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?
2 answers:
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 <em>per angle addition postulate.</em>
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 <em>by segment addition postulate.</em>
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
Answer: 6 cm
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