Chapter 15.2, Problem 12PSB

a.

To determine

To find: the restrictions on $x$ in the given figure.

Answer to Problem 12PSB

  $x<{110}^{\circ },55<x<{110}^{\circ }$ .

Explanation of Solution

Given:

The given figure is shown below:

  

Calculation:

An exterior angle of a triangle is greater than either remote interior angle.

Therefore, ${55}^{\circ }<x$

  $\begin{array}{l}{x}^{\circ }+y^{\circ }+{70}^{\circ }={180}^{\circ }\\ \Rightarrow x^{\circ }+y^{\circ }={110}^{\circ }\\ \Rightarrow x^{\circ }={110}^{\circ }-y^{\circ }\\ \therefore x<{110}^{\circ },55<x<{110}^{\circ }\end{array}$

Hence, the restrictions on $x$ is $x<{110}^{\circ },55<x<{110}^{\circ }$

b.

To determine

To find: the restrictions on $x$ in the given figure.

Answer to Problem 12PSB

  $x<{90}^{\circ }$ .

Explanation of Solution

Given:

The given figure is shown below:

  

Calculation:

As the sum of all angle of the triangle are supplementary.

Therefore, ${55}^{\circ }<x$

  $\begin{array}{l}{x}^{\circ }+x^{\circ }+y^{\circ }={180}^{\circ }\\ \Rightarrow 2x^{\circ }+y^{\circ }={180}^{\circ }\\ \Rightarrow y^{\circ }={180}^{\circ }-2x^{\circ }\\ \therefore x<{90}^{\circ }\end{array}$

Hence, the restrictions on $x$ is $x<{90}^{\circ }$

c.

To determine

To find: the restrictions on $x$ in the given figure.

Answer to Problem 12PSB

  $x<20$ .

Explanation of Solution

Given:

The given figure is shown below:

  

Calculation:

As the sum of any two sides is greater than its third side. Therefore,

  $\begin{array}{l}x<10+10\\ \Rightarrow x<20\end{array}$

Hence, the restrictions on $x$ is $x<20$

d.

To determine

To find: the restrictions on $x$ in the given figure.

Answer to Problem 12PSB

  $x=7.5\sqrt{3}$ .

Explanation of Solution

Given:

The given figure is shown below:

  

Calculation:

The triangle must be $30,60,90$ . Therefore,

  $\begin{array}{l}{\left(15\right)}^2={\left(7.5\right)}^2+x^2\left\{\text{by Pythagoras theorem}\right\}\\ \Rightarrow 225=56.25+x^2\\ \Rightarrow x^2=225-56.25=168.75\\ \Rightarrow x=\sqrt{168.75}=7.5\sqrt{3}\end{array}$

Hence, the restrictions on $x$ is $x=7.5\sqrt{3}$