[3 marks sum]

Question 13 Marks

Use the information given to find the length of $AB.$

Answer

From right $\triangle AQP$
$\tan 30^{\circ}=\frac{ AQ }{ AP } $
$ \frac{1}{\sqrt{3}}=\frac{10}{ AP }$
$ AP =10 \sqrt{3}$
Also from $\triangle PBR$
$\tan 45^{\circ}=\frac{ PB }{ BR }$
$1=\frac{ PB }{8}$
Therefore,
$AB=AP+PB=10 \sqrt{3}+8$.

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Question 23 Marks

Find $AD,$ if :

Answer

From the $\triangle ABD$ we have
$\sin B=\frac{AD}{A B}$
$\sin 30=\frac{AD}{100} ...[$Since $\angle ACD$ is the exterior angle of the $\triangle ABC]$
$\frac{1}{2}=\frac{AD}{100} $
$ AD=50\ m$

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Question 33 Marks

Find $AD,$ if :

Answer

From the right $\triangle ABE$
$\tan 45^{\circ}=\frac{ AE }{ BE }$
$1=\frac{ AE }{ BE }$
$AE = BE$
Therefore $AE = BE = 50\ m.$
Now from the rectangle $\text{BCDE}$ we have
$DE = BC = 10\ m.$
Therefore the length of $AD$ will be:
$AD = AE + DE = 50 + 10 = 60\ m.$

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Question 43 Marks

In the given figure, $AB$ and $EC$ are parallel to each other. Sides $AD$ and $BC$ are $2 \ cm$ each and are perpendicular to $AB.$

Given that $\angle AED = 60^\circ $ and $\angle ACD = 45^\circ .$ Calculate$: AB.$

Answer

From the $\triangle ADC$ we have
$\tan 45^\circ =\frac{ AD }{ DC }$
$1=\frac{2}{D C}$
$DC = 2$
Since $AD \| DC$ and $AD\perp EC,$
$\text{ABCD}$ is a parallelogram and hence opposite sides are equal.
Therefore $ AB = DC = 2 \ cm.$

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Question 53 Marks

Find the length of $AB.$

Answer

From right $\triangle = ADE$
$\tan 45^{\circ}=\frac{ AE }{ DE } $
$1=\frac{ AE }{30} $
$AE =30 \ cm$
Also, from triangle $DBE$
$\tan 60^{\circ}=\frac{ BE }{ DE }$
$\sqrt{3}=\frac{ BE }{30}$
$BE =30 \sqrt{3} \ cm$
Therefore $AB=AE+BE=30+30 \sqrt{3}$
$=30(1+\sqrt{3}) \ cm$

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MATHEMATICS [3 marks sum]

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