Question

find h to the nearest tenth.
the length of h is approximately
(round the final answer to the nearest tenth as needed. round all intermediate values to the nearest tenth as needed.)

Explanation:

Step1: Express \(AB\) and \(BD\) in terms of \(h\)

In right - triangle \(ACD\), \(\tan39^{\circ}=\frac{h}{AD}\), so \(AD = \frac{h}{\tan39^{\circ}}\). In right - triangle \(BCD\), \(\tan67^{\circ}=\frac{h}{BD}\), so \(BD=\frac{h}{\tan67^{\circ}}\). And we know that \(AD - BD=343\).

Step2: Substitute \(AD\) and \(BD\) into the equation \(AD - BD = 343\)

Substitute \(AD=\frac{h}{\tan39^{\circ}}\) and \(BD = \frac{h}{\tan67^{\circ}}\) into \(AD - BD=343\), we get \(h(\frac{1}{\tan39^{\circ}}-\frac{1}{\tan67^{\circ}})=343\).
Since \(\tan39^{\circ}\approx0.80\) and \(\tan67^{\circ}\approx2.36\), then \(\frac{1}{\tan39^{\circ}}-\frac{1}{\tan67^{\circ}}=\frac{1}{0.80}-\frac{1}{2.36}\approx1.25 - 0.42 = 0.83\).

Step3: Solve for \(h\)

From \(h(\frac{1}{\tan39^{\circ}}-\frac{1}{\tan67^{\circ}})=343\), we have \(h=\frac{343}{\frac{1}{\tan39^{\circ}}-\frac{1}{\tan67^{\circ}}}\). Substitute \(\frac{1}{\tan39^{\circ}}-\frac{1}{\tan67^{\circ}}\approx0.83\) into the formula, then \(h=\frac{343}{0.83}\approx413.3\).

Answer:

\(413.3\)