Answer:
1. 35 km
2. 1 cm to 5m
3. 6 cm
4. [tex]42^\circ[/tex]
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 [tex]\times 5[/tex]
Actual distance = 7 [tex]\times 5[/tex] = 35 km
-------------------
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
-------------------
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 [tex]\frac{1}{5}[/tex] cm on scale
30m is equivalent to [tex]\frac{1}{5} \times 30[/tex] = 6 cm on scale
-------------------
Solution 4.
Angle of elevation of A from B = [tex]42^\circ[/tex]
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 = [tex]42^\circ[/tex]
-------------------
Solution 5.
Given that:
String length, or hypotenuse of triangle BC= 25 m
Angle of string with horizontal, [tex]\angle B = 42^\circ[/tex]
Please refer to the attached image for the clear understanding of the situation.
To find:
Height, AC = ?
We can use trigonometric identity:
[tex]sin\theta = \dfrac{Perpendicular}{Hypotenuse}[/tex]
OR
[tex]sinB = \dfrac{AC}{BC}\\\Rightarrow sin42^\circ = \dfrac{AC}{25}\\\Rightarrow AC = 25 \times 0.67\\\Rightarrow AC = 16.73 m[/tex]
============
So, the answers are:
1. 35 km
2. 1 cm to 5m
3. 6 cm
4. [tex]42^\circ[/tex]
5. 16.73 m
