Exterior angle \(\displaystyle \angle 1\) forms a linear pair with its interior angle \(\displaystyle \angle B AC\). Either both are right, or one is acute and one is obtuse. From Statement 1 alone, since \(\displaystyle \angle 1\) is acute, \(\displaystyle \angle B AC\) is obtuse, and \(\displaystyle \bigtriangleup ABC\) is an obtuse triangle.

Statement 2 alone is insufficient. Every triangle has at least two acute angles, and Statement 2 only establishes that  \(\displaystyle \angle B\) and \(\displaystyle \angle C\) are both acute; the third angle, \(\displaystyle \angle B AC\), can be acute, right, or obtuse, so the question of whether \(\displaystyle \bigtriangleup ABC\) is an acute, right, or obtuse triangle is not settled.

Assume Statement 1 alone. An exterior angle of a triangle forms a linear pair with the interior angle of the triangle of the same vertex. The two angles, whose measures total \(\displaystyle 180^{\circ }\), must be two right angles or one acute angle and one obtuse angle. Since \(\displaystyle \angle 1\), \(\displaystyle \angle 2\), and \(\displaystyle \angle 3\) are all obtuse angles, it follows that their respective interior angles - the three angles of \(\displaystyle \bigtriangleup ABC\) - are all acute. This makes \(\displaystyle \bigtriangleup ABC\) an acute triangle.

Statement 2 alone provides insufficient information to answer the question. For example, if \(\displaystyle \angle A\) and \(\displaystyle \angle B\) each measure \(\displaystyle 10^{\circ }\) and \(\displaystyle \angle C\) measures \(\displaystyle 160^{\circ }\), the sum of the angle measures is \(\displaystyle 180^{\circ }\), \(\displaystyle \angle A\) and \(\displaystyle \angle B\) are congruent, and \(\displaystyle \angle C\) is an obtuse angle (measuring more than \(\displaystyle 90^{\circ }\)); this makes \(\displaystyle \bigtriangleup ABC\) an obtuse triangle. But  if \(\displaystyle \angle A\), \(\displaystyle \angle B\), and \(\displaystyle \angle C\) each measure \(\displaystyle 60^{\circ }\), the sum of the angle measures is again \(\displaystyle 180^{\circ }\), \(\displaystyle \angle A\) and \(\displaystyle \angle B\) are again congruent, and all three angles are acute (measuring less than \(\displaystyle 90^{\circ }\)); this makes \(\displaystyle \bigtriangleup ABC\) an acute triangle. 

The degree measures of the angles of a triangle add up to a total of 180, so we can set up the following equations:

From the first triangle:

\(\displaystyle X+2X+ Y = 180\)

\(\displaystyle 3X+ Y = 180\)

From the second:

\(\displaystyle (Y + 10)+( Y + 20)+ (2X-10 )= 180\)

\(\displaystyle 2X+ 2Y + 20 = 180\)

\(\displaystyle 2X+ 2Y + 20 - 20 = 180 - 20\)

\(\displaystyle 2X+ 2Y = 160\)

\(\displaystyle (2X+ 2Y)\div 2 = 160 \div 2\)

\(\displaystyle X+Y = 80\)

These equations form a system of equations that can be solved:

\(\displaystyle X+Y = 80\)

\(\displaystyle -(X+Y) = - (80)\)

\(\displaystyle -X-Y = -80\)

\(\displaystyle \underline{3X+ Y = 180}\)

\(\displaystyle 2X\)             \(\displaystyle = 100\)

\(\displaystyle X = 50\)

\(\displaystyle X+Y = 80\), so

\(\displaystyle 50 +Y = 80\)

and \(\displaystyle Y = 30\).

The height of the obtuse isosceles triangle bisects the  \(\displaystyle 120^{\circ}\)  angle and forms two congruent right triangles. The hypotenuse of each of these triangles is either side of equivalent length, and we can see that the base of either triangle makes up half of the hypotenuse of the obtuse isosceles triangle. Because we know the angle opposite each base is half of  \(\displaystyle 120^{\circ}\),  or  \(\displaystyle 60^{\circ}\),  we can use the sine of this angle to find the length of the base. As there are two congruent right triangles that make up the obtuse isosceles triangle, the length of either base makes up half of the overall hypotenuse, so we then multiply the result by  \(\displaystyle 2\)  to obtain the final answer. In the following solution,  \(\displaystyle b\)  is the length of the base of one of the right triangles,  \(\displaystyle l\)  is the length of the two equivalent sides, and  \(\displaystyle c\)  is the length of the hypotenuse:

\(\displaystyle \sin60^{\circ}=\frac{b}{l}\)

\(\displaystyle \frac{\sqrt{3}}{2}=\frac{b}{11}\rightarrow b=\frac{11}{2}\sqrt{3}\)

\(\displaystyle c=2b=2\left(\frac{11}{2}\sqrt{3}\right)=11\sqrt{3}\)

Use Heron's formula:

\(\displaystyle \small A = \sqrt{s (s-a) (s-b)(s-c)}\)

where \(\displaystyle \small a = 12, b= 13, c = 15\), and

\(\displaystyle \small s = \frac{a + b + c}{2} = \frac{12 + 13 + 15}{2} = 20\)

\(\displaystyle \small A = \sqrt{s (s-a) (s-b)(s-c)}\)

\(\displaystyle \small \small \small A = \sqrt{20 (20-12) (20-13)(20-15)}\)

\(\displaystyle \small A = \sqrt{20 \cdot 8\cdot 7\cdot 5} = \sqrt{5,600} \approx 75\)

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

In this problem, the base is 12 and the height is 6. Therefore:

\(\displaystyle A=\frac{1}{2}(12)(6)=36\)

The figure is a composite of a rectangle and a triangle, as shown:

The rectangle has area \(\displaystyle A_ {1}= LW = 16 \cdot8 = 128\)

The triangle has area \(\displaystyle A_ {2} = \frac{1}{2} bh = \frac{1}{2} \cdot 14 \cdot 16 = 112\)

The total area of the figure is \(\displaystyle A_ {1} + A_ {2} = 128 + 112 = 240\)

The only restriction on the measure of the vertex angle of an isosceles triangle is the restriction on any angle of a triangle - that it fall between \(\displaystyle 0^{\circ }\) and \(\displaystyle 180^{\circ }\), noninclusive. If \(\displaystyle N\) is any number in that range, each base angle, the two being congruent, will measure \(\displaystyle \frac{180 -N }{2}\), which will fall in the acceptable range.

Since all of these measures fall in that range, the correct response is that all are allowed.

The easiest way to solve this is to graph the three lines and to observe the dimensions of the resulting triangle. It helps to know the coordinates of the three points of intersection, which we can do as follows:

The intersection of \(\displaystyle y = 4x\) and the \(\displaystyle x\)-axis - that is, the line \(\displaystyle y = 0\) can be found with some substitution:

\(\displaystyle y = 4x\)

\(\displaystyle 0 = 4x\)

\(\displaystyle 0 = x\)

The lines intersect at \(\displaystyle (0,0)\)

The intersection of \(\displaystyle y + 2x = 18\) and the  \(\displaystyle x\)-axis can be found the same way:

\(\displaystyle y + 2x = 18\) 

\(\displaystyle 0+ 2x = 18\)

\(\displaystyle 2x = 18\)

\(\displaystyle x=9\)

These lines intersect at \(\displaystyle (9,0)\)

The intersection of \(\displaystyle y = 4x\) and \(\displaystyle y + 2x = 18\)  can be found via the substitution method:

\(\displaystyle y + 2x = 18\)

\(\displaystyle 4x + 2x = 18\)

\(\displaystyle 6x = 18\)

\(\displaystyle x=3\)

\(\displaystyle y = 4x= 4 \cdot 3 = 12\)

The lines intersect at \(\displaystyle (3,12)\)

The triangle therefore has these three vertices. It is shown below.

As can be seen, it is a triangle with base 9 and height 12, so its area is 

\(\displaystyle A = \frac{1}{2} \cdot 9 \cdot 12 = 54\)

The vertical segment connecting \(\displaystyle ( 0, 3 )\) and \(\displaystyle (0, -5)\) can be seen as the base of this triangle; this base has length \(\displaystyle 3 - (-5) = 8\). The height is the perpendicular (horizontal) distance from \(\displaystyle (6,9)\) to this segment, which is 6, the same as the \(\displaystyle x\)-coordinate of this point. The area of the triangle is therefore 

\(\displaystyle A = \frac{1}{2} \cdot 6 \cdot 8 = 24\).