Solved Examples

SOLVED EXAMPLES 1. In Triangle ABC , ∠A = 50o and AB=AC. Find ∠B and ∠C .

Sol: In Triangle ABC , we have AB=AC

Therefore ∠B = ∠C (Angles opposite equal sides are equal) By Angle Sum Property of a Triangle, we have

∠A + ∠B + ∠C = 180o or ∠A + ∠B + ∠B = 180o or 50o + 2∠B = 180o or 2∠B = 180o − 50o or 2∠B = 130o

130o or ∠B = = 65o 2

( Since ∠B = ∠C )

Thererore ∠C = ∠B = 65o Hence the angles are ∠A = 50o , ∠B = 65o , ∠C = 65o 2. In an isosceles triangle if the vertex angle is twice the sum of base angles, calculate the degree measure of all the angles of the triangle.

Sol: Let ABC be an isosceles triangle in which AB=AC and ∠A = 2(∠B + ∠C ) Since AB=AC, Therefore ∠B = ∠C (Angles opposite equal sides of a isosceles triangle are equal) Let ∠B = x o

Therefore ∠C = x o

( Since ∠B = ∠C )

Therefore ∠A = 2(∠B + ∠C ) ∠A = 2( x o + x o ) ∠A = 2(2 x o ) ∠A = 4 x o

Using Angle Sum Property of Triangles we have

∠A + ∠B + ∠C = 180o or 4 x o + x o + x o = 180o or 6 x o = 180o

180o or x = = 30o 6 o

Therefore 4 x o = 4 × 30o = 120o Hence ∠A = 120o , ∠B = 30o , ∠C = 30o 3. In figure 5-5, AB=AC and ∠ACD = 120o . Find ∠A

Sol: Since AB=AC, Therefore ∠B = ∠C (Angles opposite equal sides are equal) Now, ∠ACB + ∠ACD = 180o

(They form a linear pair)

Therefore ∠C + 120o = 180o or ∠C = 180o − 120o = 60o

Therefore ∠B = 60o ( Since ∠B = ∠C )

4. In figure 5-6, it is given that AB=CF, EF=BD and ∠AFE = ∠DBC . Prove that Triangle AFE ≅ Triangle CBD Sol: It is given that AB=CF

Therefore AB + BF = CF = BF (By adding BF on both sides) or AF=CB

… (1)

In Triangle AFE and Triangle CBD , we have AF=CB

(From 1)

∠AFE = ∠DBC

(Given)

EF=BD

(Given)

So, by SAS congruence we have Triangle AFE ≅ Triangle CBD . 5. In figure 5-7, Triangle ABC is an isosceles triangle with AB=AC. Side BA is produced to D such that AB=AD. Prove that ∠BCD = 90o ?

Since AB=AC

Therefore ∠ACB = ∠ABC (Angles opposite equal sides are equal)

… (1)

Also AB=AC and AB=AD ∴ AC=AD (Transitive Property) Thus ∠ACD = ∠ADC

… (2)

Adding (1) and (2), we have ∠ACB + ∠ACD = ∠ABC + ∠ADC = ∠BCD

… (3)

By Angle Sum Property of Triangles, we have

∠ABC + ∠ADC + ∠BCD = 180o or ∠BCD + ∠BCD = 180o

( Since ∠ABC + ∠ADC = ∠BCD )

or 2∠BCD = 180o or ∠BCD =

180o = 90o 2

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