Question

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

Explanation:

Step1: Find angles in $\triangle ABC$

Since $\triangle ABC$ is isosceles with $\angle ACB = 120^{\circ}$, then $\angle CAB=\angle CBA = 30^{\circ}$.

Step2: Analyze right - triangle $ACD$

In right - triangle $ACD$, $\angle ACD = 60^{\circ}$ and $\angle CAD = 30^{\circ}$. Given $CD = 4$ cm, by the property of a 30 - 60 - 90 triangle ($AD=\sqrt{3}CD$), so $AD = 4\sqrt{3}$ cm.

Step3: Analyze right - triangle $ADB$

In right - triangle $ADB$, $\angle DAB = 30^{\circ}$, and we want to find the length of $DH$ (the distance from $D$ to $AB$). Let $DH = x$. In right - triangle $ADH$, $\sin\angle DAH=\frac{DH}{AD}$. Since $\angle DAH = 30^{\circ}$ and $AD = 4\sqrt{3}$ cm, and $\sin30^{\circ}=\frac{1}{2}$, we have $\frac{x}{4\sqrt{3}}=\frac{1}{2}$.

Step4: Solve for $x$

Cross - multiply: $x = 2\sqrt{3}$ cm.

Answer:

$2\sqrt{3}$