Question

Activities-
(1) In the figure below, the measures of two angles are shown. Write the third angle as a sum or difference and compute its measure? (1 Marks)
(2) In the figure below, the measures of two angles are shown. Write the third angle as a sum or difference and compute its measure? (1 Marks)


(3) In the figure below, $\angle ACD=\angle BCE.$ How much is each? (3 Marks)

Answer 1

(1) $\angle DAB=\angle DAC+\angle CAB$
From the figure, $\angle DAC={40}^{\circ }$ and $\angle CAB={60}^{\circ }$
$\begin{array}{cc}& \Rightarrow \angle DAB={40}^{\circ }+{60}^{\circ }\\ & \Rightarrow \angle DAB={100}^{\circ }\end{array}$

(2) $\angle CAB=\angle DAB-\angle DAC$
Given that $\angle DAB={80}^{\circ }$ and $\angle DAC={50}^{\circ }$
$\begin{array}{cc}\therefore \angle CAB& ={80}^{\circ }-{50}^{\circ }\\ \Rightarrow & \angle CAB={30}^{\circ }\end{array}$
(3) Let us label the figure as follows for better calculation.
From the figure,
$\angle BCD=y^{\circ };\angle ECD={30}^{\circ };\angle ACE=x^{\circ }$
We need to find the value of $x^{\circ }$ and $y^{\circ }$
We know that $\angle ACE+\angle ECD+\angle BCD={180}^{\circ }$
i.e., $x^{\circ }+{30}^{\circ }+y^{\circ }={180}^{\circ }$
$\Rightarrow x^{\circ }+y^{\circ }={180}^{\circ }-{30}^{\circ }$
$\Rightarrow x^{\circ }+y^{\circ }={150}^{\circ }\dots ($ Eq. 1)
Given that $\angle ACD=\angle BCE$
From the figure,
$\begin{array}{cc}\angle ACD& =\angle ACE+\angle ECD\\ & \Rightarrow \angle ACD=x^{\circ }+{30}^{\circ }\text{and}\\ \angle BCE& =\angle ECD+\angle BCD\\ & \Rightarrow \angle BCE={30}^{\circ }+y^{\circ }\\ \text{So,}\angle ACD=\angle BCE& \\ & \Rightarrow x^{\circ }+{30}^{\circ }={30}^{\circ }+y^{\circ }\\ & \Rightarrow x^{\circ }={30}^{\circ }-{30}^{\circ }+y^{\circ }\\ & \Rightarrow x^{\circ }=y^{\circ }\\ \text{Put}& \text{the value of}{x}^{\circ }=y^{\circ }\text{in}\left(\text{Eq.}1\right)\\ & \Rightarrow x^{\circ }+x^{\circ }={150}^{\circ }\\ & \Rightarrow 2x^{\circ }={150}^{\circ }\\ & \Rightarrow x^{\circ }=\frac{{150}^{\circ }}{2}\\ & \Rightarrow x^{\circ }={75}^{\circ }\\ & \Rightarrow \text{Thus,}{y}^{\circ }=x^{\circ }={75}^{\circ }\end{array}$