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Question-16108




Question Number 16108 by Joel577 last updated on 18/Jun/17
Commented by Joel577 last updated on 18/Jun/17
How to find ∠PAD ?
$$\mathrm{How}\:\mathrm{to}\:\mathrm{find}\:\angle{PAD}\:? \\ $$
Answered by ajfour last updated on 18/Jun/17
Commented by ajfour last updated on 18/Jun/17
  2θ+2θ+θ = 180°  ⇒  θ= 36°     ?+? +(90−36°)=180°          ? = ((90°+36°)/2) = 63 ° .
$$\:\:\mathrm{2}\theta+\mathrm{2}\theta+\theta\:=\:\mathrm{180}° \\ $$$$\Rightarrow\:\:\theta=\:\mathrm{36}°\: \\ $$$$\:\:?+?\:+\left(\mathrm{90}−\mathrm{36}°\right)=\mathrm{180}° \\ $$$$\:\:\:\:\:\:\:\:?\:=\:\frac{\mathrm{90}°+\mathrm{36}°}{\mathrm{2}}\:=\:\mathrm{63}\:°\:. \\ $$
Commented by ajfour last updated on 18/Jun/17
they are both radius of the   semicircle  because i took point  P to be the centre of the semicircle.
$${they}\:{are}\:{both}\:{radius}\:{of}\:{the}\: \\ $$$${semicircle}\:\:{because}\:{i}\:{took}\:{point} \\ $$$${P}\:{to}\:{be}\:{the}\:{centre}\:{of}\:{the}\:{semicircle}. \\ $$
Commented by Joel577 last updated on 18/Jun/17
How did u know that DP  and CP   have same length?
$$\mathrm{How}\:\mathrm{did}\:\mathrm{u}\:\mathrm{know}\:\mathrm{that}\:{DP}\:\:\mathrm{and}\:{CP}\: \\ $$$$\mathrm{have}\:\mathrm{same}\:\mathrm{length}? \\ $$
Commented by Joel577 last updated on 18/Jun/17
thank you very much
$${thank}\:{you}\:{very}\:{much} \\ $$
Commented by Joel577 last updated on 23/Jun/17
one more, why  ∠P  is 2θ?
$$\mathrm{one}\:\mathrm{more},\:\mathrm{why}\:\:\angle{P}\:\:\mathrm{is}\:\mathrm{2}\theta? \\ $$

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