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.

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

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

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

Answer; a)

$\text{SecA}=\frac{\sqrt[]{113}}{7},\cot B=\frac{8}{7}$