Question

lesson = trigonometry with pythagoras theorem question 1 of 10
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an
answe
find the length of (overline{bc}) in the
figure.
hint: use the pythagorean theorem to
find (x). remember, you can use a
calculator for this lesson.

Explanation:

Step1: Find the length of \( x \)

In right - triangle \( EAB \), by the Pythagorean theorem \( x=\sqrt{7^{2}+5^{2}}=\sqrt{49 + 25}=\sqrt{74}\approx8.6\)

Step2: Find the length of \( y \)

Since \( \angle ECB = 35.5^{\circ}\) and \( \triangle EBC\) is a right - triangle (\(\angle EBC = 90^{\circ}\)), and we know that \(\sin\theta=\frac{\text{opposite}}{\text{hypotenuse}}\), \(\cos\theta=\frac{\text{adjacent}}{\text{hypotenuse}}\). Here, if we consider \( \theta = 35.5^{\circ}\) and the opposite side to \( \theta\) is \( x\) and the adjacent side is \( y\).
We know that \(\tan\theta=\frac{x}{y}\), but also, since \( EC = AD\) (opposite sides of a rectangle are equal), and \( EC=\sqrt{7^{2}+AD^{2}}\) (wait, no, better way: in right - triangle \( EAB\), \(x = \sqrt{7^{2}+5^{2}}\approx8.6\), and in right - triangle \( BCD\) (since \( EC\parallel AD\), \( \angle ECB=\angle CBD = 35.5^{\circ}\), and \( EB = CD = 7\) (opposite sides of a rectangle).
We use the trigonometric relation \(\sin35.5^{\circ}=\frac{EB}{EC}\), \(EC=\frac{EB}{\sin35.5^{\circ}}=\frac{7}{\sin35.5^{\circ}}\approx\frac{7}{0.58}\approx12.06\)

Another way:
In right - triangle \( EAB\), \(x=\sqrt{7^{2}+5^{2}}=\sqrt{49 + 25}=\sqrt{74}\approx8.6\)
In right - triangle \( EBC\), \(\sin35.5^{\circ}=\frac{x}{EC}\), \(EC=\frac{x}{\sin35.5^{\circ}}\). But also, since \( EC = AD\) and in right - triangle \( EAB\) and rectangle \( EADC\)
In right - triangle \( BCD\), \(\sin35.5^{\circ}=\frac{7}{EC}\), \(EC=\frac{7}{\sin35.5^{\circ}}\approx\frac{7}{0.58}\approx12.06\)

Answer:

\(12.06\)