Take the 2-minute tour ×
Mathematics Stack Exchange is a question and answer site for people studying math at any level and professionals in related fields. It's 100% free, no registration required.

nn

If I join the chord then I am getting the angles of the triangle are $45,45,90$ so $\theta=180-2x$ where $x$ is the angle of the other triangle whose angle is $\theta$

share|improve this question
4  
Hint: Inscribed Angle Theorem. (It may help to draw the full circle.) –  Blue Sep 26 '13 at 6:45
    
The central angle is twice the inscribed angle. en.wikipedia.org/wiki/Inscribed_angle#Theorem –  Sammy Black Sep 26 '13 at 6:46

2 Answers 2

up vote 4 down vote accepted

If you draw a line from the center of the circle to your angle, you get 2 isoscele triangles, so your angle is the sum of those 2 other angles in this quadrangle. the sum of all angles of quadrangle is $$360^{\circ}$$ so your angle is $$\frac{360^{\circ}-90^{\circ}}{2} = 135^{\circ}$$

Explanation:

image $$AD = AB = AC$$ $$\angle{ADB} = \angle{ABD} ; \angle{ADC} = \angle{ACD}$$ $$\angle{BDC} = \angle{ADB} + \angle{ADC} = \angle{ABD} + \angle{ACD}$$ $$\angle{BAC} + \angle{ABD} + \angle{BDC} + \angle{ACD} = 360^{\circ}$$ $$90^{\circ} + 2\angle{BDC} = 360^{\circ}$$ $$\angle{BDC} = \frac{360^{\circ}-90^{\circ}}{2} = 135^{\circ}$$

share|improve this answer

Draw the rest of circle (complete the circle),

Inscribed Angle =(1/2)Intercepted Arc

enter image description here $$m(\widehat{CDB})=\frac{1}{2}m( \stackrel \frown{CEB})$$ $$m(\widehat{CDB})=\frac{1}{2}270^{o}=135^{o}$$

share|improve this answer

Your Answer

 
discard

By posting your answer, you agree to the privacy policy and terms of service.

Not the answer you're looking for? Browse other questions tagged or ask your own question.