Rd Chapter 9 Triangle and its Angles Ex 9.2 Solutions
Question - 1 : - The exterior angles, obtained on producing the base of a triangle both way are 104° and 136°. Find all the angles of the triangle.
Answer - 1 : - 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, EBC is a straight line, so we get,
Further, using angle sum property in ΔABC
Therefore,
Question - 2 : - In a Δ ABC, the internal bisectors of ∠B and ∠C meet at P and the external bisectors of ∠B and ∠C meet at Q, Prove that ∠BPC + ∠BQC = 180°.
Answer - 2 : -
In the given problem, BP and CP arethe internal bisectors of 
respectively. Also, BQ and CQ arethe external bisectors of respectively. Here, we need toprove:
We know that if the bisectors of angles
and 
of ΔABC meetat a point O then
Thus, in ΔABC
……(1)
Also, using the theorem, “if the sides AB and AC of a ΔABC are produced, and theexternal bisectors of 
and 
meet at O, then 
Thus, ΔABC
∠BQC=90°−12∠A ......(2)∠BQC=90°-12∠A ......2
Adding (1) and (2), we get
Thus, 
Hence proved.
Answer - 3 : -
In the given ΔABC, 
and 
. We need to find 
.
Here,
are vertically oppositeangles. So, using the property, “vertically opposite angles are equal”, we get,
Further, BCD is a straight line. So, usinglinear pair property, we get,
Now, in ΔABC, using “the angle sum property”, we get,
Therefore,
Question - 4 : - Compute the value of x in each of the following figures:
Answer - 4 : -
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 theproperty, “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 theproperty, “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 thetransversal, so 
form a pair of alternate interiorangles. 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 theproperty, “the angles forming a linear pair are supplementary”, we get,
Also, using the property, “an exterior angle of a triangle isequal to the sum of its two opposite interior angles”
In ΔDEC, x is its exterior angle
Thus,
Therefore,
.
Question - 5 : - In the given figure, AB divides ∠DAC in the ratio 1 : 3 and AB = DB. Determinethe value of x.
Answer - 5 : -
In the given figure,
and
Since,and angles opposite to equal sidesare equal. We get,
∠BDA=∠BAD .....(1)∠BDA=∠BAD .....1
Also, EAD is a straight line. So, using theproperty, “the angles forming a linear pair are supplementary”, we get,
Further, it is given AB divides
in the ratio 1 : 3.
So, let
∠DAB=y, ∠BAC=3y∠DAB=y, ∠BAC=3y
Thus,
y+3y=∠DAC⇒4y=72°⇒y=72°4⇒y=18°y+3y=∠DAC⇒4y=72°⇒y=72°4⇒y=18°
Hence, ∠DAB=18°, ∠BAC=3×18°=54°∠DAB=18°, ∠BAC=3×18°=54°
Using (1)
Now, in ΔABC , using the property, “exterior angleof a triangle is equal to the sum of its two opposite interior angles”, we get,
∠EAC=∠ADC+x⇒108°=18°+x⇒x=90°∠EAC=∠ADC+x⇒108°=18°+x⇒x=90°
Therefore,
Question - 6 : - ABC is a triangle. The bisector of the exterior angle at B and the bisector of ∠C intersect each other at D. Prove that ∠D = 1212 ∠A.
Answer - 6 : -
In the given ΔABC, the bisectors of 
and intersect at D
We need to prove:
Now, using the exterior angle theorem,
∠ABE=∠BAC+∠ACB∠ABE=∠BAC+∠ACB .….(1)
As ∠ABE and ∠ACB are bisectedAs ∠ABE and ∠ACB are bisected
∠DCB=12∠ACB∠DCB=12∠ACB
Also,
∠DBA=12∠ABE∠DBA=12∠ABE
Further, applying angle sum property of the triangle
In ΔDCB
Question - 7 : - In the given figure, AC ⊥ CE and ∠A : ∠B : ∠C = 3 : 2 : 1, find the value of ∠ECD.
Answer - 7 : -
In the given figure,
and 
. We need to find thevalue of 
Since,
Let,
Applying the angle sum property of the triangle, in ΔABC,we get,
Thus,
Further, BCD is a straight line. So, applyingthe property, “the angles forming a linear pair are supplementary”, we get,
Therefore,
.
Question - 8 : - In the given figure, AM ⊥ BC and AN is the bisector of ∠A. If ∠B = 65° and ∠C = 33°, find ∠MAN.
Answer - 8 : -
In the given ΔABC,
,
is the bisector of 
, 
and
We need to find
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,
.