Shapes And Angles - Class 5 Maths - Extra Questions

Identify whether the following pairs of angles are complementary or supplementary: 
$$65^o,  115^o$$   

If sum of two angles is $$180^o$$ ,

 then they are called supplementary angles.

 If sum of two angles is $$90^o$$ .

 then they are called complementary angles. 
$$65^o+115^o =180^o$$  

These are supplementary angles.

In the adjoining figure:
(i) Is $$\angle 1$$ adjacent to $$\angle 2$$ ?
(II) Is $$\angle AOC$$ adjacent to $$\angle AOE$$ ?
(iii) Do $$\angle COE$$ and $$\angle EOD$$ form a linear pair?
(Iv) Are $$\angle BOD$$ and $$\angle DOA$$ supplementary?
(v) Is $$\angle 1$$  vertically opposite to $$\angle 4$$ ?
(vi) What is the vertically opposite angle of $$\angle 5$$ ? 

(i) Yes, in $$\angle AOE, OC$$ is common arm. 
(ii) No, they have no non-common arms on opposite side of common arm. 
(iii) Yes, they form linear pair. 
(iv) Yes, they are supplementary. 
(v) Yes, they are vertically opposite angles. 
(vi) Vertically opposite angles of $$\angle 5$$ is $$\angle  COB$$ .

An angle is greater than $$45^o$$ . Is its complementary angle greater than $$45^o$$ or equal to $$45^o$$ or less than $$45^o$$ ?

Let the complementary angles be  $$x$$ and $$y$$ ,

 i.e., $$x+y =90^o$$
It is given that $$x > 45^o$$  
Adding $$y$$ both sides,   

  $$x+y>45^o+y$$
$$\Rightarrow  90^o > 45^o + y$$

$$\Rightarrow  90^o -45^o > y$$

$$\Rightarrow  y < 45^o$$  
Thus, its complementary angle is less than $$45^o$$ . 

Construct the following angles using set-squares: $$45^{\circ}$$

Place $$45^°$$ set-square.

Draw two rays $$AB$$ and $$AC$$ along the edges from the vertex of  $$45^°$$  angle of the set-square.

The angle so formed is a  $$45^°$$  angle.

$$∠BAC = 45^°$$

Construct the following angles using set-squares: $$90^{\circ}$$

Place $$45^°$$ set-square as shown in the figure.

Draw two rays $$BC$$ and $$BA$$ along the edge from the vertex of $$90^°$$ angle.

The angle so formed is  $$90^°$$  angle.

$$∠ABC = 90^°$$

What is the shape of:
your instrumental box

Consider an instrumental box shown in the above image:

Hence, from the image, we find that the shape of the instrumental box is cuboid.

Construct the following angles using set-squares: $$60^{\circ}$$

Place $$30^°-60^o$$ set-square as shown in the figure.

Draw the rays $$BA$$ and $$BC$$ along the edges from the vertex of $$60^°$$ .

The angle so formed is  $$60^°$$ .

$$∠ABC = 60^°$$

You are standing in a class-room facing north. In what direction are you facing after making a quarter turn?

Right now you are standing in North so a quarter turn means 90° right so which means right to which is east

Construct the following angles using set-squares: $$105^{\circ}$$

Place $$30^°$$ set-square and make an angle of $$60^°$$ by drawing the rays $$BA$$ and $$BC$$ as shown in the above figure.

Then, place the vertex of $$45^°$$ of  the set-square on the ray $$BA$$ as shown in the figure and draw the ray $$BD$$ .

The angle so formed is $$105^°$$

Therefore, $$∠DBC = 105^°$$

State the kind of angle, in the case, formed between the following direction.
East and West.

If a bicycle wheel has $$48$$ spokes, then find the angle between a pair of adjacent spokes.

The drawing below Fig., show angles formed by the goalposts at different positions of a football player.The greater the angle, the better chance the player has of scoring a goal. For example, the player has a better chance of scoring a goal from Position $$A$$ than from POsition $$B$$ .
In Parts (a) and (b) given below it may help to trace the diagram and draw and measure angles.
Now the player are lined up as shown in Fig. Which player has the best kicking angle?

$$4^{th}$$ player has the highest degree of angle to kick. Hence, this player has best kicking angle.

Using protractor, draw an angle of measure $$108^{\circ}.$$ With this angle as given, draw an angle of $$54^{\circ}.$$

We are asked to draw an angle of $$108^°$$ , using protractor and obtain an angle of $$54^°$$ .

We will follow certain algorithm for this construction:

Steps of construction:

STEP1: Using the protractor, draw angle $$ABC$$ of measure $$108^°$$ .

STEP2: With centre $$B$$ , and taking any radius, draw an arc, intersecting the ray $$BA$$ and the ray $$BC$$ at point $$D$$ and $$E$$ respectively.

STEP 3: With centre $$D$$ and radius greater than half of $$DE$$ , draw an arc inside the angle.

STEP 4: With centre $$E$$ and taking the same radius as in STEP 3, draw an arc intersecting the arc drawn in STEP 3, at $$F$$ .

STEP 5: Draw the ray $$BF$$ . This is the angle bisector of angle $$ABC$$ . So angle $$FBC = 54^°$$ .

STEP 6: With centre $$B$$ and taking any radius, draw an arc intersecting the ray $$BF$$ and the ray $$BC$$ at $$G$$ and $$H$$ respectively.

STEP 7: Draw a ray $$MN$$ .

STEP 8: With centre $$M$$ and taking the same radius as in STEP 6, draw an arc intersecting the ray $$MN$$ at point $$P$$ .

STEP 9: With centre $$P$$ and taking radius equal to $$HG$$ , draw an arc intersecting the arc drawn in STEP 8, at point $$L$$ .

STEP 10: Draw the ray $$ML$$ .

Angle $$LMN$$ is the desired angle of measure $$54^°$$

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