Question

determine each segment length in right triangle abc. 4 4\sqrt{2} 2\sqrt{2} 6 4\sqrt{3} 2\sqrt{3} bc ad bd ab

Explanation:

Step1: Find \( BC \)

In right - triangle \( BDC \), \( \angle C = 60^{\circ}\), \( CD = 2 \).
Using the sine function \( \sin\theta=\frac{\text{opposite}}{\text{hypotenuse}}\), \(\sin60^{\circ}=\frac{BD}{BC}\), and using the cosine function \( \cos\theta=\frac{\text{adjacent}}{\text{hypotenuse}}\), \(\cos60^{\circ}=\frac{CD}{BC}\).
Since \( \cos60^{\circ}=\frac{1}{2}=\frac{CD}{BC}\), and \( CD = 2 \), then \( BC=\frac{CD}{\cos60^{\circ}}=\frac{2}{\frac{1}{2}} = 4 \).

Step2: Find \( AC \)

Given \( AD + CD=AC \), and \( CD = 2 \), \( AC=8 + 2=10 \).

Step3: Find \( AB \)

In right - triangle \( ABC \), \( \angle A = 30^{\circ}\), \( \angle C = 60^{\circ}\), \( \angle B = 90^{\circ}\). Using the Pythagorean theorem \( AB=\sqrt{AC^{2}-BC^{2}}\) or using the property of a \( 30 - 60-90 \) triangle (\( AC = 2BC \) is not correct here, better use Pythagorean). \( AB=\sqrt{10^{2}-4^{2}}=\sqrt{100 - 16}=\sqrt{84}\) (no, wrong. Wait, in \( \triangle ABC \), \( \angle A=30^{\circ}\), \( \angle B = 90^{\circ}\), \( \angle C = 60^{\circ}\), \( AC = 10 \), then \( BC=\frac{1}{2}AC\) (wrong, no. Wait, in a \( 30 - 60-90 \) triangle, the side opposite \( 30^{\circ}\) is half the hypotenuse. Here \( BC \) is opposite \( 30^{\circ}\) (since \( \angle A = 30^{\circ}\)), \( AC \) is hypotenuse. So \( BC=\frac{1}{2}AC\) (no, wait \( \angle A = 30^{\circ}\), opposite side \( BC \), hypotenuse \( AC \). So \( BC=\frac{1}{2}AC\) (wrong, \( AC = 10\), \( BC = 5\) no, wait no. Wait, we already found \( BC = 4\) from \( \triangle BDC\). Wait, \( \triangle BDC\): \( CD = 2\), \( \angle C = 60^{\circ}\), \( \angle BDC=90^{\circ}\), so \( BC = 4\) (since \( \cos60^{\circ}=\frac{CD}{BC}\)). Then in \( \triangle ABC\), \( AC=AD + DC=8 + 2=10\), \( BC = 4\), using Pythagorean \( AB=\sqrt{AC^{2}-BC^{2}}=\sqrt{10^{2}-4^{2}}=\sqrt{100 - 16}=\sqrt{84}\) (no, wrong. Wait, no, in \( \triangle ABC\), \( \angle A = 30^{\circ}\), \( \angle B=90^{\circ}\), \( \angle C = 60^{\circ}\), hypotenuse \( AC = 10\), then \( BC=\frac{1}{2}AC\) (no! Wait \( \sin A=\frac{BC}{AC}\), \( \sin30^{\circ}=\frac{BC}{AC}\), \( \frac{1}{2}=\frac{BC}{10}\), \( BC = 5\) (conflict). Wait, no, the error was in step1. Wait, \( \triangle BDC\): \( \angle BDC = 90^{\circ}\), \( \angle C=60^{\circ}\), \( CD = 2\). Then \( BC=\frac{CD}{\cos60^{\circ}}=\frac{2}{\frac{1}{2}} = 4\). Then in \( \triangle ABC\), \( AC=AD + DC=8 + 2=10\), \( BC = 4\), \( AB=\sqrt{AC^{2}-BC^{2}}=\sqrt{10^{2}-4^{2}}=\sqrt{84}\) (no, wrong. Wait, no, using trigonometry in \( \triangle ABC\): \( \tan A=\frac{BC}{AB}\), \( \tan30^{\circ}=\frac{4}{AB}\), \( AB = 4\sqrt{3}\).

Step4: Find \( BD \)

In right - triangle \( BDC \), \( \sin60^{\circ}=\frac{BD}{BC}\), since \( BC = 4\), \( BD = BC\sin60^{\circ}=4\times\frac{\sqrt{3}}{2}=2\sqrt{3}\).

Step5: Find \( AD \)

\( AD=AC - CD\), \( AC=10\), \( CD = 2\), \( AD=8\) (no, wait \( AC=AD + DC\), \( AD=AC - DC\). But \( AC\) in \( \triangle ABC\): Wait, no, \( AC\) is given as \( AD + DC\) where \( DC = 2\) and \( AD + DC\) (the length from \( A\) to \( C\) passing through \( D\)) is \( 8+2=10\). But in \( \triangle ABC\), \( AC = 10\) (hypotenuse). Wait, no, \( AD=AC - DC\), \( AC = 10\), \( DC = 2\), \( AD = 8\) (no, wrong. Wait, no, \( AD\) is from \( A\) to \( D\). Wait, in \( \triangle ABD\), \( AB = 4\sqrt{3}\), \( BD = 2\sqrt{3}\), using Pythagorean \( AD=\sqrt{AB^{2}+BD^{2}}=\sqrt{(4\sqrt{3})^{2}+(2\sqrt{3})^{2}}=\sqrt{48 + 12}=\sqrt{60}\) (no, wrong. Wait, no, \( \triangle ABD\): \( \angle ADB = 180^{\circ}-\angle BDC=90^{\circ}\) (since \( \angle BDC = 90^{\circ}\)), \( AB = 4\sqrt{3}\), \( BD = 2\sqrt{3}\). Using Pythagorean \( AD=\sqrt{AB^{2}-BD^{2}}=\sqrt{(4\sqrt{3})^{2}-(2\sqrt{3})^{2}}=\sqrt{48 - 12}=\sqrt{36}=6\).

Answer:

\( BC = 4\), \( AD = 6\), \( BD = 2\sqrt{3}\), \( AB = 4\sqrt{3}\)