Answer variants:
23.1
\(\frac{AB}{AC}\)
\(\frac{AC}{AB}\)
46.2
The storm breaks a tree into two parts. The top of the tree touches the ground, making an angle of 30\(^{\circ}\) with the ground. The top of the tree touches the ground 23.1 \(m\) away from the foot of the tree. Find the actual height of the tree.
Answer:
Let \(A\) denote the foot of the tree. \(C\) is the point at which the tree is broken and \(B\) is the point where the top of the tree touches the ground.
From the given data, we have:
\(\angle CBA = 30^{\circ}\)
\(AB = 23.1 \ m\)
In the right \(\triangle CAB\), \(tan \ \theta =\)
After applying the known values then we get, \(AC=\)\(\times \frac{1}{\sqrt{3}}\)
Now, in \(\triangle CAB\), \(cos \ \theta =\)
After applying the known values then we get, \(BC=\) \(\times \frac{1}{\sqrt{3}}\)
The height of the tree \(= AC + CB\)
Therefore, the actual height of the tree is \(m\).