Question

lesson = trigonometry with pythagoras theorem question 10 of 10
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13.3
16.4
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14
the above figure \\( \overline { a b } = \overline { b c } \\), find
the length of \\( \overline { d c } \\).

Explanation:

Step1: Find the length of \( BD \)

In right - triangle \( ABD \), \(\sin30^{\circ}=\frac{BD}{AD}\). Since \(\sin30^{\circ}=\frac{1}{2}\) and \(BD = 7\), we can also use the property of \(30^{\circ}-60^{\circ}-90^{\circ}\) triangle. In right - triangle \( DBC \), we know that \(AB = BC\) and from right - triangle \( ABD\), using \(\tan30^{\circ}=\frac{BD}{AB}\), \(\tan30^{\circ}=\frac{\sqrt{3}}{3}=\frac{7}{AB}\), so \(AB = 7\sqrt{3}\). But since \(AB = BC\), in right - triangle \( DBC\) with \(BD = 7\) and \(BC=7\sqrt{3}\), we can use the Pythagorean theorem \(DC^{2}=BD^{2}+BC^{2}\).
Another way: In right - triangle \( ABD\), \(\sin30^{\circ}=\frac{BD}{AD}\), but more simply, in right - triangle \( DBC\) (because \(AB = BC\) and \(BD\perp AC\)), we can use the fact that in right - triangle \( ABD\) (where \(\angle A = 30^{\circ}\)), \(AD = 14\) (since in a \(30^{\circ}-60^{\circ}-90^{\circ}\) triangle, the side opposite \(30^{\circ}\) is half the hypotenuse. Here \(BD\) is opposite \(30^{\circ}\) in \(\triangle ABD\) if we consider \(\angle A = 30^{\circ}\), assume \(AD\) is the hypotenuse. Wait, correct approach:
In right - triangle \( ABD\), \(\sin30^{\circ}=\frac{BD}{AD}\), given \(BD = 7\), \(\sin30^{\circ}=\frac{1}{2}=\frac{7}{AD}\), so \(AD\) (but no, wrong. Correct:
In right - triangle \( ABD\), \(\sin30^{\circ}=\frac{BD}{AD}\) is wrong. Correct formula for right - triangle \( ABD\) (assuming \(\angle A = 30^{\circ}\)): \(\sin30^{\circ}=\frac{BD}{AD}\) (no, \(\sin\theta=\frac{\text{opposite}}{\text{hypotenuse}}\). Wait, correct:
In right - triangle \( ABD\), \(\angle A=30^{\circ}\), \(\angle ABD = 90^{\circ}\), \(BD = 7\). Using \(\sin30^{\circ}=\frac{BD}{AD}\), \(AD = 14\) (because \(\sin30^{\circ}=\frac{1}{2}\), so \(AD=\frac{BD}{\sin30^{\circ}}=\frac{7}{\frac{1}{2}} = 14\)). Since \(AB = BC\) and \(BD\perp AC\), triangles \(ABD\) and \(CBD\) are congruent (by \(SAS\) since \(AB = BC\), \(\angle ABD=\angle CBD = 90^{\circ}\), \(BD = BD\)). So \(DC=AD = 14\)

Answer:

14