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Thay CP = 13,394 vào (1) ta có:
AP = 13,394.cotg 20 ° ≈ 36,801 (cm)
Thay CP = 13,394 vào (2) ta có:
BP = 13,394.cotg 30 ° ≈ 27,526 (cm)
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góc ACB=180-20-30=130 độ
Xét ΔABC có
AB/sinC=AC/sinB=BC/sinA
=>BC/sin20=AC/sin30=60/sin130
=>\(BC\simeq26,79\left(cm\right);AC\simeq39,16\left(cm\right)\)
\(S_{ABC}=\dfrac{1}{2}\cdot BC\cdot BA\cdot sinBCA\)
\(=\dfrac{1}{2}\cdot39.16\cdot26.79\cdot sin130=401.83\left(cm^2\right)\)
\(CP=2\cdot\dfrac{S_{ABC}}{AB}=\dfrac{2\cdot401.83}{60}\simeq13,39\left(cm\right)\)
Xét ΔCPA vuông tại P có
tan A=CP/AP
=>13,39/AP=tan20
=>\(AP\simeq36.79\left(cm\right)\)
PB=AB-AP=60-36,79=23,21cm
Tính góc A (= 130 độ ). tam giác ACP vuông tại P => AP = cot A .CP (1)
tam giác BCP vuông tại P => BP = cot B . CP (2)
(1) +(2) => AP + BP =cot A .CP +cot B . CP
<=> AB = CP( cot A + cot B)
<=>60= CP ( cot 130 + cot 20 )
=> CP xấp xỉ 31.4
từ đó có thể dễ dàng tính ra AP và BP
trong tam giac ABH co\(AH=AB\cdot\sin B\) \(\Rightarrow AH=60\cdot\sin30=30\)
trong tam giac AHC co \(\sin C=\frac{AH}{AC}\Rightarrow AC=\frac{30}{\sin130}\approx39\)(vi \(gocC=180-20-30=130\)
TRONG TAM GIAC APC CO\(PC=AC\cdot\sin A=39\cdot\sin20\approx13,34\)
\(AP=\cos A\cdot39\approx36,65\)
\(\Rightarrow AP+BP=AB\Rightarrow BP=60-36.65=23.35\)
Đặt AP=x suy ra BP=60-x.Ta có phương trình
xtg\(20^0\)=(60-x)tg\(30^0\)
Đ/s:AP ≈36,801cm;BP=23,119cm;CP=13,396cm
Tham khảo nha