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 $\angle B=\angle BAC=\frac{180 - 120}{2}=30^{\circ}$.

Step2: Find angles in $\triangle ACD$

In right - triangle $ACD$, $\angle ACD = 60^{\circ}$ (because $\angle ACB = 120^{\circ}$), so $\angle CAD=30^{\circ}$.

Step3: Use the property of 30 - 60 - 90 triangle in $\triangle ACD$

In right - triangle $ACD$, if $CD = 4$ cm, and in a 30 - 60 - 90 triangle, the side opposite the $30^{\circ}$ angle is half of the hypotenuse. Let $AC = 2x$, then $CD=x$. Since $x = 4$ cm, $AC = 8$ cm.

Step4: Find $AD$ using Pythagorean theorem or 30 - 60 - 90 triangle properties

In right - triangle $ACD$, $AD=\sqrt{AC^{2}-CD^{2}}=\sqrt{8^{2}-4^{2}}=\sqrt{64 - 16}=\sqrt{48}=4\sqrt{3}$ cm.

Step5: Find the distance from $D$ to $AB$

In right - triangle $ADH$, $\angle B = 30^{\circ}$, and we want to find the length of $DH$. In right - triangle $ADH$, the side opposite the $30^{\circ}$ angle is half of the hypotenuse. Since $AD = 4\sqrt{3}$ cm, then $DH=\frac{1}{2}AD = 2\sqrt{3}$ cm.

Answer:

$2\sqrt{3}$