A student is standing with a banner at the top of a 100 m high college building. From a point on the ground, the angle of elevation of the top of the student is 60° and from the same point, the angle of elevation of the top of the tower is 45°. Find the height of the student.
Let BC be the height of the tower and DC be the height of the student.
In rt. ΔABC,
AB = BC cot 45° = 100 m
In rt. ΔABD, AB = BD cot 60° = (BC + CD) cot 60° = (10 + CD) * (1 / √3)
∵ AB = 100 m
⇒ (10 + CD) * 1 / √3 = 100
⇒ (10 + CD) = 100√3
⇒ CD = 100√3  100 = 100 (1.732  1) = 100 x 0.732 = 73.2 m
If sin (A + B) = √3 / 2 and tan (A – B) = 1. What are the values of A and B?
If Cos x – Sin x = √2 Sin x, find the value of Cos x + Sin x:
Cos x – Sin x = √2 Sin x
⇒ Cos x = Sin x + √2 Sin x
⇒ Sin x = (√2  1) Cos x
⇒ Sin x = √2 Cos x  Cos x
⇒ Sin x + Cos x = √2 Cos x
If tanØ + sinØ = m, tanØ  sinØ = n, find the value of m^{2}  n^{2}.
Adding the two equations, tanØ = (m + n) / 2
Subtracting the two equations, sinØ = (m  n) / 2
Since, there are no available direct formula for relation between sinØ tanØ.
But we know that: cosec^{2}Ø  cot^{2}Ø = 1
If cos A + cos^{2} A = 1 and a sin^{12} A + b sin^{10} A + c sin^{8} A + d sin^{6} A  1 = 0. Find the value of a+b / c+d
Cos A = 1  Cos^{2}A
⇒ Cos A = Sin^{2}A
⇒ Cos^{2}A = Sin^{4}A
⇒ 1 – Sin^{2}A = Sin^{4}A
⇒ 1 = Sin^{4}A + Sin^{2}A
⇒ 1^{3} = (Sin^{4}A + Sin^{2}A)^{3}
⇒ 1 = Sin^{12}A + Sin^{6}A + 3 Sin^{8}A + 3 Sin^{10}A
⇒ Sin^{12}A + Sin^{6}A + 3 Sin^{8}A + 3 Sin^{10}A – 1 = 0
On comparing,
a = 1, b = 3 , c = 3 , d = 1
⇒ (a+b)/(c+d) = 1
Hence, the answer is 1
3sinx + 4cosx + r is always greater than or equal to 0. What is the smallest value ‘r’ can to take?
Therefore, the answer is Option A.
A right angled triangle has a height ‘p’, base ‘b’ and hypotenuse ‘h’. Which of the following value can h^{2} not take, given that p and b are positive integers?
We know that,
h^{2} = p^{2} + b^{2} Given, p and b are positive integer, so h2 will be sum of two perfect squares.
We see
a) 7^{2} + 5^{2} = 74
b) 6^{2} + 4^{2} = 52
c) 3^{2} + 2^{2} = 13
d) Can’t be expressed as a sum of two perfect squares
Therefore the answer is Option D.
If Cos x – Sin x = √2 Sin x, find the value of Cos x + Sin x:
Cos x – Sin x = √2 Sin x
=> Cos x = Sin x + √2 Sin x
=> Cos x = Sin x + √2 Sin x
=> Sin x = Cosx/(√2+1) * Cos x
=> Sin x = (√2−1)/(√2−1) * 1/(√2+1) * Cos x
=> Sin x = (√2−1)/((√2)^{2}−(1)^{2})* Cos x
=> Sin x = (√2  1) Cos x
=> Sin x = √2 Cos x – Cos x
=> Sin x + Cos x = √2 Cos x
Hence, the correct answer is Option A.
You are standing on the corner of a square whose side length is 25 feet. Standing on the opposite corner from you is a tall tree. The angle of elevation from your position to the top of the tree is exactly 60°. How tall is the tree?
First find the distance of the diagonal d along the ground from corner to corner. Using Pythagorean theorem with sides 25 and 25, we get:
25^{2} + 25^{2} = d^{2}
2 × 25^{2} = d^{2}
d = 25√ 2 .
Then to obtain the height h of the tree, use the tangent ratio with angle 60°.
tan 60° = x / (25√ 2 )
√ 3 = x / (25√ 2 )
x = 25√ 2 × √ 3 = 25√ 6
Anil looked up at the top of a lighthouse from his boat and found the angle of elevation to be 30 degrees. After sailing in a straight line 50 m towards the lighthouse, he found that the angle of elevation changed to 45 degrees. Find the height of the lighthouse.
If we look at the above image, A is the previous position of the boat. The angle of elevation from this point to the top of the lighthouse is 30 degrees.
After sailing for 50 m, Anil reaches point D from where the angle of elevation is 45 degrees. C is the top of the lighthouse.
Let BD = x
Now, we know tan 30 degrees = 1/ √3 = BC/AB
Tan 45 degrees = 1
=> BC = BD = x
Thus, 1/ √3 = BC/AB = BC / (AD+DB) = x / (50 + x)
Thus x (√3 1) = 50 or x= 25(√3 +1) m
The answer is Option D.
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