Question

15cota=8 cota=8/15
(a) seca = 17/15
​

Answer 1

Answer:

Step-by-step explanation:

Answer and explanation:

Given : 15\cot A=8

To find : The other trigonometric ratios of angle A ?

Solution :

We know that,

\cot A=\frac{B}{P}

Where, B is the base and P is the perpendicular

\cot A=\frac{8}{15}

Now, We find hypotenuse

H^2=P^2+B^2

H^2=(15)^2+(8)^2

H^2=225+64

H=\sqrt{289}

H=17

So, H=17 , B=8 , P=15

$\sin A=\frac{P}{H}=\frac{15}{17}$

$\cos A=\frac{B}{H}=\frac{8}{17}$

$\csc A=\frac{H}{P}=\frac{17}{15}$

$\sec A=\frac{H}{B}=\frac{17}{8}$

$\tan A=\frac{P}{B}=\frac{15}{8}$

Answer 2

Answer:

Answer and explanation:

Given : 15\cot A=815cotA=8

To find : The other trigonometric ratios of angle A ?

Solution :

We know that,

\cot A=\frac{B}{P}cotA=

P

B

Where, B is the base and P is the perpendicular

\cot A=\frac{8}{15}cotA=

15

8

Now, We find hypotenuse

H^2=P^2+B^2H

2

=P

2

+B

2

H^2=(15)^2+(8)^2H

2

=(15)

2

+(8)

2

H^2=225+64H

2

=225+64

H=\sqrt{289}H=

289

H=17H=17

So, H=17 , B=8 , P=15

\sin A=\frac{P}{H}=\frac{15}{17}sinA=

H

P

=

17

15

\cos A=\frac{B}{H}=\frac{8}{17}cosA=

H

B

=

17

8

\csc A=\frac{H}{P}=\frac{17}{15}cscA=

P

H

=

15

17

\sec A=\frac{H}{B}=\frac{17}{8}secA=

B

H

=

8

17

\tan A=\frac{P}{B}=\frac{15}{8}tanA=

B

P

=

8

15