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 *

(1)

(2)

Step-by-step explanation:

Using Pythagoras theorem

Question 1

Now, cosecant = 1/sin

Therefore:

Question 2

Now, secant = 1/cos

Therefore:

Step-by-step 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

x is greater than 5

step-by-step explanation:

2x+7 is greater than 17

    -7                           -7

2x is greater than 10

divide by 2 each side

x is greater than 5