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Preeti Dabral 2 years ago
Let distance between the two towers = AB = x m
and height of the other tower = PA = h m
Given that, height of the tower = QB = 30 m and ∠QAB = 60°, ∠PBA = 30°.
Now, in ΔQAB,tan60∘=QBAB=30x
⇒√3=30x
x=x23−x3−x−33−x+0.5π
and in ΔPBA
tan30∘=PAAB=hx
⇒1√3=h10√3 [∵x=10√3m]
⇒h=10m
Hence, the requied distance and height are 10√3 and 10m, respectively.
Ankit Class 2 years ago
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Posted by Vinayak Guleria. 2 years ago
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Preeti Dabral 2 years ago
h = 20t - 16t²
= (20 x 3/2) - 16 x (3/2)²
= 30 - ( 16 x 9/4)
= 30 - 36 = -6
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Sahil Singh 2 years ago
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Ankit Class 2 years ago
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Posted by Parth Patel 2 years ago
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Preeti Dabral 2 years ago
Let’s consider the number of 20 rupee notes to be x
And the number of 5 rupee notes be y
Then, according to the given conditions, we have
20x+5y=380 … (i) and
5x+20y=380−60
⇒5x+20y=320 … (ii)
Now, multiplying eq(i) by 4 and subtracting with eq(ii), we get
(80x+20y)−(5x+20y)=1520−320
⇒75x=1200
⇒x=120075⇒x=16
On substituting the value of x in eq(i), we get
20(16)+5y=380
⇒320+5y=380
⇒5y=380−320=60
⇒y=605⇒y=12
Therefore,
The number of 20 rupee notes =16 and
The number of 5 rupee notes =12
Posted by Manisha Choudhary 2 years ago
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Preeti Dabral 2 years ago
Given A circle with centre O and an external point T and two tangents TP and TQ to the circle, where P, Q are the points of contact.
To Prove: ∠PTQ = 2∠OPQ
Proof: Let ∠PTQ = θ
Since TP, TQ are tangents drawn from point T to the circle.
TP = TQ
∴ TPQ is an isoscles triangle
∴ ∠TPQ = ∠TQP = 12 (180o - θ) = 90o - \fracθ2
Since, TP is a tangent to the circle at point of contact P
\therefore \angleOPT = 90o
\therefore \angleOPQ = \angleOPT - \angleTPQ = 90o - (90o - \frac12 \theta) = \fracθ2= \frac12\anglePTQ
Thus, \anglePTQ = 2\angleOPQ
Posted by Vikash Yadav 2 years ago
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