Question

unit 4 assessment review – trigonometric ratios, angles of elevation & depression
end-of-unit assessment
you may use a scientific calculator and your reference chart. round angles to the nearest degree
and side lengths to the nearest tenth.
1 what is the area of triangle ( abc )?
( overline{ab} perp overline{cd} )
a. ( 6sqrt{3} ) square units
b. ( 18sqrt{3} ) square units
c. ( 36sqrt{3} ) square units
d. ( 72sqrt{3} ) square units

Explanation:

Step1: Analyze triangle CBD

In right triangle \(CBD\), \(\angle B = 60^\circ\), hypotenuse \(BC = 12\). We can find \(BD\) and \(CD\) using trigonometric ratios. \(\cos 60^\circ=\frac{BD}{BC}\), so \(BD = BC\times\cos 60^\circ=12\times\frac{1}{2} = 6\). \(\sin 60^\circ=\frac{CD}{BC}\), so \(CD = BC\times\sin 60^\circ=12\times\frac{\sqrt{3}}{2}=6\sqrt{3}\).

Step2: Analyze triangle ABC

Since \(\angle A=\angle B = 60^\circ\), triangle \(ABC\) is isosceles with \(AB = AD + BD\). And since \(\triangle ACD\) and \(\triangle BCD\) are congruent (by AAS, as \(\angle A=\angle B\), \(\angle ADC=\angle BDC = 90^\circ\), \(CD\) common), \(AD = BD = 6\). So \(AB=6 + 6=12\).

Step3: Calculate area of triangle ABC

The area of a triangle is \(\frac{1}{2}\times base\times height\). Here, base \(AB = 12\), height \(CD = 6\sqrt{3}\). So area \(=\frac{1}{2}\times12\times6\sqrt{3}=36\sqrt{3}\) square units.

Answer:

C. \(36\sqrt{3}\) square units