If x= 19 feet and y = 12 feet, what is the area and perimeter of the figure above?
1 answer:
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Given :
A particle moves in the xy plane starting from time = 0 second and position (1m, 2m) with a velocity of v=2i-4tj .
To Find :
A. The vector position of the particle at any time t .
B. The acceleration of the particle at any time t .
Solution :
A )
Position of vector v is given by :
B )
Acceleration a is given by :
Hence , this is the required solution .
Answer:
AB = 13.89
Measure of angle A = 59.74°
Measure of angle B = 30.26°
Step-by-step explanation:
The given parameters are;
∠C = 90°
AC = 7
BC = 12
Part 1
Hence, the question has the dimensions of the two adjacent sides of the right angle (angle 90°)
From Pythagoras theorem, we have;
A² = B² + C²
Where, A is the opposite side to the right angle, hence;
In the ΔABC,
AB ≡ A
Therefore;
AB² = AC² + BC² = 7² + 12² = 193
∴ AB = √193 = 13.89
Part 2
∠A is the side opposite side BC such that by trigonometric ratios
∴ ∠A = Arctan(1.714) or tan⁻¹(1.714) = 59.74°
Part 3
∠B is found from knowing that the sum of the angles in a triangle = 180°
∴ ∠A + ∠B + ∠C = 180° which gives
59.74° + 90° + ∠B = 180°
Hence, ∠B = 180° - (59.74° + 90°) = 180° - 149.74° = 30.26°.