Download RD Sharma Maths : Solutions for Class 9th Triangle and its Angles

RD Sharma Maths : Solutions for Class 9th Triangle
and its Angles ( Exercise 9.1, 9.2, formative
assessment_mcq, formative assessment_vsa )
Exercise 9.1 : Solutions of Questions on Page Number : 9.10
Solution 12 :
Answer :
Let us say in a triangle ABC angles
are three angles.
Now sum of two angles are less than third one which leads to
Now as we know sum of all angles in a triangle is always equal to 180.
We have given
Similarly we can say
Since all angles are less than 90 hence triangle is acute angled.
Solution 1 :
Answer :
As we know the sum of the angles of a triangle is always
Using the above property we can write it as,
Hence, measurement of angle C is
.
Solution 2 :
Answer :
Let us say angles are
As we know the sum of the angles of a triangle is always
Using the above property we can write it as,
Now putting the value of x we get,
Hence, the angles are
Solution 3 :
Answer :
Given angles are
As we know the sum of the angles of a triangle is always
Using the above property we can write it as,
Hence, the value of x is
.
Solution 4 :
Answer :
Let us say angles are
We know the sum of the angles of a triangle is always
Using the above property we can write it as,
Now putting the value of x we get,
So the angles are
Solution 5 :
Answer :
Let us say two equal angles are
, and then third angle will be
We know the sum of the angles of a triangle is always
Using the above property we can write it as,
Now putting the value of x we get,
So the angles are
Solution 6 :
Answer :
Let ABC is a triangle such that
As we know the sum of the angles of a triangle is always
Using the above property we can write it as,
Since, here one angle of a triangle is
, so the given triangle is a right angled triangle.
Solution 7 :
Answer :
Let ABC is a triangle such that
As we know the sum of the angles of a triangle is always
Using the above property we can write it as,
Hence magnitude of
.
Solution 8 :
Answer :
Let ABC be a triangle and Let BO and CO be the bisectors of the base angle
respectively.
Then
From the above relation it is very clear that if
is equals 90 then
is must equal to zero.
Now if possible let
is equals zero but on other hand it represents A, B, C will be collinear,
that is they do not form a triangle.
Obviously it leads to a contradiction hence
must not be equal to zero.
Solution 9 :
Answer :
Let ABC be a triangle and Let BO and CO be the bisectors of the base angle
respectively.
Then
Now let us put the value of
in the above equation which is 135 given.
Hence the triangle is a right angled triangle.
Solution 10 :
Answer :
Let ABC be a triangle and Let BO and CO be the bisectors of the base angle
respectively.
Then
Now let us put the value of
Now
are equal as
in the above equation which is 120 given.
given.
Since we know in a triangle sum of all angles is always 180.
Hence
Solution 11 :
Answer :
(i) Let us say in a triangle ABC two angles
all angles in a triangle is always equals 180.
are equals
. Now as we know sum of
Hence if two angles are equal
the third one will be equal to zero. Which leads that A, B, C is
collinear, or we can say ABC is not a triangle
A triangle can't have two right angles.
(ii) Let us say in a triangle ABC two angles
than
are obtuse angles or we can say more
.
Which leads that sum of only two angles will be equals more than 180.
Now as we know sum of all angles in a triangle is always equals 180
A triangle can't have two obtuse angles.
(iii) Let us say in a triangle ABC two angles
are acute angles or we can say less than
.
Which leads that sum of two angles will be less than
difference of 180 and sum of both acute angles
A triangle can have two acute angles.
.Hence third angle will be the
(iv) Let us say in a triangle ABC angles
are more than
Which leads that sum of three angles will be more than
A triangle can't have all angles more than
.
.
.
(v) Let us say in a triangle ABC angles
are less than
Which leads that sum of three angles will be less than
.
.
A triangle can't have all angles less than
(vi) Let us say in a triangle ABC angles
all are equal to
Which leads that sum of three angles will be equal to
.
A triangle can have all angles equal to
.
.
<< Previous Chapter 8 : Lines and AnglesNext Chapter 10 : Congruent Triangles >>
Exercise 9.2 : Solutions of Questions on Page Number : 9.18
Solution 1 :
Answer :
In the given problem, the exterior angles obtained on producing the base of a triangle both ways
are
and
. So, let us draw ΔABC and extend the base BC, such that:
Here, we need to find all the three angles of the triangle.
Now, since BCD is a straight line, using the property, “angles forming a linear pair are
supplementary“, we get
Similarly, EBS is a straight line, so we get,
Further, using angle sum property in ΔABC
Therefore,
.
Solution 2 :
Answer :
In the given problem, BP and CP are the internal bisectors of
respectively.
Also, BQ and CQ are the external bisectors of
respectively. Here, we need to prove:
So, here using the corollary, “if the bisectors of angles
and
ofΔABC meet at a
point O then
Thus, in ΔABC
……(1)
Also, using the theorem, “if the sides AB and AC of a ΔABC are produced, and the external
bisectors of
and
meet at O, then
“
Thus, ΔABC
…..(2)
Adding (1) and (2), we get
Thus,
Hence proved.
Solution 3 :
Answer :
In the given ΔABC,
and
. We need to find
.
Here,
are vertically opposite angles. So, using the property, “vertically
opposite angles are equal“, we get,
Further, BCD is a straight line. So, using linear pair property, we get,
Now, in ΔABC, using “the angle sum property“, we get,
Therefore,
.
Solution 4 :
Answer :
In the given problem, we need to find the value of x
(i) In the given ΔABC,
and
Now, BCD is a straight line. So, using the property, “the angles forming a linear pair are
supplementary“, we get,
Similarly, EAC is a straight line. So, we get,
Further, using the angle sum property of a triangle,
In ΔABC
Therefore,
(ii) In the given ΔABC,
and
Here, BCD is a straight line. So, using the property, “the angles forming a linear pair are
supplementary“ we get,
Similarly, EBC is a straight line. So, we get
Further, using the angle sum property of a triangle,
In ΔABC
Therefore,
(iii) In the given figure,
and
Here,
and AD is the transversal, so
form a pair of alternate interior
angles. Therefore, using the property, “alternate interior angles are equal“, we get,
Further, applying angle sum property of the triangle
In ΔDEC
Therefore,
(iv) In the given figure,
,
and
Here, we will produce AD to meet BC at E
Now, using angle sum property of the triangle
In ΔAEB
Further, BEC is a straight line. So, using the property, “the angles forming a linear pair are
supplementary“, we get,
Also, using the property, “an exterior angle of a triangle is equal to the sum of its two opposite
interior angles“
In ΔDEC, x is its exterior angle
Thus,
Therefore,
.
Solution 5 :
Answer :
In the given figure,
Since,
and
and angles opposite to equal sides are equal. We get,
……(1)
Also, EAD is a straight line. So, using the property, “the angles forming a linear pair are
supplementary“, we get,
Further, it is given AB divides
in the ratio 1:3
So, let
Thus,
Hence,
Also, using (1)
Now, in ΔABC , using the property, “exterior angle of a triangle is equal to the sum of its two
opposite interior angles“, we get,
Therefore,
.
Solution 6 :
Answer :
In the given ΔABC, the bisectors of
and
intersect at D
We need to prove:
Now, using the exterior angle theorem,
…….(1)
As,
and
is bisected
Also,
Further, applying angle sum property of the triangle
In ΔDCB
……(2)
Also, CBE is a straight line, So, using linear pair property
……(3)
So, using (3) in (2)
Using (2), we get
Hence proved.
Solution 7 :
Answer :
In the given figure,
and
. We need to find the value of
Since,
Let,
Applying the angle sum property of the triangle, in ΔABC, we get,
Thus,
Further, BCD is a straight line. So, applying the property, “the angles forming a linear pair are
supplementary“, we get,
Therefore,
.
Solution 8 :
Answer :
In the given ΔABC,
We need to find
,
is the bisector of
,
and
Now, using the angle sum property of the triangle
In ΔAMC, we get,
…….(1)
Similarly,
In ΔABM, we get,
…..(2)
So, adding (1) and (2)
Now, since AN is the bisector of
Thus,
Now,
Therefore,
.
Solution 9 :
Answer :
In the given ΔABC, AD bisects
and
. We need to prove
Let,
Also,
As AD bisects
,
…..(1)
Now, in ΔABD, using exterior angle theorem, we get,
Similarly,
[using (1)]
Further, it is given,
Adding
to both the sides
.
Thus,
Hence proved.
Solution 10 :
Answer :
In the given ΔABC,
and
.
We need prove
Here, in ΔBDC, using the exterior angle theorem, we get,
Similarly, in ΔEBC, we get,
Adding (1) and (2), we get,
Now, on using angle sum property,
In ΔABC, we get,
This can be written as,
Similarly, using angle sum property in ΔOBC, we get,
This can be written as,
Now, using the values of (4) and (5) in (3), we get,
Therefore,
.
Hence proved
Solution 11 :
Answer :
In the given problem, AE bisects
and
We need to prove
As,
is bisected by AE
=2
=2
..........(1)
Now, using the property, “an exterior angle of a triangle in equal to the sum of the two opposite
interior angles“, we get,
(
)
(using 1)
Hence, using the property, if alternate interior angles are equal, then the two lines are parallel, we
get,
Thus,
Hence proved.
Solution 12 :
Answer :
In the given problem,
We need to find
Now,
and AE is the transversal, so using the property, “alternate interior angles are
equal“, we get,
Further, applying angle sum property of the triangle
In ΔDCE
Further, ACE is a straight line, so using the property, “the angles forming a linear pair are
supplementary“, we get,
Therefore,
.
Solution 13 :
Answer :
(i) Sum of the three angles of a triangle is 180°
According to the angle sum property of the triangle
In ΔABC
Hence, the given statement is
.
(ii) A triangle can have two right angles.
According to the angle sum property of the triangle
In ΔABC
Now, if there are two right angles in a triangle
Let
Then,
(This is not possible.)
Therefore, the given statement is
.
(iii) All the angles of a triangle can be less than 60°
According to the angle sum property of the triangle
In ΔABC
Now, If all the three angles of a triangle is less than
Then,
Therefore, the given statement is
.
(iv) All the angles of a triangle can be greater than 60°
According to the angle sum property of the triangle
In ΔABC
Now, if all the three angles of a triangle is greater than
Then,
Therefore, the given statement is
.
(v) All the angles of a triangle can be equal to
According to the angle sum property of the triangle
In ΔABC
Now, if all the three angles of a triangle are equal to
Then,
Therefore, the given statement is
.
(vi) A triangle can have two obtuse angles.
According to the angle sum property of the triangle
In ΔABC
Now, if a triangle has two obtuse angles
Then,
Therefore, the given statement is
.
(vii) A triangle can have at most one obtuse angle.
According to the angle sum property of the triangle
In ΔABC
Now, if a triangle will have more than one obtuse angle
Then,
Therefore, the given statement is
.
(viii) If one angle of a triangle is obtuse, then it cannot be a right angles triangle.
According to the angle sum property of the triangle
In ΔABC
Now, if it is a right angled triangle
Then,
Also if one of the angle's is obtuse
This is not possible.
Thus, if one angle of a triangle is obtuse, then it cannot be a right angled triangle.
Therefore, the given statement is
.
(ix) An exterior angle of a triangle is less than either of its interior opposite angles
According to the exterior angle property, an exterior angle of a triangle is equal to the sum of the
two opposite interior angles.
In ΔABC
Let x be the exterior angle
So,
Now, if x is less than either of its interior opposite angles
Therefore, the given statement is
.
(x) An exterior angle of a triangle is equal to the sum of the two interior opposite angles.
According to exterior angle theorem,
Therefore, the given statement is
.
(xi) An exterior angle of a triangle is greater than the opposite interior angles.
According to exterior angle theorem,
Since, the exterior angle is the sum of its interior angles.
Thus,
Therefore, the given statement is
Solution 14 :
Answer :
.
(i) Sum of the angles of a triangle is 180°.
As we know, that according to the angle sum property, sum of all the angles of a triangle is 180°.
(ii) An exterior angle of a triangle is equal to the two interior opposite angles.
(iii) An exterior angle of a triangle is always greater than either of the interior opposite angles.
As according to the property: An exterior angle of a triangle is equal to the sum of two interior
opposite angles. Therefore, it has to be greater than either of them.
(iv) A triangle cannot have more than one right angle.
As the sum of all the angles of a triangle is 180°. So, if the triangle has more than one right angle
the sum would exceed 180 °.
(v) A triangle cannot have more than one obtuse angle
As the sum of all the angles of a triangle is 180°. So, if the triangle has more than one obtuse
angle the sum would exceed 180 °.
<< Previous Chapter 8 : Lines and AnglesNext Chapter 10 : Congruent Triangles >>
Exercise Formative assessment_mcq : Solutions of Questions on Page Number : 9.23
Solution 1 :
Answer :
In a given ΔABC we are given that the three angles are equal. So,
According to the angle sum property of a triangle, in ΔABC
Therefore, all the three angles of the triangle are equal to
So, the correct option is (c).
Solution 2 :
Answer :
In the given problem, we have a right angled triangle and the other two angles are equal.
So, In ΔABC
Now, using the angle sum property of the triangle, in ΔABC, we get,
(
)
Therefore, the correct option is (b).
Solution 3 :
Answer :
In the ΔABC, CD is the ray extended from the vertex C of ΔABC. It is given that the exterior
angle of the triangle is
So,
and
and two of the interior opposite angles are equal.
.
So, now using the property, “an exterior angle of the triangle is equal to the sum of the two
opposite interior angles“, we get.
In ΔABC
Therefore, each of the two opposite interior angles is
So, the correct option is (d).
Solution 4 :
Answer :
In the given problem, one angle of a triangle is equal to the sum of the other two angles.
Thus,
..........(1)
Now, according to the angle sum property of the triangle
In ΔABC
.........(2)
Further, using (2) in (1),
Thus,
Therefore, the correct option is (d).
Solution 5 :
Answer :
In the given problem, side BC of ΔABC has been produced to a point D. Such that
and
. Here, we need to find
Given
We get,
Now, using the property, “exterior angle of a triangle is equal to the sum of two opposite interior
angles“, we get,
In ΔABC
Also,
(Using 1)
Thus,
Therefore, the correct option is (a).
Solution 6 :
Answer :
In the given ΔABC,
. D is the ray extended from point A. AX bisects
and
Here, we need to find
As ray AX bisects
Thus,
Now, according to the property, “exterior angle of a triangle is equal to the sum of two opposite
interior angles“, we get,
Thus,
Therefore, the correct option is (c).
Solution 7 :
Answer :
In the given ΔABC,
and
Now, according to the property, “exterior angle of a triangle is equal to the sum of two opposite
interior angles“, we get,
So,
Therefore, the correct option is (c).
Solution 8 :
Answer :
In the given ΔABC, all the three sides of the triangle are produced. We need to find the sum of
the three exterior angles so produced.
Now, according to the angle sum property of the triangle
.......(1)
Further, using the property, “an exterior angle of the triangle is equal to the sum of two opposite
interior angles“, we get,
......(2)
Similarly,
.......(3)
Also,
.......(4)
Adding (2) (3) and (4)
We get,
Thus,
Therefore, the correct option is (d).
Solution 9 :
Answer :
In the given ΔABC,
, AD bisects
and
.
Here, we need to find
As, AD bisects
.
,
We get,
Now, according to angle sum property of the triangle
In ΔABD
Hence,
Therefore, the correct option is (c).
Solution 10 :
Answer :
In the given ΔABC, an exterior angle
ratio 4:5.
Let us take,
and its interior opposite angles are in the
Now using the property, “exterior angle of a triangle is equal to the sum of two opposite interior
angles“
We get,
Thus,
Also, using angle sum property in ΔABC
Thus,
Therefore, the correct option is (a).
Solution 11 :
Answer :
In the given ΔABC,
We need to find
and
. Bisectors of
and
meet at O.
Since, OB is the bisector of
.
Thus,
Now, using the angle sum property of the triangle
In ΔABC, we get,
Similarly, in ΔBOC
Hence,
Therefore, the correct option is (b).
Solution 12 :
Answer :
In the given problem, bisectors of the acute angles of a right angled triangle meet at O. We need
to find
.
Now, using the angle sum property of a triangle
In ΔABC
Now, further multiplying each of the term by
in (1)
Also, applying angle sum property of a triangle
In ΔAOC
Thus,
Therefore, the correct option is (c).
Solution 13 :
Answer :
In the given problem, line segment AB and CD intersect at O, such that
and
.
,
We need to find
As
Applying the property, “alternate interior angles are equal“, we get,
.......(1)
Now, using the angle sum property of the triangle
In ΔODB, we get,
(using 1)
Thus,
Therefore, the correct option is (b).
Solution 14 :
Answer :
In the given figure, bisects of exterior angles
We need to find
and
meet at O and
Now, according to the theorem, “if the sides AB and AC of a ΔABC are produced
to P and Q respectively and the bisectors of
and
intersect at O, therefore, we get,
Hence, in ΔABC
Thus,
Therefore, the correct option is (b).
Solution 15 :
Answer :
In the given figure, bisectors of
We need to find
and
meet at E and
Here, using the property, “an exterior angle of the triangle is equal to the sum of the opposite
interior angles“, we get,
In ΔABC with
as its exterior angle
........(1)
Similarly, in
with
as its exterior angle
(CE and BE are the bisectors of
and
)
.......(2)
Now, multiplying both sides of (1) by
We get,
......(3)
From (2) and (3) we get,
Thus,
Therefore, the correct option is (a).
Solution 16 :
Answer :
In the given problem, BC of ΔABC is produced to point D. bisectors of
side BC at L,
and
meet
Here, using the property, “exterior angle of a triangle is equal to the sum of the two opposite
interior angles“, we get,
In ΔABC
Now, as AL is the bisector of
Also,
is the exterior angle of ΔALC
Thus,
Again, using the property, “exterior angle of a triangle is equal to the sum of the two opposite
interior angles“, we get,
In
Thus,
Therefore the correct option is (b).
Solution 17 :
Answer :
In the given figure,
,
and
. We need to find
.
Here,
and CD is the transversal, so using the property, “corresponding angles are
equal“, we get
Also, using the property, “an exterior angle of a triangle is equal to the sum of the two opposite
interior angles“, in ΔOBD, we get,
Thus,
Therefore, the correct option is (b).
Solution 18 :
Answer :
In the given figure, we need to find
Here, AB and CD are straight lines intersecting at point O, so using the property, “vertically
opposite angles are equal“, we get,
Further, applying the property, “an exterior angle of a triangle is equal to the sum of the two
opposite interior angles“, in ΔAOC, we get,
Similarly, in ΔBOD
Thus,
Therefore, the correct option is (b).
Solution 19 :
Answer :
In the given figure, measures of the angles of ΔABC are in the ratio
measure of the smallest angle of the triangle.
. We need to find the
Let us take,
Now, applying angle sum property of the triangle in ΔABC, we get,
Substituting the value of x in
Since, the measure of
,
and
is the smallest
Thus, the measure of the smallest angle of the triangle is
Therefore, the correct option is (c).
Solution 20 :
Answer :
In the given figure,
We need to find the value of x.
Now, since AB and CD are straight lines intersecting at point O, using the property, “vertically
opposite angles are equal“, we get,
Further, applying angle sum property of the triangle
In ΔBOC
Then, using the property, “an exterior angle of the triangle is equal to the sum of the two
opposite interior angles“, we get,
In ΔEOC
Further solving for x, we get,
Thus,
Therefore, the correct option is (b).
Solution 21 :
Answer :
In the given ΔABC, we need to convert z in terms of x and y
Now, BC is a straight line, so using the property, “angles forming a linear pair are
supplementary“
Similarly,
Also, using the property, “vertically opposite angles are equal“, we get,
Further, using angle sum property of the triangle
Thus,
Therefore, the correct option is (b).
Solution 22 :
Answer :
In the given problem, we need to find the value of x if
Here, if
, then using the property, “if the two lines are parallel, then the alternate interior
angles are equal“, we get,
Further, applying angle sum property of the triangle
In ΔABC
Thus,
Therefore, the correct option is (d).
Solution 23 :
Answer :
In the given figure, we need to find y in terms of x
Now, using the property, “an exterior angle of the triangle is equal to the sum of the two opposite
interior angles“, we get
In ΔABC
..........(1)
Similarly, in ΔOCD
(using 1)
Thus,
Therefore, the correct option is (a).
Solution 24 :
Answer :
In the given problem,
We need to find the value of x
Here, as
, using the property, “consecutive interior angles are supplementary“, we get
..........(1)
Further, applying angle sum property of the triangle
In ΔABC
(using 1)
Now, AB is a straight line, so using the property, “angles forming a linear pair are
supplementary“, we get,
Thus,
Therefore, the correct option is (c).
Solution 25 :
Answer :
In the given figure, we need to find the value of x.
Here, DBA is a straight line, so using the property, “angles forming a linear pair are
supplementary“, we get,
Now, applying the value of y in
and
Also,
Further, applying angle sum property of the triangle
In ΔABC
Thus,
Therefore, the correct option is (d).
Solution 26 :
Answer :
In the given problem, we need to find the value of x.
Here, according to the corollary, “if bisectors of
then
In ΔRST
Further solving for x, we get,
and
of a ΔABC meet at a point O,
Thus,
Therefore, the correct option is (d).
Solution 27 :
Answer :
In the given figure, we need to find the value of x
Here, according to the angle sum property of the triangle
In ΔABD
Also, ABC is a straight line. So, using the property, “angles forming a linear pair are
supplementary“, we get,
Further, using the property, “exterior angle of a triangle is equal to the sum of two opposite
interior angles“, we get
Thus,
Therefore, the correct option is (d).
Solution 28 :
Answer :
In the given figure,
and
We need to find the measure of x
Here, we draw a line RS parallel to BP, i.e
Also, using the property, “two lines parallel to the same line are parallel to each other“
As,
Thus,
Now,
equal“
and BA is the transversal, so using the property, “alternate interior angles are
Similarly,
and AC is the transversal
........(2)
Adding (1) and (2), we get
Also, as
Using the property,“angles opposite to equal sides are equal“, we get
Further, using the property, “an exterior angle is equal to the sum of the two opposite interior
angles“
In ΔABC
Thus,
Therefore, the correct option is (c).
Solution 29 :
Answer :
In the given figure,
,
We need to find the value of x and y
,
and
Here, we draw a line ST parallel to AB, i.e
Also, using the property, “two lines parallel to the same line are parallel to each other“
As,
Thus,
Now,
and EF is the transversal, so using the property, “alternate interior angles are
equal“, we get,
Similarly,
and EF is the transversal
.......(2)
Adding (1) and (2), we get
Further,FPE is a straight line
Applying the property, angles forming a linear pair are supplementary
Also, applying angle sum property of the triangle
In ΔPRQ
Thus,
Therefore, the correct option is (a).
Solution 30 :
Answer :
In the given problem, the exterior angles obtained on producing the base of a triangle both ways
are
and
. So, let us draw ΔABC and extend the base BC, such that:
Here, we need to find
Now, since BCD is a straight line, using the property, “angles forming a linear pair are
supplementary“, we get
Similarly, EBS is a straight line, so we get,
Further, using angle sum property in ΔABC
Thus,
Therefore, the correct option is (c).
<< Previous Chapter 8 : Lines and AnglesNext Chapter 10 : Congruent Triangles >>
Exercise Formative assessment_vsa : Solutions of Questions on Page Number : 9.21
Solution 1 :
Answer :
A plane figure bounded by three lines in a plane is called a triangle. The figure below represents
a ΔABC, with AB, AC and BC as the three line segments.
Solution 2 :
Answer :
In the given problem, ΔABC is an obtuse triangle, with
as the obtuse angle.
So, according to “the angle sum property of the triangle“, for any kind of triangle, the sum of its
angles is 180°. So,
Therefore, sum of the angles of an obtuse triangle is
.
Solution 3 :
Answer :
In ΔABC,
,
and the bisectors of
We need to find the measure of
Since,BO is the bisector of
and
meet at O.
Similarly,CO is the bisector of
Now, applying angle sum property of the triangle, in ΔBOC, we get,
Therefore,
.
Solution 4 :
Answer :
In the given problem, angles of ΔABC are in the ratio 2:1:3
We need to find the measure of the smallest angle.
Let,
According to the angle sum property of the triangle, in ΔABC, we get,
Thus,
Since, the measure of
is the smallest of all the three angles.
Therefore, the measure of the smallest angle is
Solution 5 :
Answer :
In the given ΔABC,
,
and
satisfy the relation
We need to fine the measure of
As,
.
.
........(1)
Now, using the angle sum property of the triangle, we get,
(Using 1)
Therefore,
Solution 6 :
Answer :
In the given ΔABC,
, the bisectors of
and
meet
at O and
We need to find the measure of
So here, using the corollary, “if the bisectors of
then
Thus, in ΔABC
“
and
of a
meet at a point O,
Thus,
Solution 7 :
Answer :
Exterior angle theorem states that, if a side of a triangle is produced, the exterior angle so formed
is equal to the sum of the two interior opposite angles.
Thus, in ΔABC
Solution 8 :
Answer :
In the given problem, we need to find the difference between the sum of the exterior angles and
.
Now, according to the exterior angle theorem
.........(1)
Also,
.........(2)
Further, adding (1) and (2)
.........(3)
Also, according to the angle sum property of the triangle, we get,
.........(4)
Now, we need to find the difference between the sum of the exterior angles and
Thus,
(Using 4)
Therefore,
Solution 9 :
Answer :
In the given
We need to find
,
and AB is produced to D such that
.
Now, using the property, “angles opposite to equal sides are equal“
As
........(1)
Similarly,
As
........(2)
Also, using the property, “an exterior angle of the triangle is equal to the sum of the two opposite
interior angle“
In ΔBDC
(Using 2)
From (1), we get
.......(3)
Now, we need to find
That is,
(Using 3)
(Using 2)
Eliminating
from both the sides, we get 3:1
Thus, the ratio of
is
Solution 10 :
Answer :
In the given problem, the sum of two angles of a triangle is equal to its third angle.
We need to find the measure of the third angle.
Thus, it is given, in
........(1)
Now, according to the angle sum property of the triangle, we get,
(Using 1)
Therefore, the measure of the third angle is
.
Solution 11 :
Answer :
In the given figure,
,
,
and
We need to find
Here, GF and CD are straight lines intersecting at point H, so using the property, “vertically
opposite angles are equal“, we get,
Further, as
and AC is the transversal
Using the property, “alternate interior angles are equal“
Further applying angle sum property of the triangle
In ΔGHC
Hence, applying the property, “angles forming a linear pair are supplementary“
As AGC is a straight line
Therefore,
Solution 12 :
Answer :
In the given figure,
,
,
and
We need to find the value of x and y
Here, as
and BD is the transversal, so according to the property, “alternate interior
angles are equal“, we get
Similarly, as
and DF is the transversal
(Using 1)
Further, EGH is a straight line. So, using the property, angles forming a linear pair are
supplementary
Also, using the property, “an exterior angle of a triangle is equal to the sum of the two opposite
interior angles“, we get,
In
with
as its exterior angle
Thus,
Solution 13 :
Answer :
In the given figure, bisectors of
and
meet at E and
We need to find
Here, using the property: an exterior angle of the triangle is equal to the sum of the opposite
interior angles.
In ΔABC with
as its exterior angle
........(1)
Similarly, in ΔBE with
as its exterior angle
(CE and BE are the bisectors of
........(2)
Now, multiplying both sides of (1) by
We get,
and
)
........(3)
From (2) and (3) we get,
Thus,