Question

1. find dc.

Explanation:

Step1: Analyze triangle ABD

In right triangle \(ABD\), \(\angle A = 54^\circ\), hypotenuse \(AB = 20\). We can find \(BD\) using \(\sin(54^\circ)\):
\(\sin(54^\circ)=\frac{BD}{AB}\)
\(BD = AB \cdot \sin(54^\circ)=20 \cdot \sin(54^\circ)\) (calculate \(\sin(54^\circ)\approx0.8090\))
\(BD\approx20 \cdot 0.8090 = 16.18\)

Step2: Analyze triangle BDC

In right triangle \(BDC\), \(\angle C = 28^\circ\), \(BD\approx16.18\). We use \(\tan(28^\circ)=\frac{BD}{DC}\) (wait, no: \(\tan(\theta)=\frac{\text{opposite}}{\text{adjacent}}\), so \(\tan(28^\circ)=\frac{BD}{DC}\) → \(DC=\frac{BD}{\tan(28^\circ)}\)? Wait, no: \(\angle C = 28^\circ\), \(BD\) is opposite to \(\angle C\), \(DC\) is adjacent. So \(\tan(28^\circ)=\frac{BD}{DC}\) → \(DC=\frac{BD}{\tan(28^\circ)}\). Wait, or \(\sin(28^\circ)=\frac{BD}{BC}\), no, \(BD\) is height. Wait, correct: in \( \triangle BDC \), right-angled at \(D\), \(\angle C = 28^\circ\), \(BD\) is the side opposite \(\angle C\), \(DC\) is adjacent. So \(\tan(\angle C)=\frac{BD}{DC}\) → \(DC=\frac{BD}{\tan(28^\circ)}\).

First, recalculate \(BD\) with more precision: \(\sin(54^\circ)\approx0.809016994\), so \(BD = 20 \cdot 0.809016994 \approx 16.1803\).

\(\tan(28^\circ)\approx0.531709432\).

Then \(DC=\frac{16.1803}{0.531709432}\approx30.43\)? Wait, no, maybe I mixed up. Wait, maybe \(AB = 20\), \(\angle A = 54^\circ\), so \(AD = AB \cdot \cos(54^\circ)\), but we need \(BD\) as opposite. Wait, no, in \( \triangle ABD \), right-angled at \(D\), so \(BD = AB \cdot \sin(54^\circ)\), \(AD = AB \cdot \cos(54^\circ)\). Then in \( \triangle BDC \), right-angled at \(D\), \(\angle C = 28^\circ\), so \(\tan(28^\circ)=\frac{BD}{DC}\) → \(DC=\frac{BD}{\tan(28^\circ)}\).

Wait, let's check angles. The triangle \(ABC\): angles at \(A=54^\circ\), \(C=28^\circ\), so angle at \(B\) is \(180 - 54 - 28 = 98^\circ\). But \(BD\) is altitude.

Wait, maybe I made a mistake. Let's re-express:

In \( \triangle ABD \): right-angled at \(D\), \(\angle A = 54^\circ\), \(AB = 20\). So:
\(\sin(54^\circ) = \frac{BD}{AB}\) ⇒ \(BD = AB \cdot \sin(54^\circ) = 20 \cdot \sin(54^\circ)\)
\(\cos(54^\circ) = \frac{AD}{AB}\) ⇒ \(AD = AB \cdot \cos(54^\circ)\)

In \( \triangle BDC \): right-angled at \(D\), \(\angle C = 28^\circ\), \(BD\) is known. So:
\(\tan(28^\circ) = \frac{BD}{DC}\) ⇒ \(DC = \frac{BD}{\tan(28^\circ)}\)

Calculate \(BD\):
\(\sin(54^\circ) \approx 0.8090\), so \(BD \approx 20 \times 0.8090 = 16.18\)

\(\tan(28^\circ) \approx 0.5317\), so \(DC \approx \frac{16.18}{0.5317} \approx 30.43\)? Wait, that seems large. Wait, maybe \(AB = 20\) is not the hypotenuse? Wait, no, \( \triangle ABD \) is right-angled at \(D\), so \(AB\) is hypotenuse. Wait, maybe the diagram has \(AB = 20\), \( \angle A = 54^\circ \), \(BD \perp AC\), so \(BD\) is height. Then in \( \triangle BDC \), \( \angle C = 28^\circ \), so \(DC = \frac{BD}{\tan(28^\circ)}\). Alternatively, maybe I should use \( \cot(28^\circ) = \frac{DC}{BD} \), so \(DC = BD \cdot \cot(28^\circ)\). \( \cot(28^\circ) = \frac{1}{\tan(28^\circ)} \approx 1.8807\), so \(DC \approx 16.18 \times 1.8807 \approx 30.43\).

Wait, maybe the problem is that \(AB = 20\), and we need to find \(DC\) using trigonometry. Let's verify with another approach. Alternatively, maybe \(AB = 20\), \( \angle A = 54^\circ \), so \(BD = 20 \sin 54^\circ \approx 16.18\). Then in \( \triangle BDC \), \( \angle C = 28^\circ \), so \(DC = \frac{BD}{\tan 28^\circ} \approx \frac{16.18}{0.5317} \approx 30.4\).

Wait, maybe the diagram has \(AB = 20\), and \(D\) is on \(AC\), \(BD \perp AC\). So steps:

1. Find \(BD\) from \( \triangle ABD \): \(BD = AB \sin(54^\circ) \approx 20 \times 0.8090 = 16.18\).
2. Find \(DC\) from \( \triangle BDC \): \(DC = \frac{BD}{\tan(28^\circ)} \approx \frac{16.18}{0.5317} \approx 30.4\).

Answer:

\(\approx 30.4\) (or more precise: using calculator, \( \sin(54^\circ) \approx 0.809016994 \), \( BD = 20 \times 0.809016994 = 16.18033988 \); \( \tan(28^\circ) \approx 0.5317094317 \); \( DC = 16.18033988 / 0.5317094317 \approx 30.43 \))