Question

1.In a right triangle, <A and <B are acute. Find the value of csc A, when tan A = 8/15.

2.In a right triangle, <A and <B are acute. Find the value of sec A, when tan A = 8/15 *​

Answer 1

Explanation:

1)The tangent trigonometric ratio is expressed as

Tan # = opposite side/adjacent side

Where

# represents the reference angle

From the information given,

Tan A = 8/15

Opposite side = 8

Adjacent side = 15

We would determine the hypotenuse,h by applying Pythagoras theorem which is expressed as

Hypotenuse² = opposite side² + adjacent side²

h² = 8² + 15² = 289

h = √289 = 17

Sin A = opposite side/hypotenuse

Sin A = 8/17

Csc A = 1/SinA

Csc A = 17/8

2) Cos A = adjacent side/hypotenuse

Cos A = 15/17

Sec A = 1/Cos A

Sec A = 17/15

Answer 2

(1)
`csc A=(17)/(8)`

(2)
`sec A=(17)/(15)`

Explanation:

`Tan A=(8)/(15)\\$In Trigonometry, $Tan A=(Opposite)/(Adjacent)\\$Therefore:\\Opposite=8, Adjacent=15`

Using Pythagoras theorem

`Hypotenuse^2=Opposite^2+Adjacent^2\\Hypotenuse^2=8^2+15^2\\Hypotenuse^2=289\\Hypotenuse^2=17^2\\Hypotenuse=17`

Question 1

Now, cosecant = 1/sin

Therefore:

`csc A=(Hypotenuse)/(Opposite)\\csc A=(17)/(8)`

Question 2

Now, secant = 1/cos

Therefore:

`sec A=(Hypotenuse)/(Adjacent)\\sec A=(17)/(15)`