Question to Find Percentage of Discount on a Calculator Price

Topic : Sales and Price

Question : Peter bought a calcuator on sale for $40 the calculators retail price is $60 what is the percent discount?

Solution :
Discount in Price = 60 - 40 = $20
So discount percentage = discount price /Retail Price * 100
= 20/60 * 100
= 100/3
= 33.33%

Problem on Trigonometry Equation and Proof for the Identities

Topic : Trigonometry Identites

Simplify.

Problem on Trigonometric Equations

Topic : Trigonometric Equation

Problem : Solve 3Sin θ Cos θ – 2 Cos θ = 0 and 0 ≤ θ ≤ 2π

Solution :

3Sin θ Cos θ – 2 Cos θ = 0
Taking Cos θ as common
Cos θ (3Sin θ - 2) = 0
Cos θ = 0 or 3Sin θ – 2 = 0
3 Sin θ – 2 = 0
3Sin θ = 2
Sin θ = 2/3 (2/3 = 0.66 = 41.8º ~ 42º)
Sin θ = Sin 42º
θ = n π + (-1)ⁿ 42º
when n = 0 , θ = 42º
when n = 1 , θ = π - 42º = 138º

Let’s solve Cos θ = 0
Cos θ = Cos π/2
The formula for Cosine is :
If Cos x = Cos y
Then x = 2nπ ± y

So here it will be θ = 2nπ ± π/2
When n = 0, θ = 2(0)π ± π/2 = ± π/2
But only + π/2 lies in 0 ≤ θ ≤ 2π
So x = π/2

When n = 1, θ = 2(1)π ± π/2 = 2π ± π/2 = 2π ± π/2 = (4π ± π)/2 = 3π/2, 5π/2
But x = 3π/2

Now the further values of n will give angles greater than 360º

So θ will be 42º, 90º, 138º, 270º

Problem on Simple Linear Equation

Topic : Linear Equation

Problem : Solve y = 3y - 12

Solution :

A Simple Integration Problem

Topic : Integration

Problem : Integrate x /√(1 - x^4)

Solution :

Question to Find Interior angle of a Triangle

Topic : Interior angle of triangle

Question : Find the interior angle of the triangle if sum of all the angles is equal to 171º.

Answer :

Each interior angle = (n-2) *180º / n = 171º
180ºn - 360º = 171ºn
+360º = +360º
180ºn = -171ºn + 360º
-171ºn = -171ºn
---------------------
9ºn/9º = 360º/9º
n = 40º

Question on Similar Triangles

Topic : Similar Triangles

Question : Define and Mention the Properties of Similar Triangles

Solution :
Triangles are similar if they have the same shape, but can be different sizes.
Properties of Similar Triangles
1. Corresponding angles are the same
So in the figure above, the angle P=P', Q=Q', and R=R'.
2. Corresponding sides are all in the same proportion
Above, PQ is twice the length of P'Q'. Therefore, the other pairs of sides are also in that proportion. PR is twice P'R' and RQ is twice R'Q'.
Formally, in two similar triangles PQR and P'Q'R' :
PQ / P’Q’ = QR / Q’R’ = RP / R’P’