Problems on Angles - Test Questions

Question 21

Question:   Three angles of a quadrilateral are 100^o, 48^o and 92^o. Find the 4^th angle.

Answer:   


quadrilateral is 360^o]




=120^o
The 4^th angle measures 120^o.

Question 22

Question:  

Answer:   


Adding (i) and (ii)
...(iii)

...(iv) [ sum of the angles of a
quadrilateral equals 360^o]
Equating (iii) and (iv)

Question 23

Question:   ABCDE is a regular pentagon. The bisectors of of the pentagon meets

[Hint: Each angle of a regular pentagon is 108^o]

Answer:   

ABCDE is a regular pentagon.



Number of sides of the regular pentagon, n=5.
Each interior angle






=108^o


=54^o
In quadrilateral AMCD,

Question 24

Question:   In the figure, lines 'l' and 'm' intersect at O, forming angles as shown in the figure. If x=45^o, find the values of y, z and u.

Answer:    x^o= 45^o [given]
z^o= 45^o [ vertically opposite angles]
x^o = y^o = 180^o [Linear pair]
45^o + y^o = 180^o
y^o = 180^o - 45^o

y^o = 135^o
If y^o = 135^o, then x^o = 135^o [ vertically opposite angles]
Hence the values are: y=135^o, z=45^o and x=135^o

Question 25

Question:   In the adjoining figure, determine the value of y.

Answer:   
a^o = 5y^o [ vertically opposite angles]
Now 5y^o + a^o + 2y^o = 180^o
i.e., 5y^o + 5y^o + 2y^o = 180^o [ Linear pair]

12y^o = 180^o

y^o = 15^o

Question 26

Question:   In the adjoining figure, three coplanar lines intersect at a common point, forming angles as shown. Given a=50^o and b=90^o. Find the values of c, d, e and f.

Answer:    d = 50^o [ a^o=50^o, vertically opposite angles]
e = 90^o [ b^o=90^o, vertically opposite angles]
b + c + d = 180^o [ They form a linear pair]
90 + c + 50 = 180^o
140^o+ c = 180^o
c = 40^o
Similarly a + e + f = 180^o [ They form a linear pair]
50^o+ 90^o+ f = 180^o
140^o+ f = 180^o
f = 40^o [c=40^o, vertically opposite angles]
Hence c = 40^o; d = 50^o; e = 90^o and f = 40^o