To form an isosceles triangle here, we need to create a third vertex whose \(\displaystyle y\) coordinate is between \(\displaystyle -1\) and \(\displaystyle 7\).  If a vertex is placed at \(\displaystyle (2,3)\), the distance from \(\displaystyle (1,-1)\) to this point will be \(\displaystyle \sqrt{17}\). The distance from \(\displaystyle (1,7)\) to this point will be the same.

Because \(\displaystyle \bigtriangleup BDE\) is isosceles, \(\displaystyle \measuredangle DBE\) equals \(\displaystyle \left(\frac{180-58}{2}\right)^{\circ}\) or \(\displaystyle 61^{\circ}\).

We know that \(\displaystyle \measuredangle ABE, \measuredangle DBE, \measuredangle CBD\) add up to \(\displaystyle 180^{\circ}\), so \(\displaystyle \measuredangle CBD\) must equal \(\displaystyle (180-90-61)^{\circ}\) or \(\displaystyle 29^{\circ}\).

This problem can be solved when one realizes that the light beam's split has resulted in an acute isosceles triangle. The triangle as stated has two sides of \(\displaystyle 5\) feet apiece, which meets the requirement for isosceles triangles, and having one angle of \(\displaystyle 30^{\circ}\) at the vertex where the two congruent sides meet means the other two angles must be \(\displaystyle 75^{\circ}\) and \(\displaystyle 75^{\circ}\). The missing side connecting the two sensors, therefore, is opposite the \(\displaystyle 30^{\circ}\) angle.

Since we know at least two angles and at least one side of our triangle, we can use the Law of Sines to calculate the remainder. The Law of Sines says that for any triangle with angles \(\displaystyle A, B\) and \(\displaystyle C,\) and opposite sides \(\displaystyle a, b\) and \(\displaystyle c\):

\(\displaystyle \frac{\sin A}{a} = \frac{\sin B}{b} = \frac{\sin C}{c}\).

Plugging in one of our \(\displaystyle 75^{\circ}\) angles (and its corresponding \(\displaystyle 5\) ft side) into this equation, as well as our \(\displaystyle 30^{\circ}\) angle (and its corresponding unknown side) into this equation gives us:

\(\displaystyle \frac{\sin 75^{\circ}}{5} = \frac{ \sin 30^{\circ}}{x}\)

Next, cross-multiply:

\(\displaystyle \frac{\sin 75^{\circ}}{5} = \frac{\sin 30^{\circ}}{x}\) ---> \(\displaystyle x\sin 75^{\circ} = 5\sin 30^{\circ}\)

Now simplify and solve:

\(\displaystyle x = \frac{5\sin 30^{\circ}}{\sin 75^{\circ}} = 2.588\)

Rounding, we see our missing side is \(\displaystyle 2.65\textup{ feet}\) long.

\(\displaystyle A=\frac{1}{2}bh\). 

\(\displaystyle 6x=42\)

\(\displaystyle x=7\)

The area of a triangle is given by the equation: 

\(\displaystyle \small A=\frac{1}{2}bh\)

In this case, we are given the area (\(\displaystyle \small A\)) and the base (\(\displaystyle \small b\)) and are asked to solve for height (\(\displaystyle \small h\)).

To do this, we must plug in the given values for \(\displaystyle \small A\) and \(\displaystyle \small b\), which gives the following:

\(\displaystyle \small 24\;cm^2=\frac{1}{2}(4\; cm)h\)

We then must multiply the right side, and then divide the entire equation by 2, in order to solve for \(\displaystyle \small h\):

\(\displaystyle \small 24\;cm^2=2\;cm(h)\)

\(\displaystyle \small \frac{24\; cm^2}{2\; cm}=\frac{2\; cm(h)}{2\; cm}\)

\(\displaystyle \small 12\;cm=h\)

\(\displaystyle \small h=12\;cm\)

Therefore, the height of the triangle is \(\displaystyle \small 12\;cm\).

In order for two obtuse triangles to be congruent, the sum of the two smaller angles must equal the sum of the two smaller angles of the second triangle. That is, excluding the obtuse angle. 

The first triangle has an obtuse angle of \(\displaystyle 101^{\circ}\). That means the sum of the other two angles is \(\displaystyle 79^{\circ}\). The sum of the corresponding angles in triangle 2 is \(\displaystyle 80^{\circ}\). Therefore, because \(\displaystyle 79^{\circ}\) is not equal to \(\displaystyle 80^{\circ}\), the two obtuse triangles cannot be congruent.