Question

Find Sec A and Cot B exactly if a=8 and b=7

Answer 1

The given triangle is right angle triangle with side a = 8 and b 7

Apply pythagoras theorem for the side c;

In a right-angled triangle, the square of the hypotenuse side is equal to the sum of squares of the other two sides.

Hypotenuse² = Perpendicular² + Base²

Here, perpendicular, a =8 and Base b = 7

`\begin{gathered} c^(2)=a^(2)+b^(2) \\ c^(2)=8^(2)+7^(2) \\ c^2=113 \\ c=\sqrt[]{113} \end{gathered}`

The trignometric ratio for sec of an angle define as the ratio of the hypotenuse to the side adjacent to a given angle in a right triangle.

`\begin{gathered} \sec A=\frac{Hypotenuse}{Adjacent\text{ side of angle A}} \\ \sec A=(c)/(b) \\ \sec A=\frac{\sqrt[]{113}}{7} \end{gathered}`

The trignometric ratio for cosine of angle define as the ratio of the adjacent side to the the opposite side of the angle,

`\begin{gathered} CotB=\frac{\text{Adjacent side to angle B}}{\text{Opposite side to angle B}} \\ CotB=(a)/(b) \\ CotB=(8)/(7) \end{gathered}`

Answer; a)

`\text{SecA}=\frac{\sqrt[]{113}}{7},\cot B=(8)/(7)`