Question

Instructions:Type the correct answer in each box. Use numerals instead of words. If necessary, use / for the fraction bar(s).

Answer 1

Given

∠ABC is an inscribed angle  

  Find out the  m∠BDA  and m∠BCA .

To proof

First find the value  of the central angle ( intercepted arc measure BA)

∠BOA = 360° - 250°

          =  110 °

Thus the intercepted arc AB is of measure 110°

FORMULA

$inscribed\ angle =\frac{1}{2} in tercepted\ arc \ measure$

thus putting the value in the above equation

we get

$\angle BDA=\frac{1}{2}(110^{\circ})$

∠BDA = 55°

Now find out ∠BCA  

In the quadilateral AOBC

As shown in the diagram AC & BD  are tangent

thus

∠CAO = 90°

∠CBO = 90°

As we know the sum of a quadilateral is 360°.

thus

∠AOB + ∠CAO + ∠CBO +∠ BCA  = 360°

Put the value as mentioned above

110° +90° + 90° +∠BCA = 360°

∠BCA = 360° - 290°

∠BCA =  70°

Hence proved