Question #142687
Angle bisectors of angles BAC and BCA of triangle ABC intersect its altitude BH at points P and
Q respectively. It is known that point H lies on side AC, P is situated between H and Q,
BQ = 54,5, QP = 17,5, PH = 36. Find the area S of triangle ABC. As the answer,
indicate the value of S√3 .
1
Expert's answer
2020-11-08T18:42:52-0500
BH=BQ+QP+PH=BH=BQ+QP+PH=

=54.5+17.5+36=108=54.5+17.5+36=108

The right triangle APHAPH

Let AH=x,PAH=α.AH=x, \angle PAH=\alpha. Then tan(α)=PHAH=36x\tan(\alpha)=\dfrac{PH}{AH}=\dfrac{36}{x}

The right triangle ABHABH

tan(2α)=BHAH=108x\tan(2\alpha)=\dfrac{BH}{AH}=\dfrac{108}{x}

tan(2α)=2tan(α)1tan2(α)=108x\tan(2\alpha)=\dfrac{2\tan(\alpha)}{1-\tan^2(\alpha)}=\dfrac{108}{x}

108x=236x1(36x)2\dfrac{108}{x}=\dfrac{2\cdot\dfrac{36}{x}}{1-(\dfrac{36}{x})^2 }1(36x)2=231-(\dfrac{36}{x})^2=\dfrac{2}{3}

x=363x=36\sqrt{3}

α=30°\alpha=30\degree

The right triangle QCHQCH

Let CH=y,QCH=β.CH=y, \angle QCH=\beta. Then tan(β)=QHCH=53.5y\tan(\beta)=\dfrac{QH}{CH}=\dfrac{53.5}{y}

The right triangle BCHBCH

tan(2β)=BHCH=108y\tan(2\beta)=\dfrac{BH}{CH}=\dfrac{108}{y}

tan(2β)=2tan(β)1tan2(β)=108y\tan(2\beta)=\dfrac{2\tan(\beta)}{1-\tan^2(\beta)}=\dfrac{108}{y}

108y=253.5y1(53.5y)2\dfrac{108}{y}=\dfrac{2\cdot\dfrac{53.5}{y}}{1-(\dfrac{53.5}{y})^2 }1(53.5y)2=1071081-(\dfrac{53.5}{y})^2=\dfrac{107}{108}

(y53.5)2=108(\dfrac{y}{53.5})^2=108

y=3213y=321\sqrt{3}

Then

AC=AH+CH=363+3213=3573AC=AH+CH=36\sqrt{3}+321\sqrt{3}=357\sqrt{3}

SABC=12BHAC=S_{ABC}=\dfrac{1}{2}BH\cdot AC=

=12(108)(3573)=192783=\dfrac{1}{2}(108)(357\sqrt{3})=19278\sqrt{3}

S3=57834S\sqrt{3}=57834


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