WAEC Mathematics Question 2015 ( Essay Question Ten 10)

QUESTION 10

10(a)without using mathematical tables or calculators;

Simplify: 

10(b)From an aeroplane in the air at a horizontal distance of 1050m, the angle of depression of the base of a control tower at an instance are 36 and 41 respectively.

Calculate, correct to the nearest meter, the;

(i)                 Height of the control tower.

(ii)               Shortest distance between the aeroplane and the base of the control tower.

 

Solution

10a. We solve using equilateral triangle

Dividing the triangle into two equal parts.

Note: BC = BD + CD

Where;

BD = CD

BC = 2

So therefore, 2 = CD + CD

2 = 2CD

CD =  = 1

We solve for the unknown side (AD) by applying Pythagoras theorem.

|AC|²= |AD|² + |CD|²

|AD|²= |AC|² - |CD|²

|AD|²= 2² - 1²

|AD|²= 4 - 1

|AD|²= 3

AD = 

 The diagram becomes

From the above diagram, we consider SOHCAHTOA

tan60 = 

tan60 = 

 

 Also, 

Substituting the value of tan60, cos30 and sin60 into the equation   respectively,

The LCM of the numerator is 2

 

= 5 ANSWER

 

 

10b.

Central Tower

Let the height of the tower be h1 and the distance between the top of the central tower and the horizontal distance of the aeroplane, be h2. Hence, total distance from the base of the central tower and the base of the aeroplane be h.

So therefore, h = h1 + h2

To solve for h1, we consider      ABD and apply SOHCAHTOA

Using tanA =

tan36 =

h1 = 1050tan36

h1 = 762.9m

Also, to solve for h, ie. h1 + h2, we consider    ABC

Using tanA =

tan41 =

h = 1050tan41

So therefore, h = 912.8m

Hence, the height of the central tower h2 can be calculated using the formula

h = h1 + h2

h2 = h – h1

h2 = 912.8 – 762.9

h2 = 149.9m

Note: The shortest distance between the aeroplane and the base of the control tower is the distance between AC

cosA =

cosA =

AC =

AC = 1391.3m