Question

6 practice 7 (from unit 4, lesson 6)
right triangle ( abc ) is shown.
select all expressions which are equal to the length of side ( bc ).
a ( sqrt{4.9^2 + 6^2} ) b ( sqrt{6^2 - 4.9^2} )
c ( 4.9sin(55^circ) ) d ( \frac{4.9}{sin(55^circ)} )
e ( 4.9\tan(55^circ) ) f ( \frac{4.9}{\tan(55^circ)} )
g ( 6cos(55^circ) ) h ( \frac{6}{cos(55^circ)} )

Explanation:

Step1: Recall Pythagorean theorem

In right triangle \(ABC\) (right - angled at \(C\)), by Pythagorean theorem, \(AB^{2}=AC^{2}+BC^{2}\), so \(BC = \sqrt{AB^{2}-AC^{2}}\). Given \(AB = 6\) and \(AC=4.9\), then \(BC=\sqrt{6^{2}-4.9^{2}}\), so option B is correct and option A is incorrect.

Step2: Use trigonometric ratios

We know that in right - triangle \(ABC\) (right - angled at \(C\)):

• \(\sin(55^{\circ})=\frac{BC}{AB}\), so \(BC = AB\sin(55^{\circ})\)? No, wait, \(\sin(55^{\circ})=\frac{BC}{AB}\) is wrong. Wait, angle at \(B\) is \(55^{\circ}\). So \(\sin(55^{\circ})=\frac{AC}{AB}\)? No, let's define the angles properly. In right - triangle \(ABC\), right - angled at \(C\), angle at \(B\) is \(55^{\circ}\).
• \(\sin(55^{\circ})=\frac{AC}{AB}\)? No, \(\sin(\theta)=\frac{\text{opposite}}{\text{hypotenuse}}\). For angle \(B = 55^{\circ}\), the opposite side to angle \(B\) is \(AC\), and the adjacent side is \(BC\), and the hypotenuse is \(AB\). Wait, no, angle at \(B\) is \(55^{\circ}\), so:
• \(\sin(55^{\circ})=\frac{AC}{AB}\), \(\cos(55^{\circ})=\frac{BC}{AB}\), \(\tan(55^{\circ})=\frac{AC}{BC}\)
• From \(\cos(55^{\circ})=\frac{BC}{AB}\), we get \(BC = AB\cos(55^{\circ})\). Since \(AB = 6\), \(BC = 6\cos(55^{\circ})\), so option G is correct.
• From \(\tan(55^{\circ})=\frac{AC}{BC}\), we can re - arrange to get \(BC=\frac{AC}{\tan(55^{\circ})}\). Since \(AC = 4.9\), \(BC=\frac{4.9}{\tan(55^{\circ})}\), so option F is correct.
• Also, from Pythagorean theorem, we had \(BC=\sqrt{6^{2}-4.9^{2}}\) (option B). Let's check other options:
• For option C: \(4.9\sin(55^{\circ})\): \(\sin(55^{\circ})=\frac{AC}{AB}\), \(4.9\sin(55^{\circ})=AC\times\frac{AC}{AB}=\frac{4.9^{2}}{6}

eq BC\), so C is incorrect.

• For option D: \(\frac{4.9}{\sin(55^{\circ})}\): \(\sin(55^{\circ})=\frac{AC}{AB}\), \(\frac{4.9}{\sin(55^{\circ})}=\frac{AC}{\frac{AC}{AB}}=AB = 6

eq BC\), so D is incorrect.

• For option E: \(4.9\tan(55^{\circ})\): \(\tan(55^{\circ})=\frac{AC}{BC}\), \(4.9\tan(55^{\circ})=AC\times\frac{AC}{BC}=\frac{4.9^{2}}{BC}

eq BC\), so E is incorrect.

• For option H: \(\frac{6}{\cos(55^{\circ})}\): \(\cos(55^{\circ})=\frac{BC}{AB}\), \(\frac{AB}{\cos(55^{\circ})}=\frac{6}{\cos(55^{\circ})}\) is the hypotenuse divided by \(\cos(55^{\circ})\), which is not equal to \(BC\), so H is incorrect.

Answer:

B, F, G