Problem on Simple Differentiation

Topic : Simple Differentiation
Problem : Solve y = x² + 4e^x
Solution :
y = x² + 4e^x
dy/dx = d(x²)/dx + 4d(e^x)/dx
dy/dx = 2x + 4e^x

Word Problem on Sine Rule

Topic : Sine Rule

Question : Lookout station B is located 7 miles due east of station A. The bearing of a fire from A is S 8degree50’ W and the bearing from B is S 28degree20’ W. Determine the distance from the fire to B(to the nearest tenth of a mile)

Solution :

Figure is as shown in the figure according the question

From Sine Rule

a / Sin A = b / Sin B = c / Sin C

a = c Sin A / Sin C
a = 7 * Sin (90º + 8º 50’) / Sin (90º - 28º 20’)
a = 7 * Sin (98º 50’) / Sin (19º 30’)
a = 20.72miles

Question on Ratio/Proportionality

Topic : Ratio /Proportionality

Question : Are the ratios 25g: 30g and 40Kg: 48Kg in proportion?

Solution :
25g: 30 g = 25/30 = 5:6
40Kg: 48Kg = 40/48 = 5:6
So 25: 30 = 40:48
Therefore, the ratios are in proportion
25:30: : 40:48.

Question on Linear Equation

Topic : Linear Equation

Question : 9 added to thrice a whole number gives 45. Find the number.

Solution :
Step 1. Let the required number be X.
Step 2 Thrice the whole number is 3X
Step 3 Nine added to 3X is 3X + 9
Step 4 3X + 9 = 45
Step 5 3X = 45 -- 9
Step 6 3X = 36
Step 7 X = 36/3
Step 8 X = 12
The required number is 12

Question on Finding Resultant and Direction of a Moving Article

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.

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.