1. 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?

1. 35 km

2. 1 cm to 5m

3. 6 cm

4.

5. 16.73 m

Step-by-step explanation:

Solution 1.

Reading of map scale = 1cm:5km

i.e. 1 cm is equivalent to 5 km

Measurement of map  = 7 cm

Actual distance = Measurement of map

Actual distance = 7 = 35 km

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Solution 2.

Length of dining hall = 35 m

Measurement of map = 7 cm

Scale = Measurement of map : Length of dining hall (i.e. ratio)

Scale = 7 cm :35 m = 1 cm : 5 m

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Solution 3.

Length of building = 30 m

Scale = 1 cm to 5m

5 m is equivalent to 1 cm on scale

1 m is equivalent to cm on scale

30m is equivalent to = 6 cm on scale

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Solution 4.

Angle of elevation of A from B =

Angle of depression of B from A = ?

Please refer to the image attached, we can clearly observe that both the angles (i.e. angle of elevation from A to B and angle of depression from B to A )will be equal.

Angle of depression of B from A =

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Solution 5.

Given that:

String length, or hypotenuse of triangle BC= 25 m

Angle of string with horizontal,

Please refer to the attached image for the clear understanding of the situation.

To find:

Height, AC = ?

We can use trigonometric identity:

OR

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So, the answers are:

1. 35 km

2. 1 cm to 5m

3. 6 cm

4.

5. 16.73 m

341

step-by-step explanation:

all you have to do is 9x31 (which equals 279) and then 20x31 (which equals 620). then you have to subtract. 620-279=341.

-2

7 - 9 = -2

reversed:

-2 + 9 = 7