225k views
5 votes
Help please need answer

Help please need answer-example-1

1 Answer

1 vote

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.

User Eron Wright
by
8.3k points

No related questions found

Welcome to QAmmunity.org, where you can ask questions and receive answers from other members of our community.