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=4cm?

1. consider right triangle adc. in this triangle cd=4 cm and m∠acd=180°-m∠acb=180°-120°=60°. then m∠cad=180°-m∠adc-m∠acd=180°-90°-60°=30°. the leg that is opposite to the angle with measure 30° is half of the hypotenuse, then ac=8 cm.

2. consider isosceles triangle abc. ab is its base, the angles adjacent to the base of isosceles triangle are congruent, then

also ac=cb=8 cm.

3. consider right triangle adb. draw the altitude dh to the hypotenuse ab. this altitutde divides right triangle adb into two right triangles adh and bdh.

4. consider right triangle dbh. in this triangle db is hypotenuse and db=cd+cb=4+8=12 cm. the leg dh is opposite to the angle b with measure 30°, then dh is half of the hypotenuse db. thus, dh=12: 2=6 cm.

answer: 6 cm

its 0.1 miles per minute

step-by-step explanation:

just plug in each number for x and see which ones work and don't