The length of the rope is 20 m, and the angle made by the rope with the ground level is 30°.
Given: AC = 20 m and angle C = 30°
To find: Height of the pole
Let AB be the vertical pole
In the right $\triangle$ABC, using the sine formula
sin 30° =$ \dfrac{AB}{AC}$
Using the value of sin 30 degrees is $\dfrac{1}{2}$, we have
$\dfrac{1}{2}$ = $\dfrac{AB}{20}$
AB = $\dfrac{20}{2} $
AB = 10
Therefore, the height of the pole is 10 m.
Using the given instructions, draw a figure. Let AC be the broken part of the tree. Angle C = 30°
BC = 8 m
To find: Height of the tree, which is AB
From figure: Total height of the tree is the sum of AB and AC, i.e., AB+AC
In the right $\triangle$ABC,
Using Cosine and tangent angles,
cos 30° = $\dfrac{BC}{AC}$
We know that, cos 30° = $\dfrac{\sqrt{3}}{2}$
$\dfrac{\sqrt{3}}{2}$ = $\dfrac{8}{AC}$
AC = $\dfrac{16}{\sqrt{3}}$ …(1)
Also,
tan 30° = $\dfrac{AB}{BC}$
$ \dfrac{1}{\sqrt{3}}$ = $\dfrac{AB}{8}$
AB = $\dfrac{8}{\sqrt{3}}$ ….(2)
Therefore, total height of the tree = AB + AC
= $\dfrac{16}{\sqrt{3}} + \dfrac{8}{\sqrt{3}}$
= $\dfrac{24}{\sqrt{3}}$
= $\dfrac{8}{\sqrt{3}}$ m.
As per the contractor’s plan,
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Let ABC is the slide inclined at 30° with length AC, and PQR be the slide inclined at 60° with length PR.
To Find: AC and PR
In the right $\triangle$ABC,
sin 30° = $\dfrac{AB}{AC}$
$\dfrac{1}{2}$ = $ \dfrac{1.5}{AC}$
AC = 3
Also,
In the right $\triangle$PQR,
sin 60° = $\dfrac{PQ}{PR}$
⇒ $\dfrac{\sqrt{3}}{2}$ = $\dfrac{3}{PR}$
⇒ PR = $2\sqrt{3}$
Hence, the length of the slide below 5 = 3 m and
Length of the slide for elders children =$ 2\sqrt{3}$ m
Let AB be the height of the tower and C be the point elevation, which is 30 m away from the foot of the tower.
To find: AB (height of the tower)
In the right ABC
tan 30° = $\dfrac{AB}{BC}$
$\dfrac{1}{\sqrt{3}}$ =$ \dfrac{AB}{30}$
⇒ AB = $10\sqrt{3}$
Thus, the height of the tower is $10\sqrt{3}$ m.
Draw a figure based on the given instructions,
Let BC = Height of the kite from the ground, BC = 60 m
AC = Inclined length of the string from the ground and
A is the point where the string of the kite is tied.
To find: Length of the string from the ground, i.e., the value of AC
From the above figure,
sin 60° = $\dfrac{BC}{AC}$
⇒ $\dfrac{\sqrt{3}}{2}$ = $\dfrac{60}{AC}$
⇒ AC = $40\sqrt{3} m$
Thus, the length of the string from the ground is $40\sqrt{3}m.$
Let the boy initially stand at point Y with an inclination 30°, and then he approaches the building to point X with an inclination 60°.
To find: The distance boy walked towards the building, i.e., XY
From figure,
XY = CD.
Height of the building = AZ = 30 m
AB = AZ – BZ = 30 – 1.5 = 28.5
The measure of AB is 28.5 m
In the right $\triangle$ABD,
tan 30° = $ \dfrac{AB}{BD}$
$ \dfrac{1}{\sqrt{3}}$ = $ \dfrac{28.5}{BD}$
BD = 28.5$\sqrt{3}m$
Again,
In the right $\triangle$ABC,
tan 60° = $\dfrac{AB}{BC}$
$\sqrt{3}$ = $\dfrac{28.5}{BC}$
BC = $\dfrac{28.5}{\sqrt{3}}$
= $\dfrac{28.5\sqrt{3}}{3}$
Therefore, the length of BC is $\dfrac{28.5\sqrt{3}}{3} m.$
XY = CD = BD – BC
= $(28.5\sqrt{3}- \dfrac{28.5\sqrt{3}}{3})$
= $28.5\sqrt{3}(1-\dfrac{1}{3}) $
= $28.5\sqrt{3} × \dfrac{2}{3} $
= $\dfrac{57}{\sqrt{3}} m.$
Thus, the distance boy walked towards the building is $\dfrac{57}{\sqrt{3}} m.$
Let BC be the 20 m high building.
D is the point on the ground from where the elevation is taken.
Height of transmission tower = AB = AC – BC
To find: AB, the height of the tower
From the figure, in the right $\triangle$BCD,
tan 45° = $\dfrac{BC}{CD}$
1 = $\dfrac{20}{CD}$
CD = 20
Again,
In the right $\triangle$ACD,
tan 60° = $\dfrac{AC}{CD}$
$\sqrt{3}$=$ \dfrac{AC}{20}$
AC = $20\sqrt{3}$
Now, AB = AC – BC
=$ (20\sqrt{3}-20)$
= $20(\sqrt{3}-1)$
Height of transmission tower = $20(\sqrt{3} – 1) m.$
Let AB be the height of the statue.
D is the point on the ground from where the elevation is taken.
To find: The height of pedestal = BC = AC-AB
From figure,
In the right triangle BCD,
tan 45° =$ \dfrac{BC}{CD}$
1 = $\dfrac{BC}{CD}$
BC = CD …..(1)
Again,
In the right $\triangle$ACD,
tan 60° =$\dfrac{AC}{CD}$
$\sqrt{3} $= $\dfrac{( AB+BC)}{CD}$
$\sqrt{3}CD$ = 1.6 + BC
$\sqrt{3}BC$ = 1.6 + BC (using equation (1)
$\sqrt{3}BC – BC$ = 1.6
$BC(\sqrt{3}-1)$ = 1.6
BC = $\dfrac{1.6}{(\sqrt{3}-1)} m$
BC = $0.8(\sqrt{3}+1)$
Thus, the height of the pedestal is $0.8(\sqrt{3}+1) m.$
Let CD be the height of the tower.
AB be the height of the building. BC be the distance between the foot of the building and the tower.
The elevation is 30 degrees and 60 degrees from the tower and the building, respectively.
In the right $\triangle$BCD,
tan 60° = $\dfrac{CD}{BC}$
$\sqrt{3}$ =$ \dfrac{50}{BC}$
BC = $\dfrac{50}{\sqrt{3}}$ …(1)
Again,
In the right $\triangle$ABC,
tan 30° = $\dfrac{AB}{BC}$
⇒$ \dfrac{1}{\sqrt{3}} $=$ \dfrac{AB}{BC}$
Use the result obtained in equation (1)
AB =$ \dfrac{50}{3}$
Thus, the height of the building is $\dfrac{50}{3}$ m.
Let AB and CD be the poles of equal height.
O is the point between them from where the height of elevation is taken. BD is the distance between the poles.
As per the above figure, AB = CD,
OB + OD = 80 m
Now,
In the right $\triangle$CDO,
tan 30° = $\dfrac{CD}{OD}$
$\dfrac{1}{\sqrt{3}}$ = $\dfrac{CD}{OD}$
CD = $\dfrac{OD}{\sqrt{3}}$ … (1)
Again,
In the right $\triangle$ABO,
tan 60° = $\dfrac{AB}{OB}$
$\sqrt{3}$ = $\dfrac{AB}{(80-OD)}$
AB = $\sqrt{3}(80-OD)$
AB = CD (Given)
$\sqrt{3}(80-OD)$ =$ \dfrac{OD}{\sqrt{3}}$ (Using equation (1))
3(80-OD) = OD
240 – 3 OD = OD
4 OD = 240
OD = 60
Putting the value of OD in equation (1)
CD = $\dfrac{OD}{\sqrt{3}}$
CD = $\dfrac{60}{\sqrt{3}}$
CD = $20\sqrt{3} m$
Also,
OB + OD = 80 m
⇒ OB = (80-60) m = 20 m
Thus, the height of the poles is $20\sqrt{3} m$, and the distance from the point of elevation is 20 m and 60 m, respectively.