5 Mark Question

Question 15 Marks

Calculate the value of x in the given figure.

Answer

Produce CD to cut AB at E.

Now, in $\triangle\text{BDE},$ we have, Exterior $\angle\text{"CDB}=\angle\text{CEB}+\angle\text{DBE}$$\Rightarrow\text{x}^\circ=\angle\text{CEB}+45^\circ\ ....(\text{i)}$
In $\triangle\text{AEC}$ we have, Exterior $\angle\text{CEB}=\angle\text{CAB}+\angle\text{ACE}$$55^\circ+30^\circ=85^\circ$
Putting $\angle\text{CEB}=85^\circ$ in (i), we get,$\text{x}^\circ=85^\circ+45^\circ=130^\circ$
$\therefore\text{x}=130$

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Question 25 Marks

In $\triangle\text{ABC},$ if $3\angle\text{A}=4\angle\text{B}=6\angle\text{C},$ calculate $\angle\text{A},\angle\text{B}$ and $\angle\text{C}.$

Answer

Let $3\angle\text{A}=4\angle\text{B}=6\angle\text{C}=\text{x}$ (say) Then, $3\angle\text{A}=\text{x}$$\Rightarrow\angle\text{A}=\frac{\text{x}}{3}$
$4\angle\text{B}=\text{x}$
$\Rightarrow\angle\text{B}=\frac{\text{x}}{4}$
and $6\angle\text{C}=\text{x}$$\Rightarrow\angle\text{C}=\frac{\text{x}}{6}$
As $\angle\text{A}+\angle\text{B}+\angle\text{C}=180^\circ$$\Rightarrow\frac{\text{x}}{3}+\frac{\text{x}}{4}+\frac{\text{x}}{6}=180$
$\Rightarrow\frac{4\text{x }+\ 3\text{x }+2\text{x}}{12}=180$
$\Rightarrow9\text{x}=180\times12$
$\Rightarrow\text{x}=\frac{180\times12}{9}=240$
$\therefore\angle\text{A}\frac{\text{x}}{3}=\frac{240}{3}=80^\circ$
$\angle\text{B}=\frac{\text{x}}{4}=\frac{240}{4}=60^\circ$
$\angle\text{C}=\frac{\text{x}}{6}=\frac{240}{6}=60^\circ$

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Question 35 Marks

In the given figure, AD divides $\angle\text{BAC}$ in the ratio 1 : 3 and AB = DB. Determine the value of x.

Answer

The angle $\angle\text{BAC}$ is divided by AD in the ratio 1 : 3. Let $\angle\text{BAD}$ and $\angle\text{DAC}$ be y and 3y, respectively. As BAE is a straight line,$\angle\text{BAC}=\angle\text{CAE}=180^\circ $ [linear pair]
$\Rightarrow\angle\text{BAD}+\angle\text{DAC}+\angle\text{CAE}=180^\circ$
$\Rightarrow\text{y}+3\text{y}+108^\circ=180^\circ$
$\Rightarrow4\text{y}=180^\circ-108^\circ=72^\circ$
$\Rightarrow\text{y}=\frac{72^\circ}{4}=18^\circ$
Now, in $\triangle\text{ABC},$$\angle\text{ABC}+\angle\text{BCA}+\angle\text{BAC}=180^\circ$
$\text{y}+\text{x}+4\text{y}=180^\circ$
$[$ Since, $\angle\text{ABC}=\angle\text{BAD}$ (given AD = DB) and $\angle\text{BAC}=\text{y}+3\text{y}=4\text{y}]$
$\Rightarrow5\text{y}+\text{x}=180$
$\Rightarrow5\times18+\text{x}=180$
$\Rightarrow90+\text{x}=180$
$\Rightarrow\text{x}=180-90=90$

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