Question

A MAP SCALE READS 1CM:5KM. IF A DISTANCE ON THE MAP MEASURES 7CM, THE ACTUAL DISTANCE WILL BE?. 2. A SCHOOL DINING HALL IS 35M LONG. IF ON A PLAN, THIS LENGTH IS REPRESENTED BY 7CM, WHAT IS THE SCALE?. 3. THE LENGTH OF A BUILDING IS 30 METRES. A SCALE DIAGRAM OF THE BUILDING IS BEING DRAWN TO A SCALE OF 1CM TO 5 METRES. THE LENGTH OF THE BUILDING ON THE SCALE DIAGRAM IS?. 4. IF THE ANGLE OF ELEVATION OF A FROM B IS 42°, WHAT IS THE ANGLE OF DEPRESSION OF B FROM A?. 5. A BOY IS FLYING A KITE, THE STRING IS 25M LONG AND IS AT AN ANGLE OF 42° WITH THE HORIZONTAL, USING A SCALE DIAGRAM, FIND HOW HIGH THE KITE IS ABOVE THE BOY’S HEAD?

Answer 1

Actual distance will be 35 km

Answer 2

Answer:

Step-by-step explanation:

1.

  1 CM = 5 KM

  7 CM = 7*5

           =35 KM Ans.

2.

35 M is represented by 7 CM

1 CM = 35/7

1 CM = 5 M (Scale used) Ans.

3.

Length of building = 30 M

Scale used, 5 M = 1 CM

1 M = 1/5

30 M = 30/5

= 6 CM length of building on scale diagram will be used Ans.

4.

As in a triangle sum of all angles = 90

So,

Angle of elevation + 90 + Angle of depression = 180

Angle of elevation + 90 +42 = 180

Angle of elevation + 132 = 180

Angle of elevation = 180-132

Angle of elevation = 48

5.

Note: check this part answer only if angle is 45 as i think you wrote 42 by fault.

String of kite is hypotenuse = 25 M

We need to find perpendicular height,

Sin A = P/H

Sin(45) = P/25

$\frac{1}{\sqrt{2} } = \frac{P}{25}$

P = 25$\sqrt{2}$

P = 25 *1.414

P = 35.35 M Ans.

Hope you understand all answers