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Question Number 11475 by tawa last updated on 26/Mar/17

Answered by sm3l2996 last updated on 26/Mar/17

Question 105:(B) Question 106: ΔAFD=(x^2 /2)=((25)/2)=12.5 (C)

$$\mathrm{Question}\:\mathrm{105}:\left(\mathrm{B}\right) \\ $$$$\mathrm{Question}\:\mathrm{106}: \\ $$$$\Delta\mathrm{AFD}=\frac{\mathrm{x}^{\mathrm{2}} }{\mathrm{2}}=\frac{\mathrm{25}}{\mathrm{2}}=\mathrm{12}.\mathrm{5}\:\left(\mathrm{C}\right) \\ $$

Commented by tawa last updated on 26/Mar/17

please workings

$$\mathrm{please}\:\mathrm{workings} \\ $$

Commented by sm3l2996 last updated on 26/Mar/17

AF=BE=(√(x^2 +x^2 ))=x(√2) and AD=BC=x so DF=(√(AF^2 −AD^2 ))=(√(2x^2 −x^2 ))=x ΔAFD=((DF×AD)/2)=(x^2 /2)

$$\mathrm{AF}=\mathrm{BE}=\sqrt{\mathrm{x}^{\mathrm{2}} +\mathrm{x}^{\mathrm{2}} }=\mathrm{x}\sqrt{\mathrm{2}} \\ $$$$\mathrm{and}\:\mathrm{AD}=\mathrm{BC}=\mathrm{x} \\ $$$$\mathrm{so}\:\mathrm{DF}=\sqrt{\mathrm{AF}^{\mathrm{2}} −\mathrm{AD}^{\mathrm{2}} }=\sqrt{\mathrm{2x}^{\mathrm{2}} −\mathrm{x}^{\mathrm{2}} }=\mathrm{x} \\ $$$$\Delta\mathrm{AFD}=\frac{\mathrm{DF}×\mathrm{AD}}{\mathrm{2}}=\frac{\mathrm{x}^{\mathrm{2}} }{\mathrm{2}} \\ $$

Commented by tawa last updated on 27/Mar/17

i grab. please working for (105)

$$\mathrm{i}\:\mathrm{grab}.\:\mathrm{please}\:\mathrm{working}\:\mathrm{for}\:\left(\mathrm{105}\right) \\ $$

Commented by tawa last updated on 27/Mar/17

God bless you

$$\mathrm{God}\:\mathrm{bless}\:\mathrm{you} \\ $$$$ \\ $$

Commented by sm3l2996 last updated on 27/Mar/17

Commented by tawa last updated on 27/Mar/17

I really appreciate sir. God bless you.

$$\mathrm{I}\:\mathrm{really}\:\mathrm{appreciate}\:\mathrm{sir}.\:\mathrm{God}\:\mathrm{bless}\:\mathrm{you}. \\ $$

Commented by sma3l2996 last updated on 27/Mar/17

you welcome

$${you}\:{welcome} \\ $$

Answered by mrW1 last updated on 27/Mar/17

Q105 let′s say someone goes from west to east. he turns z° to the left, then 180−x° to the right,.... since he finally goes to east again, it means that he totally turned 0° z−(180−x)+(180−x)−90+y=0 ⇒z=90−y=90−60=30°

$${Q}\mathrm{105} \\ $$$${let}'{s}\:{say}\:{someone}\:{goes}\:{from}\:{west}\:{to} \\ $$$${east}.\:{he}\:{turns}\:{z}°\:{to}\:{the}\:{left},\:{then} \\ $$$$\mathrm{180}−{x}°\:{to}\:{the}\:{right},....\:{since}\:{he} \\ $$$${finally}\:{goes}\:{to}\:{east}\:{again},\:{it}\:{means}\:{that} \\ $$$${he}\:{totally}\:{turned}\:\mathrm{0}° \\ $$$${z}−\left(\mathrm{180}−{x}\right)+\left(\mathrm{180}−{x}\right)−\mathrm{90}+{y}=\mathrm{0} \\ $$$$\Rightarrow{z}=\mathrm{90}−{y}=\mathrm{90}−\mathrm{60}=\mathrm{30}° \\ $$$$ \\ $$

Commented by mrW1 last updated on 27/Mar/17

Commented by tawa last updated on 27/Mar/17

Thanks sir. God bless you.

$$\mathrm{Thanks}\:\mathrm{sir}.\:\mathrm{God}\:\mathrm{bless}\:\mathrm{you}. \\ $$

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