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

Answer:

(1)$csc A=\dfrac{17}{8}$

(2)$sec A=\dfrac{17}{15}$

Step-by-step explanation:

$Tan A=\dfrac{8}{15}\\ In Trigonometry, Tan A=\dfrac{Opposite}{Adjacent}\\ Therefore:\\Opposite=8, Adjacent=15$

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:

$csc A=\dfrac{Hypotenuse}{Opposite}\\csc A=\dfrac{17}{8}$

Question 2

Now, secant = 1/cos

Therefore:

$sec A=\dfrac{Hypotenuse}{Adjacent}\\sec A=\dfrac{17}{15}$

Answer 2

Answer:

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