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Question Number 19709 by NECC last updated on 14/Aug/17

In the cyclic quadrilateral ABCD AB=7,BC=8,CD=8,DA=15. Calculate the angle ADC and the length ofAC.

$${In}\:{the}\:{cyclic}\:{quadrilateral}\:{ABCD} \\ $$$${AB}=\mathrm{7},{BC}=\mathrm{8},{CD}=\mathrm{8},{DA}=\mathrm{15}. \\ $$$${Calculate}\:{the}\:{angle}\:{ADC}\:{and} \\ $$$${the}\:{length}\:{ofAC}. \\ $$

Answered by Tinkutara last updated on 15/Aug/17

Commented by Tinkutara last updated on 15/Aug/17

∠B = π − ∠D ⇒ cos B + cos D = 0 ((7^2 + 8^2 − AC^2 )/(2×7×8)) + ((15^2 + 8^2 − AC^2 )/(2×15×8)) = 0 ((113 − AC^2 )/7) + ((289 − AC^2 )/(15)) = 0 3718 = 22AC^2 AC = 13 cos ∠D = ((15^2 + 8^2 − 13^2 )/(2×15×8)) = (1/2) ⇒ ∠ADC = 60°

$$\angle{B}\:=\:\pi\:−\:\angle{D} \\ $$$$\Rightarrow\:\mathrm{cos}\:{B}\:+\:\mathrm{cos}\:{D}\:=\:\mathrm{0} \\ $$$$\frac{\mathrm{7}^{\mathrm{2}} \:+\:\mathrm{8}^{\mathrm{2}} \:−\:{AC}^{\mathrm{2}} }{\mathrm{2}×\mathrm{7}×\mathrm{8}}\:+\:\frac{\mathrm{15}^{\mathrm{2}} \:+\:\mathrm{8}^{\mathrm{2}} \:−\:{AC}^{\mathrm{2}} }{\mathrm{2}×\mathrm{15}×\mathrm{8}}\:=\:\mathrm{0} \\ $$$$\frac{\mathrm{113}\:−\:{AC}^{\mathrm{2}} }{\mathrm{7}}\:+\:\frac{\mathrm{289}\:−\:{AC}^{\mathrm{2}} }{\mathrm{15}}\:=\:\mathrm{0} \\ $$$$\mathrm{3718}\:=\:\mathrm{22}{AC}^{\mathrm{2}} \\ $$$${AC}\:=\:\mathrm{13} \\ $$$$\mathrm{cos}\:\angle{D}\:=\:\frac{\mathrm{15}^{\mathrm{2}} \:+\:\mathrm{8}^{\mathrm{2}} \:−\:\mathrm{13}^{\mathrm{2}} }{\mathrm{2}×\mathrm{15}×\mathrm{8}}\:=\:\frac{\mathrm{1}}{\mathrm{2}} \\ $$$$\Rightarrow\:\angle{ADC}\:=\:\mathrm{60}° \\ $$

Commented by NECC last updated on 15/Aug/17

thank you sir.

$${thank}\:{you}\:{sir}. \\ $$

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