Question

lesson = trigonometry with pythagoras theorem question 2 of 10
back principle clue solution tutorials ok
select
an
answer:
9.42
13.32
7.32
13.58
14.92
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 \( a^{2}+b^{2}=c^{2} \), where \( a = 6 \), \( b = 8 \). Then \( x=\sqrt{6^{2}+8^{2}}=\sqrt{36 + 64}=\sqrt{100}=10 \).

Step2: Find the length of \( y \)

Since \( \angle ECB=36.9^{\circ} \) and \( \triangle EBC \) is a right - triangle (\( \angle EBC = 90^{\circ} \)), and we know that \( \cos(36.9^{\circ})=\frac{x}{y} \) (adjacent over hypotenuse). Given \( x = 10 \), and \( \cos(36.9^{\circ})\approx0.8 \). Then \( y=\frac{x}{\cos(36.9^{\circ})}=\frac{10}{0.8}=12.5 \) (This step is wrong, we should use another way. Since \( \triangle EAB\cong\triangle BDC \) (by AAS, \( \angle EAB=\angle BDC = 90^{\circ} \), \( \angle AEB=\angle DBC \) (because \( \angle AEB+\angle ABE = 90^{\circ} \) and \( \angle DBC+\angle ABE = 90^{\circ} \)), \( EB = BC \) (hypotenuse - leg in right - triangles)). Wait, no. Let's use trigonometry correctly.

In rectangle \( AEDC \), \( ED=AC \). \( AC=\sqrt{6^{2}+8^{2}} = 10 \). Since \( \triangle EBC \) is a right - triangle and \( \angle ECB = 36.9^{\circ} \), and \( \sin(36.9^{\circ})=\frac{EB}{EC} \), \( \cos(36.9^{\circ})=\frac{BC}{EC} \). Also, \( EC=\sqrt{6^{2}+8^{2}}=10\) (wrong, \( EC \) is the hypotenuse of the rectangle's right - triangle. Wait, \( AE = 8 \), \( AB = 6 \), \( EB=\sqrt{6^{2}+8^{2}}=10 \). In right - triangle \( EBC \), \( \sin(36.9^{\circ})=\frac{EB}{EC}\), \( EC=\frac{EB}{\sin(36.9^{\circ})}\approx\frac{10}{0.6}=16.67 \) (wrong).

Correct way: Since \( \triangle EAB\) is a right - triangle with \( EA = 8 \), \( AB = 6 \), then \( EB=\sqrt{6^{2}+8^{2}}=10 \). In right - triangle \( EBC \), \( \sin(36.9^{\circ})=\frac{EB}{EC}\), \( EC=\frac{EB}{\sin(36.9^{\circ})}\approx\frac{10}{0.6}=16.67 \) (wrong). Wait, no. The figure is a rectangle \( AEDC \), so \( ED = AC \), \( AC=\sqrt{6^{2}+8^{2}}=10 \). Also, \( \triangle EAB\) and \( \triangle BDC \) are similar. But better:

Since \( \triangle EAB\) is a right - triangle \( EB=\sqrt{6^{2}+8^{2}} = 10 \). In right - triangle \( EBC \), \( \cos(36.9^{\circ})=\frac{EB}{EC}\) (no, \( \cos(36.9^{\circ})=\frac{BC}{EC}\)). Wait, \( \tan(36.9^{\circ})=\frac{EB}{BC}\), \( BC=\frac{EB}{\tan(36.9^{\circ})}\). Since \( \tan(36.9^{\circ})\approx0.75 \), \( EB = 10 \), \( BC=\frac{10}{0.75}\approx13.33\approx13.32 \)

Answer:

13.32