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# Rd Chapter 9- Triangle and its Angles Ex-9.2 Interview Questions Answers

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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°.

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 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°−12A        ......(2)BQC=90°-12A        ......2

Adding (1) and (2), we get Thus, Hence proved.

Question 3 : In the given figure, the sides BC, CA and AB of a Δ ABC have been produced to D, E and F respectively. If ∠ACD = 105° and ∠EAF = 45°, find all the angles of the Δ ABC. 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:

(i) (ii) (iii) (iv) 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 ΔDECx 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. In the given figure, and  Since, and angles opposite to equal sidesare equal. We get,

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=yBAC=3yDAB=y, BAC=3y

Thus,

y+3y=DAC4y=72°y=72°4y=18°y+3y=DAC4y=72°y=72°4y=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,