square root learning task 2 A.tell whether each of the following is a rational or irrational if it is rational give the principal root
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Answer:
Suppose the tower be line AB.
therefore, height of the tower = AB
It is given that, from the point on the other bank directly opposite the tower, angle of elevation of the top of the tower is 60°.
therefore,\displaystyle{\sf{ \angle ACB = 60°}}∠ACB=60°
Again, angle of the elevation from the point D to the top of the tower = 30°
therefore,\displaystyle{\sf{ \angle ADB = 30°}}∠ADB=30°
and DC = 20 m.
To Find : the height of the tower AB and the width of the canal BC.
because the tower is perpendicular to the ground.
\displaystyle{\sf{ \angle ABD = 90°}}∠ABD=90°
In right angled triangle ACB,
\displaystyle{\sf{tan\:C = \dfrac{side\:opposite\:to\:angle\:C}{side\:adjacent\:to\:angle\:C}}}tanC=
sideadjacenttoangleC
sideoppositetoangleC
:\Rightarrow{\sf{tan\:60°= \dfrac{AB}{BC}}}⇒tan60°=
BC
AB
:\Rightarrow{\sf{ \sqrt{3} = \dfrac{AB}{BC}}}⇒
3
=
BC
AB
:\Rightarrow{\sf{ \sqrt{3}BC = AB\:\:\:\:...(1)}}⇒
3
BC=AB...(1)
In right triangle ADB,
\displaystyle{\sf{tan\:D = \dfrac{side\:opposite\:to\:angle\:D}{side\:adjacent\:to\:angle\:D}}}tanD=
sideadjacenttoangleD
sideoppositetoangleD
:\Rightarrow{\sf{tan\:30°= \dfrac{AB}{BD}}}⇒tan30°=
BD
AB
:\Rightarrow{\sf{ \dfrac{1}{ \sqrt{3}}= \dfrac{AB}{BD}}}⇒
3
1
=
BD
AB
:\Rightarrow{\sf{ \dfrac{1}{ \sqrt{3}}= \dfrac{AB}{BD}}}⇒
3
1
=
BD
AB
:\Rightarrow{\sf{ \dfrac{BD}{ \sqrt{3}}=AB\:\:\:...(2)}}⇒
3
BD
=AB...(2)
From (1) and (2),we get
:\Rightarrow{\sf{ \sqrt{3}BC = \dfrac{BD}{ \sqrt{3}}}}⇒
3
BC=
3
BD
:\Rightarrow{\sf{ \sqrt{3}BC \times \sqrt{3}= BD}}⇒
3
BC×
3
=BD
:\Rightarrow{\sf{ 3BC = BD}}⇒3BC=BD
:\Rightarrow{\sf{ 3BC = BC+CD}}⇒3BC=BC+CD
:\Rightarrow{\sf{ 3BC = BC+20}}⇒3BC=BC+20
:\Rightarrow{\sf{ 3BC - BC= 20}}⇒3BC−BC=20
:\Rightarrow{\sf{ 2BC= 20}}⇒2BC=20
:\Rightarrow{\sf{ BC= \dfrac{20}{2}}}⇒BC=
2
20
:\Rightarrow{\sf{ BC= 10}}⇒BC=10
\large{\boxed{\sf{\red{Hence,\:the\:width\:of\:the\:canal\:is\:10m}}}}
Hence,thewidthofthecanalis10m
From (1),
:\Rightarrow{\sf{ \sqrt{3}BC = AB}}⇒
3
BC=AB
:\Rightarrow{\sf{ \sqrt{3}10 = AB}}⇒
3
10=AB
:\Rightarrow{\sf{AB=10 \sqrt{3}}}⇒AB=10
3
\large{\boxed{\sf{\red{Hence,\:the\:height\:of\:the\:tower\:is\:10 \sqrt{3}m}}}}
Hence,theheightofthetoweris10
3
m