1.1 Angles and Triangles

Because the sum of the angles is ${180}^{\circ }$, we know that $x+2x+2x-15=180$. Solving this equation gives $$x={39}^{\circ },2x={78}^{\circ },2x-15={63}^{\circ }$$

No, the third angle must be ${100}^{\circ }$.

The third angle is $x={60}^{\circ }$, so the triangle is equilateral and $y=8,z=8$.

The vertex angle is ${180}^{\circ }-({20}^{\circ }+{20}^{\circ })={140}^{\circ }$, and the triangle is isosceles so $y=9$.

Angle $\alpha$ is supplementary to ${50}^{\circ }$ so $\alpha ={130}^{\circ }$. Then $\beta$ is supplementary to $\alpha$ and $\gamma$ is supplementary to $\beta$, giving us $\beta ={50}^{\circ },\gamma ={130}^{\circ }.$

$\alpha ={40}^{\circ }$ because of vertical angles, and $\beta$ is its supplement at ${40}^{\circ }$. $\delta ={65}^{\circ }$, the supplement of ${115}^{\circ }$, and $\gamma =180-(65+40)={75}^{\circ }$.

Note that angles 2 and 6 are supplementary because they form a straight angle. Angle 1 equals angle 2 because they are alternate interior angles, and similarly angle 4 equals angle 5. Angle 5 equals angle 6 because they are corresponding angles. Thus, angle 4 equals angle 6, and angle 1 equals angle 2. So angles 4 and 1 are supplementary because 2 and 6 are.

$$\theta ={108.8}^{\circ }$$

$$\alpha ={29}^{\circ }$$

$$\beta ={77}^{\circ }$$

$$\alpha ={12}^{\circ }$$

$$\theta ={65}^{\circ }$$

$$\theta ={12}^{\circ }$$

$$\psi ={73}^{\circ }$$

$$\varphi ={88}^{\circ }$$

1. $\varphi ={120}^{\circ }$
2. $\varphi ={160}^{\circ }$
3. $\varphi =\alpha +\beta$
4. An exterior angle is equal to the sum of the opposite interior angles.

$$\theta ={72}^{\circ },\varphi ={54}^{\circ }$$

$$\theta ={100}^{\circ },\varphi ={30}^{\circ }$$

1. ${180}^{\circ }$
2. ${90}^{\circ }$
3. a right triangle

1. They are base angles of an isosceles triangle.
2. They are base angles of an isosceles triangle.
3. $\mathrm{\angle }OAB$ corresponds to $\theta$ of Problem 29, and $\mathrm{\angle }OBC$ corresponds to $\varphi$ of Problem 29.

$$\alpha ={30}^{\circ },\beta ={60}^{\circ }$$

$$x={47}^{\circ },y={133}^{\circ }$$

$$x={60}^{\circ },y={15}^{\circ }$$

$$x={100}^{\circ },y={16}^{\circ }$$

$$x={90}^{\circ },y={55}^{\circ }$$

$$x={50}^{\circ },y={80}^{\circ }$$

1. $\mathrm{\angle }1=\mathrm{\angle }4,\mathrm{\angle }3=\mathrm{\angle }5$
2. ${180}^{\circ }$
3. In the equation $\mathrm{\angle }4+\mathrm{\angle }2+\mathrm{\angle }5={180}^{\circ },$ substitute $\mathrm{\angle }1$ for $\mathrm{\angle }4$, and substitute $\mathrm{\angle }3$ for $\mathrm{\angle }5$ to conclude that the sum of the angles in the triangle is ${180}^{\circ }$.

$\mathrm{\angle }1={130}^{\circ }$ because vertical angles are equal. $\mathrm{\angle }2={50}^{\circ }$ because it makes a straight angle with a ${130}^{\circ }$ angle. $\mathrm{\angle }3={65}^{\circ }$ because it is a base angle of an isosceles triangle whose vertex angle is ${50}^{\circ }$. $\mathrm{\angle }4={65}^{\circ }$ for the same reason. $\mathrm{\angle }5={25}^{\circ }$ because it is complementary to $\mathrm{\angle }4$.