Question

Help please need answer

Answer 1

17Given:

The smaller triangle is,

Find the length of the hypotenuse,

`\begin{gathered} \text{hypotenuse}^2=4^2+1^2 \\ =16+1 \\ =17 \\ \text{hypotenuse}=\sqrt[]{17} \end{gathered}`

The larger traingle is,

`\begin{gathered} \text{Hypotenuse}^2=12^2+3^2 \\ =144+9 \\ =153 \\ \text{Hypotenuse}=3\sqrt[]{17} \end{gathered}`

Now, find all the trigonometric functions for a smaller triangle.

`\begin{gathered} \sin \theta=\frac{opposite\text{ side}}{\text{hypotenuse}}=\frac{1}{\sqrt[]{17}} \\ \cos \theta=\frac{Adjacent\text{ side}}{\text{hypotenuse}}=\frac{4}{\sqrt[]{17}} \\ \tan \theta=\frac{opposite\text{ side}}{\text{adjacent side}}=(1)/(4) \\ co\sec \theta=\frac{\text{hypotenuse}}{opposite\text{ side}}=\frac{\sqrt[]{17}}{1}=\sqrt[]{17} \\ \sec \theta=\frac{\text{hypotenuse}}{adjacent\text{ side}}=\frac{\sqrt[]{17}}{4} \\ \cot \theta=\frac{adjecent\text{ side}}{\text{opposite side}}=(4)/(1)=4 \end{gathered}`

For the larger triangle,

`\begin{gathered} \sin \theta=\frac{opposite\text{ side}}{\text{hypotenuse}}=\frac{3}{3\sqrt[]{17}}=\frac{1}{\sqrt[]{17}} \\ \cos \theta=\frac{Adjacent\text{ side}}{\text{hypotenuse}}=\frac{12}{3\sqrt[]{17}}=\frac{4}{\sqrt[]{17}} \\ \tan \theta=\frac{opposite\text{ side}}{\text{adjacent side}}=(3)/(12)=(1)/(4) \\ co\sec \theta=\frac{\text{hypotenuse}}{opposite\text{ side}}=\frac{3\sqrt[]{17}}{3}=\sqrt[]{17} \\ \sec \theta=\frac{\text{hypotenuse}}{adjacent\text{ side}}=\frac{3\sqrt[]{17}}{12}=\frac{\sqrt[]{17}}{4} \\ \cot \theta=\frac{adjecent\text{ side}}{\text{opposite side}}=(12)/(3)=4 \end{gathered}`

Therefore, the value of the function is the same because the triangles are similar so, corresponding sides are proportional.