Question

use your knowledge of triangles to solve problems 5 - 7. 5. determine which groups of three angles could form triangles. 30°, 30°, 90°; 50°, 80°, 50°; 45°, 60°, 75°; 120°, 30°, 45°; 50°, 60°, 70°. 6. what is the longest side in triangle cde? why? 7. what is the largest angle in triangle mno? how do you know? use algebra to solve for the unknown angle(s). 8. what is the measure of ∠u? 9. ∠c is twice the measure of ∠a. if ∠b measures 120°, what are the measures of ∠a and ∠c? 10. in the isosceles triangle vwx, ∠w measures 50°. what are the measures of ∠v and ∠x? 11. investigate the rectangle below is divided into two triangles. based on your knowledge of triangles, what is the sum of the angles in a rectangle? 12. triangle bcd is an isosceles triangle. if sides bc and cd are equal in length, then ∠b must be equal to which other angle? 13. draw can there be a right angle in an isosceles triangle? create a drawing that proves your answer.

Explanation:

Step1: Recall triangle - angle sum property

The sum of interior angles of a triangle is 180°.

Step2: Check each group of angles for triangle - formation

For 30°, 30°, 90°:

$30 + 30+90=150
eq180$, so it cannot form a triangle.

For 50°, 80°, 50°:

$50 + 80+50 = 180$, so it can form a triangle.

For 45°, 60°, 75°:

$45+60 + 75=180$, so it can form a triangle.

For 120°, 30°, 45°:

$120+30 + 45=195
eq180$, so it cannot form a triangle.

For 50°, 60°, 70°:

$50+60 + 70=180$, so it can form a triangle.

Step3: For question 6

In $\triangle CDE$, $\angle C = 40^{\circ}$, $\angle D=60^{\circ}$, $\angle E = 80^{\circ}$. The longest side is opposite the largest angle. So the longest side is $\overline{CD}$ since the largest angle is $\angle E$.

Step4: For question 7

In $\triangle MNO$ with side lengths $MN = 2.6$ cm, $NO=3$ cm, $MO = 4$ cm. By the triangle - side relationship (the largest angle is opposite the longest side), the largest angle is $\angle N$ since the longest side is $MO$.

Step5: For question 8

In right - triangle $STU$ with $\angle S = 90^{\circ}$ and $\angle T=50^{\circ}$, using the angle - sum property of a triangle ($\angle S+\angle T+\angle U=180^{\circ}$), we have $90 + 50+\angle U=180$. Then $\angle U=180-(90 + 50)=40^{\circ}$.

Step6: For question 9

Let $\angle A=x$, then $\angle C = 2x$. Given $\angle B = 120^{\circ}$. Using the angle - sum property of a triangle ($\angle A+\angle B+\angle C=180^{\circ}$), we get $x + 120+2x=180$. Combining like terms: $3x=180 - 120=60$. Solving for $x$, $x = 20^{\circ}$. So $\angle A = 20^{\circ}$ and $\angle C=40^{\circ}$.

Step7: For question 10

In isosceles $\triangle VWX$ with $\angle W = 50^{\circ}$.
Case 1: If $\angle V=\angle X$, then $\angle V=\angle X=\frac{180 - 50}{2}=65^{\circ}$.
Case 2: If $\angle V=\angle W = 50^{\circ}$, then $\angle X=180-(50 + 50)=80^{\circ}$.

Step8: For question 11

A rectangle is divided into two triangles. Since the sum of the angles in a triangle is $180^{\circ}$, and a rectangle is composed of two non - overlapping triangles, the sum of the angles in a rectangle is $360^{\circ}$.

Step9: For question 12

In isosceles $\triangle BCD$ with $BC = CD$, $\angle B=\angle D$ (angles opposite equal sides are equal).

Step10: For question 13

Yes, there can be a right - angle in an isosceles triangle. A right - isosceles triangle has angles 90°, 45°, 45°. You can draw a right - isosceles triangle with two equal sides forming the right - angle.

Answer:

1. The groups of angles that can form triangles are 50°, 80°, 50°; 45°, 60°, 75°; 50°, 60°, 70°.
2. The longest side in $\triangle CDE$ is $\overline{CD}$ because it is opposite the largest angle $\angle E$.
3. The largest angle in $\triangle MNO$ is $\angle N$ as it is opposite the longest side $MO$.
4. $\angle U = 40^{\circ}$
5. $\angle A = 20^{\circ}$, $\angle C = 40^{\circ}$
6. $\angle V=\angle X = 65^{\circ}$ or $\angle V = 50^{\circ},\angle X=80^{\circ}$
7. $360^{\circ}$
8. $\angle D$
9. Yes, a right - isosceles triangle has angles 90°, 45°, 45°.