Let's label the vertices of the quadrilateral and focus on the two triangles.
Using these newly added point labels, we can summarize what we know about triangles â–³ ADC and â–³ ABC.
Claim
Justification
DC< BC
14<15
AD≅AB
Segments of the same length are congruent.
Since AC is a common side of triangles â–³ ADC and â–³ ABC, these triangles have two pairs of congruent sides.
This means that we can use the Converse of the Hinge Theorem.
DC< BC
⇓
m∠DAC< m∠BAC
Let's substitute the value and the expression given on the diagram so that we may solve the resulting inequality for x.
Combining these inequalities, we have a lower and upper bound for the value of x.
2.8
Extra
Can x be any value within these two bounds?
In our solution, we found an upper and lower bound for x.
However, we can actually do more than this.
By constructing a scaled diagram, we can find that only one value of x is possible.
Let's use the labels from the solution and start the construction by drawing segment AB with a length of 12.
To continue, let's use a protractor to measure 46^(∘) and draw a ray in the direction of vertex C.
We can open our compass to 15 and use it to mark the position of C on the ray.
We can use our compass opened to 12 and 14 to draw two arcs to mark the position of vertex D.
This construction shows that there is only one way to draw a quadrilateral with measurements given on the figure.
We can now use our protractor to measure the angle marked as (5x-14)^(∘).
