Question

question 7 of 26 a viewfinder has a triangular lens. some of the measurements of the lens are shown below. which of the following best represents the length of a? triangle not drawn to scale. a. 5.2 in b. 3.9 in c. 4.6 in d. 5.9 in

Explanation:

Step1: Find angle at B

In triangle \(ABC\), the sum of angles is \(180^\circ\). So, \(\angle B = 180^\circ - 38^\circ - 32^\circ = 110^\circ\)? Wait, no, wait. Wait, the side opposite angle \(A\) is \(a\), side opposite angle \(C\) is \(c\), and side \(b\) is opposite angle \(B\)? Wait, no, the given side is 7 in, which is side \(b\) (opposite angle \(B\))? Wait, no, let's label correctly. In triangle \(ABC\), angle \(A = 38^\circ\), angle \(C = 32^\circ\), side \(b\) (opposite angle \(B\))? Wait, no, the side given is 7 in, which is side \(b\)? Wait, no, the side adjacent to angle \(A\) and angle \(C\) is 7 in? Wait, no, the diagram: point \(A\), \(B\), \(C\). Side \(AC\) is 7 in, angle at \(A\) is \(38^\circ\), angle at \(C\) is \(32^\circ\), and side \(a\) is \(BC\), side \(c\) is \(AB\). So we can use the Law of Sines: \(\frac{a}{\sin A} = \frac{AC}{\sin B}\). First, find angle \(B\): \(\angle B = 180^\circ - 38^\circ - 32^\circ = 110^\circ\)? Wait, no, that can't be. Wait, maybe I mislabeled. Wait, the side \(AC\) is 7 in, which is side \(b\) (opposite angle \(B\)). Then angle \(A = 38^\circ\), angle \(C = 32^\circ\), so angle \(B = 180 - 38 - 32 = 110^\circ\). Then by Law of Sines: \(\frac{a}{\sin A} = \frac{b}{\sin B}\), so \(a = \frac{b \cdot \sin A}{\sin B}\). Wait, but \(b = 7\) in, \(\sin A = \sin 38^\circ\), \(\sin B = \sin 110^\circ\). Wait, but maybe I made a mistake. Wait, maybe the side 7 in is side \(c\) (opposite angle \(C\))? Wait, no, the diagram: \(A\) to \(C\) is 7 in, so side \(AC = 7\) in, which is side \(b\) (opposite angle \(B\)). Then angle \(A = 38^\circ\), angle \(C = 32^\circ\), angle \(B = 110^\circ\). Then \(a\) is side \(BC\), opposite angle \(A\). So Law of Sines: \(\frac{a}{\sin 38^\circ} = \frac{7}{\sin 110^\circ}\). Let's calculate \(\sin 38^\circ \approx 0.6157\), \(\sin 110^\circ = \sin (70^\circ) \approx 0.9397\). Then \(a = \frac{7 \times 0.6157}{0.9397} \approx \frac{4.31}{0.9397} \approx 4.6\) in. Wait, but let's check again. Wait, maybe the side 7 in is side \(a\)'s opposite? No, the question is to find \(a\). Wait, maybe I mixed up the angles. Wait, angle at \(A\) is \(38^\circ\), angle at \(C\) is \(32^\circ\), so side \(a\) is opposite angle \(A\)? No, in standard notation, side \(a\) is opposite angle \(A\), side \(b\) opposite angle \(B\), side \(c\) opposite angle \(C\). So if angle \(A = 38^\circ\), angle \(C = 32^\circ\), then angle \(B = 180 - 38 - 32 = 110^\circ\). Side \(b\) (opposite angle \(B\)) is 7 in? Wait, no, the side given is 7 in, which is side \(b\)? Then \(a\) is opposite angle \(A\), so \(\frac{a}{\sin A} = \frac{b}{\sin B}\), so \(a = \frac{7 \times \sin 38^\circ}{\sin 110^\circ}\). Calculating: \(\sin 38^\circ \approx 0.6157\), \(\sin 110^\circ \approx 0.9397\), so \(a \approx \frac{7 \times 0.6157}{0.9397} \approx \frac{4.31}{0.9397} \approx 4.6\) in. So the answer is C.

Step1: Calculate angle at B

Sum of angles in triangle: \(180^\circ\).
\(\angle B = 180^\circ - 38^\circ - 32^\circ = 110^\circ\).

Step2: Apply Law of Sines

Law of Sines: \(\frac{a}{\sin A} = \frac{b}{\sin B}\), where \(b = 7\) in, \(A = 38^\circ\), \(B = 110^\circ\).
Solve for \(a\):
\(a = \frac{b \cdot \sin A}{\sin B}\)

Step3: Substitute values

\(\sin 38^\circ \approx 0.6157\), \(\sin 110^\circ \approx 0.9397\), \(b = 7\) in.
\(a = \frac{7 \times 0.6157}{0.9397} \approx \frac{4.31}{0.9397} \approx 4.6\) in.

Answer:

C. 4.6 in