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 \(AB\) in terms of \(h\)

In right - triangle \(BCD\), \(\tan B=\frac{h}{BD}\), so \(BD = \frac{h}{\tan59^{\circ}}\).
In right - triangle \(ACD\), \(\tan A=\frac{h}{AD}\), and \(AD=AB + BD\). Since \(AD = 339\), we have \(339=\frac{h}{\tan15^{\circ}}-\frac{h}{\tan59^{\circ}}\).
Factor out \(h\): \(h=\frac{339}{\frac{1}{\tan15^{\circ}}-\frac{1}{\tan59^{\circ}}}\).

Step2: Calculate the values of \(\tan15^{\circ}\) and \(\tan59^{\circ}\)

We know that \(\tan15^{\circ}=2 - \sqrt{3}\approx0.268\), \(\tan59^{\circ}\approx1.664\).
Then \(\frac{1}{\tan15^{\circ}}\approx3.731\), \(\frac{1}{\tan59^{\circ}}\approx0.601\).

Step3: Calculate \(h\)

\(h=\frac{339}{3.731 - 0.601}=\frac{339}{3.13}\approx108.3\)

Answer:

\(108.3\)