Question

if sinA/3=sinB/4=1/5 and A,B are angles in the second quadrant then prove that 4cosA+3cosB=-5​

Answer 1

Answer:

$4\:cosA+3\:cosB=5$

Step-by-step explanation:

Given:

$\frac{sinA}{3}=\frac{sinB}{4}=\frac{1}{5}$

$\implies\:\frac{sinA}{3}=\frac{1}{5}$

$\implies\:sinA=\frac{3}{5}$

similarly,

$\implies\:sinB=\frac{4}{5}$

Now,

$cos^2A=1-sin^2A$

$cos^2A=1-(\frac{3}{5})^2$

$\implies\:cos^2A=1-\frac{9}{25}$

$\implies\:cos^2A=\frac{25-9}{25}$

$\implies\:cos^2A=\frac{16}{25}$

$\implies\:cosA=\pm\frac{4}{5}$

since A lies in second quadrant,

$cosA=\frac{-4}{5}$

$cos^2B=1-sin^2B$

$cos^2B=1-(\frac{4}{5})^2$

$\implies\:cos^2B=1-\frac{16}{25}$

$\implies\:cos^2B=\frac{25-16}{25}$

$\implies\:cos^2B=\frac{9}{25}$

$\implies\:cosB=\pm\frac{3}{5}$

since B lies in second quadrant,

$cosA=\frac{-3}{5}$

$4\:cosA+3\:cosB$

$=4(\frac{-4}{5})+3(\frac{-3}{5})$

$=\frac{-16}{5}-\frac{9}{5}$

$=\frac{-16-9}{5}$

$=\frac{-25}{5}$

$=-5$

$\implies\:4\:cosA+3\:cosB=5$

Answer 2

Step-by-step explanation:

Answer:

4\:cosA+3\:cosB=54cosA+3cosB=5

Step-by-step explanation:

Given:

\frac{sinA}{3}=\frac{sinB}{4}=\frac{1}{5}

3

sinA

=

4

sinB

=

5

1

\implies\:\frac{sinA}{3}=\frac{1}{5}⟹

3

sinA

=

5

1

\implies\:sinA=\frac{3}{5}⟹sinA=

5

3

similarly,

\implies\:sinB=\frac{4}{5}⟹sinB=

5

4

Now,

cos^2A=1-sin^2Acos

2

A=1−sin

2

A

cos^2A=1-(\frac{3}{5})^2cos

2

A=1−(

5

3

)

2

\implies\:cos^2A=1-\frac{9}{25}⟹cos

2

A=1−

25

9

\implies\:cos^2A=\frac{25-9}{25}⟹cos

2

A=

25

25−9

\implies\:cos^2A=\frac{16}{25}⟹cos

2

A=

25

16

\implies\:cosA=\pm\frac{4}{5}⟹cosA=±

5

4

since A lies in second quadrant,

cosA=\frac{-4}{5}cosA=

5

−4

cos^2B=1-sin^2Bcos

2

B=1−sin

2

B

cos^2B=1-(\frac{4}{5})^2cos

2

B=1−(

5

4

)

2

\implies\:cos^2B=1-\frac{16}{25}⟹cos

2

B=1−

25

16

\implies\:cos^2B=\frac{25-16}{25}⟹cos

2

B=

25

25−16

\implies\:cos^2B=\frac{9}{25}⟹cos

2

B=

25

9

\implies\:cosB=\pm\frac{3}{5}⟹cosB=±

5

3

since B lies in second quadrant,

cosA=\frac{-3}{5}cosA=

5

−3

4\:cosA+3\:cosB4cosA+3cosB

=4(\frac{-4}{5})+3(\frac{-3}{5})=4(

5

−4

)+3(

5

−3

)

=\frac{-16}{5}-\frac{9}{5}=

5

−16

−

5

9

=\frac{-16-9}{5}=

5

−16−9

=\frac{-25}{5}=

5

−25

=-5=−5

\implies\:4\:cosA+3\:cosB=5⟹4cosA+3cosB=5