In a triangle ABC,the altitude AD and the med | vedclass

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(A) Let $\angle BAD = \angle DAE = \angle EAC = 	heta$. Thus,$\angle A = 3	heta$.Since $AD$ is the altitude,$	riangle ABD$ and $	riangle ADC$ are

Vedclass · Accepted Answer

$38$

Vedclass · Answer

(A) Let $\angle BAD = \angle DAE = \angle EAC = 	heta$. Thus,$\angle A = 3	heta$.Since $AD$ is the altitude,$	riangle ABD$ and $	riangle ADC$ are right-angled at $D$.In $	riangle ABD$,$	an 	heta = \frac{BD}{AD}$.In $	riangle ADE$,$	an 	heta = \frac{DE}{AD}$.Since $	an 	heta = 	an 	heta$,we have $BD = DE$. Let $BD = DE = x$.Since $AE$ is the median,$BE = EC = 14$. Thus,$DE = BE - BD = 14 - x$.Equating the two expressions for $DE$: $x = 14 - x$ $\Rightarrow 2x = 14$ $\Rightarrow x = 7$.So,$BD = 7$ and $DE = 7$.In $	riangle ADC$,$\angle DAC = 2	heta$,so $	an 2	heta = \frac{DC}{AD} = \frac{DE+EC}{AD} = \frac{7+14}{AD} = \frac{21}{AD}$.In $	riangle ABD$,$	an 	heta = \frac{BD}{AD} = \frac{7}{AD}$.Dividing the two equations: $\frac{	an 2	heta}{	an 	heta} = \frac{21}{7} = 3$.Using the identity $	an 2	heta = \frac{2	an 	heta}{1-	an^2 	heta}$,we get $\frac{2}{1-	an^2 	heta} = 3$.$2 = 3 - 3	an^2 	heta$ $\Rightarrow 3	an^2 	heta = 1$ $\Rightarrow 	an 	heta = \frac{1}{\sqrt{3}}$.Thus,$	heta = 30^{\circ}$.In $	riangle ABD$,$\sin 30^{\circ} = \frac{BD}{AB} = \frac{7}{AB} \Rightarrow AB = \frac{7}{1/2} = 14$.In $	riangle ADC$,$\angle DAC = 60^{\circ}$,so $\sin 60^{\circ} = \frac{DC}{AC} = \frac{21}{AC}$ $\Rightarrow AC = \frac{21}{\sqrt{3}/2} = \frac{42}{\sqrt{3}} = 14\sqrt{3}$.$AB + AC = 14 + 14\sqrt{3} = 14(1 + 1.732) = 14(2.732) = 38.248$.The nearest integer is $38$.

In a $\triangle ABC$,the altitude $AD$ and the median $AE$ divide $\angle A$ into three equal parts. If $BC=28$,then the nearest integer to $AB+AC$ is

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