44115.From a tower of 80 m high, the angle of depression of a bus is 30°. How far is the bus from the tower?
40 m
138.4 m
46.24 m
160 m
Explanation:
Let AC be the tower and B be the position of the bus.
Then BC = the distance of the bus from the foot of the tower.
Given that height of the tower, AC = 80 m and the angle of depression, $\angle$DAB = 30°
$\angle$ABC = $\angle$DAB = 30° (because DA || BC)
tan30°=$\dfrac{AC}{BC}$
=>tan30°=$\dfrac{80}{BC}$
=>BC=$\dfrac{80}{tan30°}=\dfrac{80}{\left(\dfrac{1}{\sqrt{3}}\right)}$
=80×1.73=138.4 m
i.e., Distance of the bus from the foot of the tower = 138.4 m
Let AC be the tower and B be the position of the bus.
Then BC = the distance of the bus from the foot of the tower.
Given that height of the tower, AC = 80 m and the angle of depression, $\angle$DAB = 30°
$\angle$ABC = $\angle$DAB = 30° (because DA || BC)
tan30°=$\dfrac{AC}{BC}$
=>tan30°=$\dfrac{80}{BC}$
=>BC=$\dfrac{80}{tan30°}=\dfrac{80}{\left(\dfrac{1}{\sqrt{3}}\right)}$
=80×1.73=138.4 m
i.e., Distance of the bus from the foot of the tower = 138.4 m
44121.Two ships are sailing in the sea on the two sides of a lighthouse. The angle of elevation of the top of the lighthouse is observed from the ships are 30° and 45° respectively. If the lighthouse is 100 m high, the distance between the two ships is:
173 m
200 m
273 m
300 m
Explanation:
Let AB be the lighthouse and C and D be the positions of the ships.
Then, AB = 100 m, $\angle$ACB = 30° and $\angle$ADB = 45°.
$\dfrac{AB}{AC}$=tan 30°=$\dfrac{1}{\sqrt{3}}\Rightarrow$ AC = AB x $\sqrt{3}$ = 100$\sqrt{3}$ m.
$\dfrac{AB}{AD}$=tan 45°=1 $\Rightarrow$AD = AB = 100 m.
$\therefore$ CD = (AC + AD) = (100$\sqrt{3}$ + 100) m
= 100($\sqrt{3}$ + 1)
= (100 x 2.73) m
= 273 m.
Let AB be the lighthouse and C and D be the positions of the ships.
Then, AB = 100 m, $\angle$ACB = 30° and $\angle$ADB = 45°.
$\dfrac{AB}{AC}$=tan 30°=$\dfrac{1}{\sqrt{3}}\Rightarrow$ AC = AB x $\sqrt{3}$ = 100$\sqrt{3}$ m.
$\dfrac{AB}{AD}$=tan 45°=1 $\Rightarrow$AD = AB = 100 m.
$\therefore$ CD = (AC + AD) = (100$\sqrt{3}$ + 100) m
= 100($\sqrt{3}$ + 1)
= (100 x 2.73) m
= 273 m.
44123.The angle of elevation of a ladder leaning against a wall is 60º and the foot of the ladder is 4.6 m away from the wall. The length of the ladder is:
2.3 m
4.6 m
7.8 m
9.2 m
Explanation:
Let AB be the wall and BC be the ladder.
Then, $\angle$ACB = 60º and AC = 4.6 m.
$\dfrac{AB}{BC}$=cos 60°=1/2
$\Rightarrow$ BC = 2 x AC
= (2 x 4.6) m
= 9.2 m.
Let AB be the wall and BC be the ladder.
Then, $\angle$ACB = 60º and AC = 4.6 m.
$\dfrac{AB}{BC}$=cos 60°=1/2
$\Rightarrow$ BC = 2 x AC
= (2 x 4.6) m
= 9.2 m.
44125.A balloon leaves the earth at a point A and rises vertically at uniform speed. At the end of 2 minutes, John finds the angular elevation of the balloon as 60°. If the point at which John is standing is 150 m away from point A, what is the speed of the balloon?
0.63 meter/sec
2.16 meter/sec
3.87 meter/sec
0.72 meter/sec
Explanation:
Let C be the position of John. Let A be the position at which balloon leaves the earth and B be the position of the balloon after 2 minutes.
Given that CA = 150 m, $\angle$BCA = 60°
tan60°=$\dfrac{BA}{CA}$
=>√3=$\dfrac{BA}{150}$
BA=150√3
i.e, the distance travelled by the balloon =150√3 meters
time taken = 2 min = 2 × 60 = 120 seconds
Speed =$\dfrac{Distance}{Time}$
=$\dfrac{150\sqrt{3}}{120}$=1.25√3
=1.25×1.73=2.16 meter/second
Let C be the position of John. Let A be the position at which balloon leaves the earth and B be the position of the balloon after 2 minutes.
Given that CA = 150 m, $\angle$BCA = 60°
tan60°=$\dfrac{BA}{CA}$
=>√3=$\dfrac{BA}{150}$
BA=150√3
i.e, the distance travelled by the balloon =150√3 meters
time taken = 2 min = 2 × 60 = 120 seconds
Speed =$\dfrac{Distance}{Time}$
=$\dfrac{150\sqrt{3}}{120}$=1.25√3
=1.25×1.73=2.16 meter/second
44126.An observer 1.6 m tall is 203 away from a tower. The angle of elevation from his eye to the top of the tower is 30º. The heights of the tower is:
21.6 m
23.2 m
24.72 m
None of these
Explanation:
Let AB be the observer and CD be the tower.
Draw BE $\bot$ CD.
Then, CE = AB = 1.6 m,
BE = AC = 20$\sqrt{3}$ m.
$\dfrac{DE}{BE}$ = tan 30°=1/$\sqrt{3}$
$\Rightarrow \dfrac{20\sqrt{3}}{\sqrt{3}}$m=20 m
$\therefore$ CD = CE + DE = (1.6 + 20) m = 21.6 m.
Let AB be the observer and CD be the tower.
Draw BE $\bot$ CD.
Then, CE = AB = 1.6 m,
BE = AC = 20$\sqrt{3}$ m.
$\dfrac{DE}{BE}$ = tan 30°=1/$\sqrt{3}$
$\Rightarrow \dfrac{20\sqrt{3}}{\sqrt{3}}$m=20 m
$\therefore$ CD = CE + DE = (1.6 + 20) m = 21.6 m.
44134.The angle of elevation of the sun, when the length of the shadow of a tree 3 times the height of the tree, is:
30º
45º
60º
90º
Explanation:
Let AB be the tree and AC be its shadow.
Let $\angle$ACB = $\theta$
$\dfrac{AC}{AB}$= $\sqrt{3}$=$\Rightarrow$ cot$\theta$= $\sqrt{3}$
$\therefore \theta$ = 30º.
Let AB be the tree and AC be its shadow.
Let $\angle$ACB = $\theta$
$\dfrac{AC}{AB}$= $\sqrt{3}$=$\Rightarrow$ cot$\theta$= $\sqrt{3}$
$\therefore \theta$ = 30º.
44140.A ladder 10 m long just reaches the top of a wall and makes an angle of 60° with the wall.Find the distance of the foot of the ladder from the wall (√3=1.73)
4.32 m
17.3 m
5 m
8.65 m
Explanation:
Let BA be the ladder and AC be the wall as shown above.
Then the distance of the foot of the ladder from the wall = BC
Given that BA = 10 m , $\angle$BAC = 60°
sin 60°=$\dfrac{BC}{BA}$
$\dfrac{\sqrt{3}}{2}=\dfrac{BC}{10}$
BC = 10×$\dfrac{\sqrt{3}}{2}$
=5×1.73
=8.65 m
Let BA be the ladder and AC be the wall as shown above.
Then the distance of the foot of the ladder from the wall = BC
Given that BA = 10 m , $\angle$BAC = 60°
sin 60°=$\dfrac{BC}{BA}$
$\dfrac{\sqrt{3}}{2}=\dfrac{BC}{10}$
BC = 10×$\dfrac{\sqrt{3}}{2}$
=5×1.73
=8.65 m
44141.Find the angle of elevation of the sun when the shadow of a pole of 18 m height is 6√3m long?
30°
60°
45°
None of these
Explanation:
Let RQ be the pole and PQ be the shadow
Given that RQ = 18 m and PQ =6√3 m
Let the angle of elevation, $\angle$RPQ = θ
From the right $\triangle$ PQR,
tanθ=$\dfrac{RQ}{PQ}=\dfrac{18}{6\sqrt{3}}=\dfrac{3}{\sqrt{3}}$=√3
⇒θ=tan−1(√3)=60°
Let RQ be the pole and PQ be the shadow
Given that RQ = 18 m and PQ =6√3 m
Let the angle of elevation, $\angle$RPQ = θ
From the right $\triangle$ PQR,
tanθ=$\dfrac{RQ}{PQ}=\dfrac{18}{6\sqrt{3}}=\dfrac{3}{\sqrt{3}}$=√3
⇒θ=tan−1(√3)=60°
44142.A man standing at a point P is watching the top of a tower, which makes an angle of elevation of 30º with the man s eye. The man walks some distance towards the tower to watch its top and the angle of the elevation becomes 60º. What is the distance between the base of the tower and the point P?
4$\sqrt{3}$ units
8 units
12 units
Data inadequate
Explanation:
One of AB, AD and CD must have given.
So, the data is inadequate.
One of AB, AD and CD must have given.
So, the data is inadequate.
44143.From a point P on a level ground, the angle of elevation of the top tower is 30º. If the tower is 100 m high, the distance of point P from the foot of the tower is:
149 m
156 m
173 m
200 m
Explanation:
Let AB be the tower.
Then, $\angle$APB = 30º and AB = 100 m.
$\dfrac{AB}{AB}$=$\dfrac{1}{\sqrt{3}}$
$\Rightarrow$ (AB x $\sqrt{3}$) m
= 100$\sqrt{3}$ m
= (100 x 1.73) m
= 173 m.
Let AB be the tower.
Then, $\angle$APB = 30º and AB = 100 m.
$\dfrac{AB}{AB}$=$\dfrac{1}{\sqrt{3}}$
$\Rightarrow$ (AB x $\sqrt{3}$) m
= 100$\sqrt{3}$ m
= (100 x 1.73) m
= 173 m.