[tex]\huge\mathfrak{\underline{answer:}}[/tex]
[tex]\large\bf{\angle ACD = 105°}[/tex]
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[tex]\large\bf{\underline{Here:}}[/tex]
- BCD is an isosceles right triangle , right angled at D
- ABC is an equilateral triangle
[tex]\large\bf{\underline{To\:find:}}[/tex]
[tex]\large\bf{In\: triangle\:ABC:}[/tex]
❒ Sum of all angles is 180° , since it is an equilateral triangle all the three angles would be same
[tex]\large\bf{\underline{So}}[/tex]
[tex]\bf{⟼\angle ACB = \frac{180}{3}}[/tex]
[tex]\bf{⟼\angle ACB = 60°}[/tex]
[tex]\large\bf{In\: triangle\:BDC}[/tex]
[tex]\bf{⟼\angle D = 90°}[/tex]
[tex]\bf{⟼\angle DBC =\angle DCB }[/tex](Isosceles triangle)
[tex]\bf{⟼\angle DBC + \angle BDC + \angle DCB = 180° }[/tex]
[tex]\bf{⟼2\angle DCB + 90° = 180°}[/tex]
[tex]\bf{⟼2\angle DCB = 180 -90}[/tex]
[tex]\bf{⟼\angle DCB = \frac{90}{2}}[/tex]
[tex]\bf{⟼\angle DCB = 45°}[/tex]
[tex]\large\bf{\underline{Therefore,}}[/tex]
[tex]\bf{⟹\angle ACD = \angle ACB + \angle DCB}[/tex]
[tex]\bf{⟹\angle ACD = 60+45}[/tex]
[tex]\large\bf{⟹\angle ACD = 105°}[/tex]
