p}^{2} + {b}^{2}$

$\sf \to \: {13}^{2} = {p}^{2} + {12}^{2}$

$\sf \to \: 169 = {p}^{2} + 144$

$\sf \to \: {p}^{2} = 25$

$\sf \to \: p = 5$

$\sf \to \: p = 5,h = 13 \: \: and \: b \: = 12$

Now we get

$\sf \to \: sinB = \dfrac{5}{13}$

We get

$\sf \to \: sin(A + B) = sinA cosB + cosA sinB\\ \sf \to \: cos(A + B) = cosA cosB − sinA sinB$

Put the value on

$\sf \to \: sin(A + B) = sinA cosB + cosA sinB$

$\sf \to \: sinA = \dfrac{3}{5} , cos \: A = \dfrac{4}{5} \: \: and \: \: cosB = \dfrac{12}{13} , sinB = \dfrac{5}{13}$

$\sf \to \: \dfrac{3}{5} \times \dfrac{12}{13} + \dfrac{4}{5} \times \dfrac{5}{13}$

$\sf \to \: \dfrac{36}{65} + \dfrac{20}{65}$

$\sf \to \: \dfrac{56}{65}$

Now take

$\sf \to \: cos(A + B) = cosA cosB − sinA sinB$

$\sf \to \: sinA = \dfrac{3}{5} , cos \: A = \dfrac{4}{5} \: \: and \: \: cosB = \dfrac{12}{13} , sinB = \dfrac{5}{13}$

$\sf \to \: \dfrac{4}{5} \times \dfrac{12}{13} + \dfrac{3}{5} \times \dfrac{5}{13}$

$\sf \to \: \dfrac{48}{65} + \dfrac{15}{65}$

$\sf \to \: \dfrac{63}{65}$

Answer 2

Answer:

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Step-by-step explanation:

Solution:-

Given:-

\sf \to \: sinA = \dfrac{3}{5} \: \: and \: \: cosB = \dfrac{12}{13}→sinA=

5

3

andcosB=

13

12

To find :-

\begin{gathered} \sf \to \: sin(A + B) \\ \sf \to \: cos(A + B)\end{gathered}

→sin(A+B)

→cos(A+B)

First of All we have to Find

\sf \to \: sinB \: and \: \: cosA→sinBandcosA

We know That

\sf \to \: sinA = \dfrac{3}{5} = \dfrac{p}{h} \: \: and \: \: cosB = \dfrac{12}{13} = \dfrac{b}{h}→sinA=

5

3

=

h

p

andcosB=

13

12

=

h

b

Now Take

\sf \to \: sinA = \dfrac{3}{5} = \dfrac{p}{h}→sinA=

5

3

=

h

p

We Have

\sf \to \: p = 3,h = 5 \: \: and \: b \: = x→p=3,h=5andb=x

Now using Pythagoras theorem

\sf \to \: {h}^{2} = {p}^{2} + {b}^{2}→h

2

=p

2

+b

2

\sf \to \: {5}^{2} = {3}^{2} + {b}^{2}→5

2

=3

2

+b

2

\sf \to \: 25 = 9 + {b}^{2}→25=9+b

2

\sf \to \: {b}^{2} = 16→b

2

=16

\sf \to \: b = 4→b=4

So we get

\sf \to \: p = 3,h = 5 \: \: and \: b \: = 4→p=3,h=5andb=4

so

\sf \to \: cos \: A = \dfrac{4}{5}→cosA=

5

4

Now find SinB

\sf \to cosB = \dfrac{12}{13} = \dfrac{b}{h}→cosB=

13

12

=

h

b

\sf \to \: p = x,h = 13 \: \: and \: b \: = 12→p=x,h=13andb=12

Now using Pythagoras theorem

\sf \to \: {h}^{2} = {p}^{2} + {b}^{2}→h

2

=p

2

+b

2

\sf \to \: {13}^{2} = {p}^{2} + {12}^{2}→13

2

=p

2

+12

2

\sf \to \: 169 = {p}^{2} + 144→169=p

2

+144

\sf \to \: {p}^{2} = 25→p

2

=25

\sf \to \: p = 5→p=5

\sf \to \: p = 5,h = 13 \: \: and \: b \: = 12→p=5,h=13andb=12

Now we get

\sf \to \: sinB = \dfrac{5}{13}→sinB=

13

5

We get

\begin{gathered} \sf \to \: sin(A + B) = sinA cosB + cosA sinB\\ \sf \to \: cos(A + B) = cosA cosB − sinA sinB \end{gathered}

→sin(A+B)=sinAcosB+cosAsinB

→cos(A+B)=cosAcosB−sinAsinB

Put the value on

\sf \to \: sin(A + B) = sinA cosB + cosA sinB→sin(A+B)=sinAcosB+cosAsinB

\sf \to \: sinA = \dfrac{3}{5} , cos \: A = \dfrac{4}{5} \: \: and \: \: cosB = \dfrac{12}{13} , sinB = \dfrac{5}{13}→sinA=

5

3

,cosA=

5

4

andcosB=

13

12

,sinB=

13

5

\sf \to \: \dfrac{3}{5} \times \dfrac{12}{13} + \dfrac{4}{5} \times \dfrac{5}{13}→

5

3

×

13

12

+

5

4

×

13

5

\sf \to \: \dfrac{36}{65} + \dfrac{20}{65}→

65

36

+

65

20

\sf \to \: \dfrac{56}{65}→

65

56

Now take

\sf \to \: cos(A + B) = cosA cosB − sinA sinB →cos(A+B)=cosAcosB−sinAsinB

\sf \to \: sinA = \dfrac{3}{5} , cos \: A = \dfrac{4}{5} \: \: and \: \: cosB = \dfrac{12}{13} , sinB = \dfrac{5}{13}→sinA=

5

3

,cosA=

5

4

andcosB=

13

12

,sinB=

13

5

\sf \to \: \dfrac{4}{5} \times \dfrac{12}{13} + \dfrac{3}{5} \times \dfrac{5}{13}→

5

4

×

13

12

+

5

3

×

13

5

\sf \to \: \dfrac{48}{65} + \dfrac{15}{65}→

65

48

+

65

15

\sf \to \: \dfrac{63}{65}→

65

63