Question

in the figure below, the length of segment cb is 64 units and the length of segment bg is 137 units. what is the length of segment ea? diagram: points c, b, a, g, e with right angles at b (cb ⊥ bg) and g (eg ⊥ bg); angle 45° at a (between cb and ab), angle 60° at a (between ea and ag); cb = 64, bg = 137; options: 128, 73√3, 146, 64√2

Explanation:

Step1: Analyze triangle CBA

In right triangle \( CBA \), \( \angle CAB = 45^\circ \), \( \angle CBA = 90^\circ \), so it's an isosceles right triangle. Thus, \( CB = BA = 64 \) units.

Step2: Find length of AG

Given \( BG = 137 \) units and \( BA = 64 \) units, then \( AG = BG - BA = 137 - 64 = 73 \) units.

Step3: Analyze triangle EAG

In right triangle \( EAG \), \( \angle EAG = 60^\circ \), \( \angle EGA = 90^\circ \). We know that in a 30 - 60 - 90 triangle, if the shorter leg (opposite \( 30^\circ \)) is \( x \), the longer leg (opposite \( 60^\circ \)) is \( x\sqrt{3} \), and the hypotenuse is \( 2x \). Here, \( AG \) is the shorter leg (opposite \( 30^\circ \) if we consider the angles, wait, actually \( \angle AEG = 30^\circ \), so \( AG \) is opposite \( 30^\circ \), so \( EA \) is the hypotenuse? Wait, no, \( \angle EAG = 60^\circ \), so the side opposite \( 60^\circ \) is \( EG \), and the side adjacent is \( AG \), and hypotenuse is \( EA \). Wait, using trigonometry: \( \cos(60^\circ)=\frac{AG}{EA} \), so \( EA=\frac{AG}{\cos(60^\circ)} \). Since \( \cos(60^\circ)=\frac{1}{2} \), then \( EA = \frac{73}{\frac{1}{2}}=146 \)? Wait, no, wait, maybe I mixed up the angles. Wait, in triangle \( EAG \), right - angled at \( G \), \( \angle EAG = 60^\circ \), so \( \sin(60^\circ)=\frac{EG}{EA} \), \( \cos(60^\circ)=\frac{AG}{EA} \). We have \( AG = 73 \), \( \cos(60^\circ)=\frac{1}{2} \), so \( EA=\frac{AG}{\cos(60^\circ)}=\frac{73}{\frac{1}{2}} = 146 \)? Wait, but let's check again. Wait, maybe the triangle is a 30 - 60 - 90 triangle where the side opposite \( 30^\circ \) is \( AG \), so hypotenuse \( EA = 2\times AG \)? Wait, no, \( \angle AEG = 30^\circ \), so the side opposite \( 30^\circ \) is \( AG \), so hypotenuse \( EA = 2\times AG \). Since \( AG = 73 \), then \( EA = 2\times73 = 146 \). Wait, but let's confirm the first step. In triangle \( CBA \), right - angled at \( B \), \( \angle CAB = 45^\circ \), so \( \tan(45^\circ)=\frac{CB}{BA} \), and \( \tan(45^\circ)=1 \), so \( CB = BA = 64 \). Then \( AG = BG - BA = 137 - 64 = 73 \). Then in triangle \( EAG \), right - angled at \( G \), \( \angle EAG = 60^\circ \), so \( \angle AEG = 30^\circ \), so the side opposite \( 30^\circ \) is \( AG \), so hypotenuse \( EA = 2\times AG = 2\times73 = 146 \).

Answer:

146