Question
The angle of elevation of an aeroplane from a point on
the ground is 60°. After 15 seconds flight the elevation changes to 30°, if the aeroplane is flying at a height of 2400√3 m. find the speed of the aeroplane.Solution
Distance between two place in which angle change is D = n(cotθ₁ + cotθ₂) Distance = 2400√3 (cot30° + cot60°) = 2400√3 × √3 - 1/√3 = 2400√3 × 2/√3 = 4800 m So, distance covered by plane in 15 sec = 4800 m Speed of plane = 4800/15 = 320 m/sec.
If p = 24 - q - r and pq + r(q + p) = 132, then find the value of (p² + q² + r²).
((99.9 - 20.9)² + (99.9 + 20.9)² )/(99.9 x 99.9 + 20.9 x 20.9) = ?
...
Find the value of the given expression-
(4x+4 -5× 4x+2) / 15×4x – 22×4x
If 4x² + y² = 40 and x y = 6, then find the value
of 2x + y?
If p = 40 - q - r and pq + r(q + p) = 432, then find the value of (p² + q² + r²).
47.98 × 4.16 + √325 × 12.91 + ? = 79.93 × 5.91
If x + y = 4 and (1/x) + (1/y) = 24/7, then the value of (x3 + y3).
- If p = 20 - q - r and pq + r(p + q) = 154, then find the value of (p² + q² + r²).
If a = (√2 - 1)1/3, then the value of (a-1/a)3 +3(a-1/a) is:
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