In the figure, $C$ and $D$ are the points on the semicircle inscribed on $B A$ as diameter. Given $∠$ $B A D$ $= 70^{o}$ and $∠$ $D B C$ $= 30^{o}$ , calculate $∠$ $A B D$ and $∠$ $B D C$ .

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Solution

We have $∠ A D B = 90^{o}$ ....... (Angles formed in a semicircle is a right angle)
Again $∠ A B D = 180^{o} - (∠ A D B + ∠ B A D)$
$= 180^{o} - 90^{o} - 70^{o} = 20^{o}$ ;
$∠ A D C = 180^{o} - ∠ A B C$ ........ (opposite angles of a cyclic quadrilateral)
$= 180^{o} - (∠ A B D + ∠ C B D)$
$= 180^{o} - 50^{o} = 130^{o}$
$∠ B D C = ∠ A D C - ∠ B D A = 130^{o} - 90^{o} = 40^{o}$ .

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In the figure, C and D are the points on the semicircle inscribed on BA as diameter. Given ∠ BAD =70o and ∠ DBC =30o, calculate ∠ ABD and ∠ BDC.

We have ∠ADB=90o ....... (Angles formed in a semicircle is a right angle)Again ∠ABD=180o−(∠ADB+∠BAD) =180o−90o−70o=20o;∠ADC=180o−∠ABC ........ (opposite angles of a cyclic quadrilateral) =180o−(∠ABD+∠CBD) =180o−50o=130o ∠BDC=∠ADC−∠BDA=130o−90o=40o.

In the figure, $C$ and $D$ are the points on the semicircle inscribed on $B A$ as diameter. Given $∠$ $B A D$ $= 70^{o}$ and $∠$ $D B C$ $= 30^{o}$ , calculate $∠$ $A B D$ and $∠$ $B D C$ .