5 Mark Question

Question 15 Marks

In the figure given alongside, x : y = 2 : 3 and $\angle\text{ACD}=130^\circ$. Find the values of x, y and z.

Answer

In $\triangle\text{ABC},$ sides BC is produced to D forming exterior $\angle\text{ACD}$

$\angle\text{ACD} = 130^\circ,\angle\text{A} = \text{y}^\circ,\angle\text{B} = \text{x}^\circ$ and $\angle\text{ACB} = \text{z}^\circ$
x : y = 2 : 3
Now, in $\triangle\text{ABC},$
Exterior $\angle\text{ACD} =\angle\text{A}+\angle\text{B}$
$\Rightarrow\angle\text{A}+\angle\text{B}= 130^\circ$
But $\angle\text{A} : \angle\text{B}= 2:3$
$\therefore\angle\text{B}=\frac{130^\circ\times2}{2+3}$
$=\frac{130^\circ\times2}{5}$
$=26^\circ\times2$
$=52^\circ$
and $\angle\text{A}=\frac{130^\circ\times3}{2+3}$
$=\frac{130^\circ\times3}{5}$
$=26^\circ\times3$
$=78^\circ$
But, $\angle\text{A} +\angle\text{B}++\angle\text{ACB} = 180^\circ$ (sum of angles of a triangle)
$\Rightarrow78^\circ+52^\circ+\angle\text{ACB} = 180^\circ$
$\Rightarrow130^\circ+\angle\text{ACB}=180^\circ$
$\Rightarrow\angle\text{ACB}=180^\circ-130^\circ=50^\circ$
Hence, $\angle\text{x}=52^\circ,\angle\text{y}=78^\circ$and $\angle\text{z}=50^\circ$

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Question 25 Marks

In the given figure. DE || BC. If $\angle\text{C}=65^\circ$ and $\angle\text{B}=55^\circ$, find

$\angle\text{ADE}$
$\angle\text{AED}$
$\angle\text{C}$

Answer

DE || BC

$\therefore\angle\text{ABC}=\angle\text{ADE}=55^\circ$ (corresponding angles)

Sum of the angles of any triangle is $180^\circ$.

$\therefore\angle\text{A}+\angle\text{B}+\angle\text{C}=180^\circ$

$\angle\text{C}=180^\circ-(65^\circ-55^\circ)=60^\circ$

DE || BC

$\therefore \angle\text{AED}=\angle\text{ACB}=60^\circ$ (corresponding angles)

We have found in point (ii) that $\angle\text{C}=60^\circ$

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