Topic : Trigonometry
Question : Airplane is moving with speed 400knot due East and wind 50knot due North East . Find the resultant and direction .
Solution :
400 ->E
50 ->NE
Angle between East and North direction is 45º
Hence angle between 400 and 50 is 45º
We have a formula for the resultant of two forces (velocities) P and Q
R² = P² + Q² +2PQCosθ, θ being the angle between P and Q
P = 400; Q = 50; θ = 45
R² = 400² + 50² + 2(400)(50)Cos45
= 160000 + 2500 + 40000 * 0.7071
= 190784
So resultant R = √(190784) = 436.788 = 436.8
To find the direction :
The angle Φ is in between R and P
Q Sin θ
So tan Φ = --------------------
P + Q Cos θ
50 Sin 45
tan Φ = ------------------------- = 0.0812 => Φ = tanˉ ¹ 0.0812 => 4.65
400 + 50 Cos 45
Here Φ is the angle between R and P
and Φ = 4.65 , if resultant motion is away due East.
That is same as
90 – 4.65 due North
So 85.35 ≈ 85.4 due North
Hence Solution is 436.8 due 85.4º North
Hope the Answer and Explanation will be helpful for you.
Wednesday, February 25
Thursday, February 5
Question to Find Area of Triangle
Topic : Area of Triangle
Question: In any ∆ ABC a²Sin2B+b²Sin2C =
A) Area of the triangle
B) Twice the Area of the triangle
C) Thrice the Area of the triangle
D) Four times of the Area of the triangle
Solution :
In the data any triangle is mentioned.
Therefore let us take equilateral triangle having side 1 unit.
That is a=b=c=1, A=B=C=60º.
Area of the equilateral triangle is = [(√3)/4]a²=[(√3)/4]x1=(√3)/4.
value of the expression given =1²xSin120+1²xsin120=[(√3)/2]+[(√3)/2]= √3------(1)
Option A): Area of the Triangle=(√3)/4, but value of expression = √3 These two are not equal. Hence Option A is not correct
Option B): Twice the Area of the triangle=2x[(√3)/4]=(√3)/2, but the value of expression is √3. Hence option B is incorrect
Option C):Thrice the Area of the triangle = 3x[(√3)/4], but the value of the expression is √3. hence option C also incorrect.
Option D): Four times of the Area of the triangle=4x[(√3)/4]= √3, this is also equal to the value of the expression shown in ---(1).
Hence Option D is the correct choice of the above question.
Hope the Answer and Explanation will be helpful for you.
Question: In any ∆ ABC a²Sin2B+b²Sin2C =
A) Area of the triangle
B) Twice the Area of the triangle
C) Thrice the Area of the triangle
D) Four times of the Area of the triangle
Solution :
In the data any triangle is mentioned.
Therefore let us take equilateral triangle having side 1 unit.
That is a=b=c=1, A=B=C=60º.
Area of the equilateral triangle is = [(√3)/4]a²=[(√3)/4]x1=(√3)/4.
value of the expression given =1²xSin120+1²xsin120=[(√3)/2]+[(√3)/2]= √3------(1)
Option A): Area of the Triangle=(√3)/4, but value of expression = √3 These two are not equal. Hence Option A is not correct
Option B): Twice the Area of the triangle=2x[(√3)/4]=(√3)/2, but the value of expression is √3. Hence option B is incorrect
Option C):Thrice the Area of the triangle = 3x[(√3)/4], but the value of the expression is √3. hence option C also incorrect.
Option D): Four times of the Area of the triangle=4x[(√3)/4]= √3, this is also equal to the value of the expression shown in ---(1).
Hence Option D is the correct choice of the above question.
Hope the Answer and Explanation will be helpful for you.
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