Question

question 12 of 26
michael is looking up at an object. michael points one arm at the object and
the other arm at connor. the angle formed by his outstretched arms
measures 40 degrees. connor is also looking up at the object. he points one
arm at the object and the other arm at michael. the angle formed by his
outstretched arms measures 30 degrees. the distance between the two men
is 80 feet. which of the following best represents the distance between
michael and the object?

a. 42.6 ft

b. 48.9 ft

c. 60.1 ft

d. 53.4 ft

Explanation:

Step1: Identify the triangle type

We can model this situation as a triangle where the two men (Michael and Connor) and the object form a triangle. Let's denote:
- Let \( M \) be Michael's position, \( C \) be Connor's position, and \( O \) be the object's position.
- The distance between \( M \) and \( C \) is \( MC = 80 \) feet.
- The angle at \( M \) (Michael) is \( \angle OMC = 40^\circ \)
- The angle at \( C \) (Connor) is \( \angle OCM = 30^\circ \)
- We need to find the distance \( MO \) (distance between Michael and the object).

First, we find the angle at \( O \) using the fact that the sum of angles in a triangle is \( 180^\circ \).
\( \angle MOC=180^\circ - 40^\circ - 30^\circ=110^\circ \)

Step2: Apply the Law of Sines

The Law of Sines states that for a triangle with sides \( a, b, c \) opposite angles \( A, B, C \) respectively, \( \frac{a}{\sin A}=\frac{b}{\sin B}=\frac{c}{\sin C} \)

In our triangle:
- Side opposite \( \angle OCM = 30^\circ \) is \( MO \) (let's call it \( x \))
- Side opposite \( \angle OMC = 40^\circ \) is \( CO \)
- Side opposite \( \angle MOC = 110^\circ \) is \( MC = 80 \) feet

Using the Law of Sines:
\( \frac{MO}{\sin \angle OCM}=\frac{MC}{\sin \angle MOC} \)

Substitute the known values:
\( \frac{x}{\sin 30^\circ}=\frac{80}{\sin 110^\circ} \)

We know that \( \sin 30^\circ = 0.5 \) and \( \sin 110^\circ=\sin(70^\circ)\approx 0.9397 \)

So,
\( x=\frac{80\times\sin 30^\circ}{\sin 110^\circ} \)

Step3: Calculate the value of \( x \)

Substitute the values of sines:
\( x = \frac{80\times0.5}{0.9397} \)
\( x=\frac{40}{0.9397}\approx 42.6 \) (Wait, no, wait. Wait, maybe I mixed up the angles. Wait, let's re - check.

Wait, the angle at Michael is \( \angle OMC = 40^\circ \), angle at Connor is \( \angle OCM = 30^\circ \), side \( MC = 80 \). We want to find \( MO \). Wait, maybe I assigned the angles wrong. Let's re - define:

Let’s denote:
- \( \angle M = 40^\circ \) (at Michael), \( \angle C=30^\circ \) (at Connor), \( MC = 80 \), we need to find \( MO \) (let's call it \( a \)), \( CO \) (let's call it \( b \)), and \( MO \) is opposite \( \angle C = 30^\circ \), \( CO \) is opposite \( \angle M=40^\circ \), and \( MC \) is opposite \( \angle O = 180 - 40 - 30=110^\circ \)

Wait, no, if we want to find \( MO \), which is the distance from Michael to the object, then in triangle \( MOC \), side \( MO \) is opposite angle \( C = 30^\circ \), side \( MC = 80 \) is opposite angle \( O=110^\circ \)

So by Law of Sines:

\( \frac{MO}{\sin C}=\frac{MC}{\sin O} \)

\( MO=\frac{MC\times\sin C}{\sin O} \)

\( MC = 80 \), \( \sin C=\sin 30^\circ = 0.5 \), \( \sin O=\sin 110^\circ\approx0.9397 \)

\( MO=\frac{80\times0.5}{0.9397}=\frac{40}{0.9397}\approx42.6 \)

Wait, but let's check again. Wait, maybe the angle at Michael is the angle between his arm to the object and his arm to Connor, so the triangle is such that:

Wait, another way: Let's consider the triangle with vertices \( M \) (Michael), \( C \) (Connor), \( O \) (Object). The angle at \( M \) is \( 40^\circ \), angle at \( C \) is \( 30^\circ \), side \( MC = 80 \). We need to find \( MO \).

Using Law of Sines:

\( \frac{MO}{\sin \angle C}=\frac{MC}{\sin \angle O} \)

\( \angle O=180-(40 + 30)=110^\circ \)

\( MO=\frac{80\times\sin 30^\circ}{\sin 110^\circ}=\frac{80\times0.5}{\sin(70^\circ)}\approx\frac{40}{0.9397}\approx42.6 \)

Wait, but let's check the options. Option A is 42.6 ft. But wait, maybe I made a mistake in angle assignment.

Wait, maybe the angle at Michael is the angle between his line of sight to the object and his line to Connor, and the angle at Connor is the angle between his line of sight to the object and his line to Michael. So the triangle is \( \triangle MOC \), with \( \angle M = 40^\circ \), \( \angle C=30^\circ \), \( MC = 80 \). Then the side opposite \( \angle C \) is \( MO \), side opposite \( \angle M \) is \( CO \), side opposite \( \angle O \) is \( MC \).

So Law of Sines: \( \frac{MO}{\sin C}=\frac{MC}{\sin O} \)

\( \sin C = \sin 30^\circ=0.5 \), \( \sin O=\sin(180 - 40 - 30)=\sin 110^\circ\approx0.9397 \)

\( MO=\frac{80\times0.5}{0.9397}\approx42.6 \)

So the answer should be A. 42.6 ft.

Wait, but let's re - calculate:

\( \sin(110^\circ)=\sin(70^\circ)\approx0.9396926 \)

\( 80\times\sin(30^\circ)=80\times0.5 = 40 \)

\( 40\div0.9396926\approx42.57\approx42.6 \)

Yes.

Answer:

A. 42.6 ft