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# Use function notation to describe the way the second variable (DV) depends upon the first variable (IV). Determine the domain and range for each, determine if there is a positive, negative, or no relationship, and explain your answers. A)IV: an acute angle V in a right triangle: DV: the area B of the triangle if the hypotenuse is a fixed length G. B)IV: one leg P of a right triangle: DV: the hypotenuse G of the right triangle if the other leg is 2 C)IV: the hypotenuse G of a right triangle: DV: the other leg P of the right triangle is one leg is 5. # Use function notation to describe the way the second variable (DV) depends upon the first variable (IV). Determine the domain and range for each, determine if there is a positive, negative, or no relationship, and explain your answers. A)IV: an acute angle V in a right triangle: DV: the area B of the triangle if the hypotenuse is a fixed length G. B)IV: one leg P of a right triangle: DV: the hypotenuse G of the right triangle if the other leg is 2 C)IV: the hypotenuse G of a right triangle: DV: the other leg P of the right triangle is one leg is 5.

Question
Right triangles and trigonometry asked 2020-10-18
Use function notation to describe the way the second variable (DV) depends upon the first variable (IV). Determine the domain and range for each, determine if there is a positive, negative, or no relationship, and explain your answers.
A)IV: an acute angle V in a right triangle: DV: the area B of the triangle if the hypotenuse is a fixed length G.
B)IV: one leg P of a right triangle: DV: the hypotenuse G of the right triangle if the other leg is 2
C)IV: the hypotenuse G of a right triangle: DV: the other leg P of the right triangle is one leg is 5.

## Answers (1) 2020-10-19
A) In this part, we are given an independent variable – an acute angle V in a right triangle, dependent variable - area B of the triangle and a constant – hypotenuse G has a fixed length.
We can first express the two legs of the right triangle in terms of hypotenuse G and acute angle V as shown below:
Height $$\displaystyle={G}{\sin{{\left({V}\right)}}}$$
Base $$\displaystyle={G}{\cos{{\left({V}\right)}}}$$
Use the formula for area of triangle as shown below:
Area $$\displaystyle=\frac{{1}}{{2}}$$ *Base*Height
Area $$\displaystyle=\frac{{1}}{{2}}\cdot{\left({G}{\cos{{\left({V}\right)}}}\right)}\cdot{\left({G}{\sin{{\left({V}\right)}}}\right)}$$
Area $$\displaystyle=\frac{{1}}{{2}}{G}^{{2}}{\sin{{\left({V}\right)}}}{\cos{{\left({V}\right)}}}$$
Area $$\displaystyle=\frac{{1}}{{4}}{G}^{{2}}{\sin{{\left({2}{V}\right)}}}$$
Domain of this function is $$\displaystyle{\left({0},\frac{\pi}{{2}}\right)}$$ and range is $$\displaystyle{\left({0},\frac{{G}^{{2}}}{{4}}\right)}$$.
Since area function is oscillating function, therefore, the relation between two variables is neither positive and nor negative.
B) In this part, we are given an independent variable – a leg P of a right triangle, dependent variable - the hypotenuse G of the right triangle, and constant – second leg of length 2.
We can use Pythagorean theorem to express the relationship between P, G and 2 as shown below:
$$\displaystyle{P}^{{2}}+{2}^{{2}}$$
$$\displaystyle{G}^{{2}}={P}^{{2}}+{4}$$
$$\displaystyle{G}=\sqrt{{{P}^{{2}}+{4}}}$$
In reference to the given question, since P represents a leg of a right triangle, it can take any real number greater than 0. Therefore, domain of this function is $$\displaystyle{\left({0},\infty\right)}$$.
Range of this function is all real numbers greater than 2, that is, $$\displaystyle{\left({2},\infty\right)}$$.
Since value of hypotenuse increases as the length of leg increases, therefore, there is a positive relationship between the two variables.
C) In this part, we are given an independent variable – the hypotenuse G of a right triangle, dependent variable - the leg P of the right triangle, and constant – second leg of length 5.
We can use Pythagorean theorem to express the relationship between P, G and 2 as shown below:
$$\displaystyle{P}^{{2}}+{5}^{{2}}={G}^{{2}}$$
$$\displaystyle{P}^{{2}}={G}^{{2}}-{25}$$
$$\displaystyle{P}=\sqrt{{{G}^{{2}}-{25}}}$$
G can take any real number greater than 5 in order for this function to exist. Therefore, domain of this function is $$\displaystyle{\left({5},\infty\right)}$$.
Range of this function is all positive real numbers, that is, $$\displaystyle{\left({0},\infty\right)}$$.
Since value of leg increases as the length of hypotenuse increases, therefore, there is a positive relationship between the two variables.

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