Question

How do you solve this? If you can answer on paper

Answer 1

Answer:

Question 19: b. 19 cm.

Question 20: d. B, A, C.

Question 21: b. BC, AB, AC.

Step-by-step explanation:

19.

The triangle inequality states that

• The sum of two sides of a triangle is greater than the length of the third side;
• The difference between the length of the two sides of a triangle is smaller than the length of the third side.

Apply this inequality to this problem:

• The sum of the two given sides is $\rm 20\; cm$;
• The difference between the two given sides is $\rm 4\; cm$.

As a result, the length of the third side should be between these two numbers, excluding the endpoints.

Hence the only valid choice is b. 19 cm. Note that 20 cm will not work for the two three sides will form a line, not a 2D triangle.

20.

Here's one way to draw a circle given the length of its three sides, with the help of a compass.

1. Start by drawing a segment with the length of the longest side of the triangle. Label the two endpoints with the appropriate letters. In this case, draw a segment labeled "BC" with a length of 20 (e.g., centimeters.)
2. Draw two circle centered at the two endpoints of the segments. The radius of the each circle is the same as length of each side of the triangle. In this case, draw a circle of radius 12 centered at B and a circle of radius 18 centered at C.
3. The two circles shall intersect twice if the three sides indeed make a triangle. Connect the two endpoints of the first segment with one of the two intersects. In this case, connect point B and point C with the intersection of the two circles centered at B and C.

If the angles are close in sizes, the sine rule might help.

The sine rule states that

• In a triangle, the length of each side is proportional to the sine of the angle opposite to it.

For example, in triangle $\triangle \rm ABC$,

$\displaystyle \frac{\rm AB}{\sin{C}} = \frac{\rm BC}{\sin{A}} = \frac{\rm AC }{\sin{B}}$.

Additionally, the sine angle is increasing for acute angles. Additionally, at least two of the angles in a triangle are acute. The problem is that in case one of the angle is obtuse, simply applying the sine rule might give misleading results. Therefore, it is recommended that when the length of the sides are given, start by sketching the triangle (e.g., with a compass.) Determine the largest angle. The sine rule could help comparing the size of the other two smaller and acute angles.

In this case, draw a normal at the vertex of the angle that appears to be the largest. For example, in question 20 angle B appears to be the largest. Draw a normal to segment AB or BC at point B to tell whether angle B is acute or obtuse. If it's obtuse, it is always going to be the largest in this triangle regardless of its sine value. In this case, angle B is acute. Apply the sine rule to deduce that

$\hat{B} > \hat{A} > \hat{C}$.

21.

When working backwards from the measure of the angles, directly apply the sine rule. Keep in mind that for $90^{\circ} < \theta < 180^{\circ}$, $\sin{\theta} = \sin{(\theta - 90^{\circ})}$. If one angle is obtuse, apply this identity to make it acute. After that, the length of the opposite sides can be compared using their equivalent acute angles.

In this question, all three angles are acute.

$90^{\circ}>\hat{A} > \hat{C} > \hat{B}$,

$\implies \sin{\hat{A}} >\sin{\hat{C}}> \sin{\hat{B}}$

$\implies\rm BC > AB> AC$.