Question

in the isosceles △abc, m∠acb = 120° and ad is an altitude to leg bc. what is the distance from d to base ab if cd = 4 cm? answer: 2√3 cm

Explanation:

Step1: Find angles in $\triangle ABC$

Since $\triangle ABC$ is isosceles with $\angle ACB = 120^{\circ}$, then $\angle CAB=\angle CBA = 30^{\circ}$. In right - triangle $ADC$, $\angle ACD = 60^{\circ}$, $\angle CAD = 30^{\circ}$.

Step2: Find length of $AC$ in $\triangle ADC$

In right - triangle $ADC$, given $CD = 4$ cm and $\angle CAD=30^{\circ}$. We know that in a $30 - 60-90$ triangle, if the side opposite the $30^{\circ}$ angle is $x$, the hypotenuse is $2x$. So $AC = 8$ cm.

Step3: Find length of $AD$ in $\triangle ADC$

Using the Pythagorean theorem in $\triangle ADC$, $AD=\sqrt{AC^{2}-CD^{2}}$. Substituting $AC = 8$ cm and $CD = 4$ cm, we get $AD=\sqrt{64 - 16}=\sqrt{48}=4\sqrt{3}$ cm.

Step4: Find the distance from $D$ to $AB$

Let the distance from $D$ to $AB$ be $DH$. In right - triangle $ADH$, $\angle DAH = 30^{\circ}$. In a $30 - 60-90$ triangle, if the hypotenuse is $AD = 4\sqrt{3}$ cm, the side opposite the $30^{\circ}$ angle $DH=\frac{1}{2}AD$. So $DH = 2\sqrt{3}$ cm.

Answer:

$2\sqrt{3}$ cm