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Isosceles Triangles — MATHEMATICS STD 9 — Question

ICSE BoardEnglish MediumSTD 9MATHEMATICSIsosceles Triangles4 Marks
Question
Calculate :$(i)\angle ADC(ii)\angle ABC(iii)\angle BAC$
✓

Answer

Given: $ACE = 130^\circ ; AD = BD = CD$
Proof:
$(i) \angle ACD + \angle ACE = 18^\circ \dots....... [\text{DCE}$ is a st. line $]$
$\Rightarrow \angle ACD = 180^\circ - 130^\circ $
$\Rightarrow \angle ACD = 50^\circ $
Now $, CD = AD$
$\Rightarrow \angle ACD = \angle DAC = 50^\circ\dots ..... (i)[$ Since angels opposite to equal sides are equal$]$
In $\triangle ADC,$
$\angle ACD = \angle DAC = 50^\circ $
$\angle ACD + \angle DAC + \angle ADC = 180^\circ $
$50^\circ + 50^\circ + \angle ADC = 180^\circ $
$\angle ADC =180^\circ − 100^\circ $
$\angle ADC = 80^\circ $
$(ii) \angle ADC = \angle ABD + \angle DAB\dots ....[$Exterior angle is equal to sum of opp. interor angle$]$
But $AD = BD$
$\therefore \angle DAB = \angle ABD$
$\Rightarrow 80^\circ = \angle ABD + \angle ABD$
$\Rightarrow 2\angle BD = 80^\circ $
$\Rightarrow \angle ABD = 40^\circ = \angle DAB\dots .....(ii)$
$(iii)\angle BAC = \angle DAB + \angle DAC$
substituting the value from $(i)$ and $(ii)$
$\angle BAC = 40^\circ + 50^\circ $
$\Rightarrow \angle BAC = 90^\circ $

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