Question

2 bryce wants to replace the net on his basketball hoop. the hoop is 9 feet high. bryce places his ladder 3 feet from the base of the hoop. how long must the ladder be to reach the hoop? 3. solve for x. 4. solve for x. 5. find the length of segment ou and write the trigonometric ratios for the given angles. reduce your fractions.

Explanation:

Step1: Solve problem 2

Use Pythagorean theorem \(a^{2}+b^{2}=c^{2}\), where \(a = 9\) (height of hoop), \(b = 3\) (distance of ladder from base).
\[c=\sqrt{9^{2}+3^{2}}=\sqrt{81 + 9}=\sqrt{90}=3\sqrt{10}\approx9.49\]

Step2: Solve problem 3

Use geometric - mean theorem in right - triangle. If the altitude \(h\) to the hypotenuse of a right - triangle divides the hypotenuse into segments \(m\) and \(n\), then \(h^{2}=m\times n\). Here \(x^{2}=7\times10\), so \(x=\sqrt{70}\approx8.37\)

Step3: Solve problem 4

Using the geometric - mean theorem, if the altitude \(x\) to the hypotenuse of a right - triangle divides the hypotenuse into segments \(m\) and \(n\), then \(x^{2}=m\times n\). Here \(m = 9\), \(n=25 - 9=16\), so \(x=\sqrt{9\times16}=12\)

Step4: Solve problem 5

First, find \(OU\) using Pythagorean theorem. \(OU=\sqrt{34^{2}-16^{2}}=\sqrt{(34 + 16)(34 - 16)}=\sqrt{50\times18}=\sqrt{900}=30\)
\(\sin S=\frac{OU}{SU}=\frac{30}{34}=\frac{15}{17}\), \(\sin U=\frac{OS}{SU}=\frac{16}{34}=\frac{8}{17}\)
\(\cos S=\frac{OS}{SU}=\frac{16}{34}=\frac{8}{17}\), \(\cos U=\frac{OU}{SU}=\frac{30}{34}=\frac{15}{17}\)
\(\tan S=\frac{OU}{OS}=\frac{30}{16}=\frac{15}{8}\), \(\tan U=\frac{OS}{OU}=\frac{16}{30}=\frac{8}{15}\)

Answer:

Problem 2: \(3\sqrt{10}\approx9.49\)
Problem 3: \(\sqrt{70}\approx8.37\)
Problem 4: \(12\)
Problem 5: \(OU = 30\), \(\sin S=\frac{15}{17}\), \(\sin U=\frac{8}{17}\), \(\cos S=\frac{8}{17}\), \(\cos U=\frac{15}{17}\), \(\tan S=\frac{15}{8}\), \(\tan U=\frac{8}{15}\)