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?

If ΔACB is an isosceles triangle, then ∠A ≅ ∠B and AC ≅ CB

Since ∠C = 120° and ∠A + ∠B + ∠C = 180°, then ∠A = 30° and ∠B = 30°

Next, look at ΔADB.  ∠A + ∠D + ∠B = 180°, so ∠A + 90° + 30° = 180° ⇒ ∠A = 30°

Now look at ΔADC.  Since ∠A = 30° in ΔACB, and ∠A = 60° in ΔADB, then ∠A = 30° in ΔADC per angle addition postulate.

Now that we have shown that ΔADB and ΔADC are 30-60-90 triangles, we can use that formula to calculate the side lengths.

CD = 4 cm (given) so AC = 2(4 cm) = 8 cm

Since AC ≅ BC, then BC = 8 cm. Therefore, BD = 4 + 8 = 12 by segment addition postulate.

Lastly, look at ΔBHD.  Since ∠B = 30° and ∠H = 90°, then ∠D = 60°. So, ΔBHD is also a 30-60-90 triangle.

BD = 12 cm, so HD = = 6 cm

6 cm

-2x + 7 = 7

subtract 7 from both sides.

-2x = 0

divide both sides by -2.

x = 0

hi there!

first we need to find out how much they are adding to each number to make the next number.

ex: 4 + 4 = 8, 8 + 8 = 16, and so on.

so continuing with 32,

32 + 32 = 64

64 + 64 = 128

128 + 128 = 256

256 + 256 = 512

4, 8, 16, 32, 64, 128, 256, 512

the 8th term of the sequence will be 512

hope this ! ~