Question

1. In this figure find the marked unknown angles
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Answer 1

we know that sum of all sides of triangle is 180°

we know that sum of all sides of triangle is 180°therefore,

we know that sum of all sides of triangle is 180°therefore, In ∆ABC

we know that sum of all sides of triangle is 180°therefore, In ∆ABC angle ACB+45+60 = 180°

we know that sum of all sides of triangle is 180°therefore, In ∆ABC angle ACB+45+60 = 180°angle ACB = 180°- 105°

we know that sum of all sides of triangle is 180°therefore, In ∆ABC angle ACB+45+60 = 180°angle ACB = 180°- 105°angle ACB = 75°

we know that sum of all sides of triangle is 180°therefore, In ∆ABC angle ACB+45+60 = 180°angle ACB = 180°- 105°angle ACB = 75°Now, angle ACB and angle ACD form straight angle

we know that sum of all sides of triangle is 180°therefore, In ∆ABC angle ACB+45+60 = 180°angle ACB = 180°- 105°angle ACB = 75°Now, angle ACB and angle ACD form straight angle therefore, angle ACD = 180°- angle ACB

we know that sum of all sides of triangle is 180°therefore, In ∆ABC angle ACB+45+60 = 180°angle ACB = 180°- 105°angle ACB = 75°Now, angle ACB and angle ACD form straight angle therefore, angle ACD = 180°- angle ACBangle ACD = 180°- 75°

we know that sum of all sides of triangle is 180°therefore, In ∆ABC angle ACB+45+60 = 180°angle ACB = 180°- 105°angle ACB = 75°Now, angle ACB and angle ACD form straight angle therefore, angle ACD = 180°- angle ACBangle ACD = 180°- 75°angle ACD = 105°

we know that sum of all sides of triangle is 180°therefore, In ∆ABC angle ACB+45+60 = 180°angle ACB = 180°- 105°angle ACB = 75°Now, angle ACB and angle ACD form straight angle therefore, angle ACD = 180°- angle ACBangle ACD = 180°- 75°angle ACD = 105° In ∆ ADC.

we know that sum of all sides of triangle is 180°therefore, In ∆ABC angle ACB+45+60 = 180°angle ACB = 180°- 105°angle ACB = 75°Now, angle ACB and angle ACD form straight angle therefore, angle ACD = 180°- angle ACBangle ACD = 180°- 75°angle ACD = 105° In ∆ ADC. x + angle DAC + angle ACD = 180°

we know that sum of all sides of triangle is 180°therefore, In ∆ABC angle ACB+45+60 = 180°angle ACB = 180°- 105°angle ACB = 75°Now, angle ACB and angle ACD form straight angle therefore, angle ACD = 180°- angle ACBangle ACD = 180°- 75°angle ACD = 105° In ∆ ADC. x + angle DAC + angle ACD = 180° (sum of all sides of triangle)

we know that sum of all sides of triangle is 180°therefore, In ∆ABC angle ACB+45+60 = 180°angle ACB = 180°- 105°angle ACB = 75°Now, angle ACB and angle ACD form straight angle therefore, angle ACD = 180°- angle ACBangle ACD = 180°- 75°angle ACD = 105° In ∆ ADC. x + angle DAC + angle ACD = 180° (sum of all sides of triangle)x + 20° + 105° = 180°

we know that sum of all sides of triangle is 180°therefore, In ∆ABC angle ACB+45+60 = 180°angle ACB = 180°- 105°angle ACB = 75°Now, angle ACB and angle ACD form straight angle therefore, angle ACD = 180°- angle ACBangle ACD = 180°- 75°angle ACD = 105° In ∆ ADC. x + angle DAC + angle ACD = 180° (sum of all sides of triangle)x + 20° + 105° = 180°x = 180° - 125°

we know that sum of all sides of triangle is 180°therefore, In ∆ABC angle ACB+45+60 = 180°angle ACB = 180°- 105°angle ACB = 75°Now, angle ACB and angle ACD form straight angle therefore, angle ACD = 180°- angle ACBangle ACD = 180°- 75°angle ACD = 105° In ∆ ADC. x + angle DAC + angle ACD = 180° (sum of all sides of triangle)x + 20° + 105° = 180°x = 180° - 125° x = 55°. Ans.

Answer 2

The cavaliers of x is 55 degree