100025 求三角形各角度已知三邊
阿新 • • 發佈：2022-12-01
<?php header('Content-Type: text/html; charset=utf-8'); define ('ROOT', $_SERVER['DOCUMENT_ROOT']); include ROOT.'/assets/php/head.php'; $tit= '求三角形各角度已知三邊'; /** * a為30 ,b為40，c為43 * a為30 ,b為40，c為50 * a為30 ,b為40，c為61 */ $val='30 40 50'; $img=''; //呼叫方法 mill($val,$img,$tit); /** * mill 是磨粉機的方法 * $val 傳值過來計算，以空格分割成數值 * $img 自定義圖片名，預設以檔名為圖片名 * $tit 標題名 * */ function mill($val,$img,$tit){ //初始化 include ROOT.'/assets/php/init.php'; imgt($img,$tit); //設定小數點保留位數 bcscale (2); //$pi = round(pi(),2); //以空格分割成數值 $vals = expl($val); $a = evev($vals[0]); $b = evev($vals[1]); $c = evev($vals[2]); //已知條件 $know = array(); array_push($know, $val); $v0 = eveq($vals[0]); array_push($know, "a邊長：$v0"); $v1 = eveq($vals[1]); array_push($know, "b邊長：$v1"); $v2 = eveq($vals[2]); array_push($know, "a邊長：$v2"); //計算步驟 $step = array(); array_push($step, "求三角形的{$math['ang']}A："); array_push($step, "{$math['eq']} arccos{$math['lsqb']}{$math['lpar']}$c{$math['sup2']}{$math['add']}$b{$math['sup2']}{$math['sub']}$a{$math['sup2']}{$math['rpar']}{$math['div']}{$math['lpar']}2{$math['mul']}$b{$math['mul']}$c{$math['rpar']}{$math['rsqb']}"); $aa = bcmul($a, $a); $bb = bcmul($b, $b); $cc = bcmul($c, $c); array_push($step, "{$math['eq']} arccos{$math['lsqb']}{$math['lpar']}$cc{$math['add']}$bb{$math['sub']}$aa{$math['rpar']}{$math['div']}{$math['lpar']}2{$math['mul']}$b{$math['mul']}$c{$math['rpar']}{$math['rsqb']}"); $ccbb = bcadd($cc, $bb); $bc = bcmul($b, $c); array_push($step, "{$math['eq']} arccos{$math['lsqb']}{$math['lpar']}$ccbb{$math['sub']}$aa{$math['rpar']}{$math['div']}{$math['lpar']}2{$math['mul']}$bc{$math['rpar']}{$math['rsqb']}"); $ccbbaa = bcsub($ccbb, $aa); $bc2 = bcmul($bc ,2); array_push($step, "{$math['eq']} arccos{$math['lpar']}$ccbbaa{$math['div']}$bc2{$math['rpar']}"); $ccbbaabc2 = $ccbbaa/$bc2; array_push($step, "{$math['eq']} arccos{$math['lpar']}$ccbbaabc2{$math['rpar']}"); $acosccbbaabc2 = acos($ccbbaabc2); array_push($step, "{$math['eq']} 弧度值：{$acosccbbaabc2}"); $ao = rad2deg($acosccbbaabc2); array_push($step, "{$math['eq']} 角度值：{$ao}{$math['o']}"); array_push($step, "求三角形的{$math['ang']}B："); array_push($step, "{$math['eq']} arccos{$math['lsqb']}{$math['lpar']}$a{$math['sup2']}{$math['add']}$c{$math['sup2']}{$math['sub']}$a{$math['sup2']}{$math['rpar']}{$math['div']}{$math['lpar']}2{$math['mul']}$a{$math['mul']}$c{$math['rpar']}{$math['rsqb']}"); $aa = bcmul($a, $a); $cc = bcmul($c, $c); $bb = bcmul($b, $b); array_push($step, "{$math['eq']} arccos{$math['lsqb']}{$math['lpar']}$aa{$math['add']}$cc{$math['sub']}$aa{$math['rpar']}{$math['div']}{$math['lpar']}2{$math['mul']}$a{$math['mul']}$c{$math['rpar']}{$math['rsqb']}"); $aacc = bcadd($aa, $cc); $ac = bcmul($a, $c); array_push($step, "{$math['eq']} arccos{$math['lsqb']}{$math['lpar']}$aacc{$math['sub']}$aa{$math['rpar']}{$math['div']}{$math['lpar']}2{$math['mul']}$ac{$math['rpar']}{$math['rsqb']}"); $aaccbb = bcsub($aacc, $bb); $ac2 = bcmul($ac ,2); array_push($step, "{$math['eq']} arccos{$math['lpar']}$aaccbb{$math['div']}$ac2{$math['rpar']}"); $aaccbbac2 = $aaccbb/$ac2; array_push($step, "{$math['eq']} arccos{$math['lpar']}$aaccbbac2{$math['rpar']}"); $acosaaccbbac2= acos($aaccbbac2); array_push($step, "{$math['eq']} 弧度值：{$acosaaccbbac2}"); $bo = rad2deg($acosaaccbbac2); array_push($step, "{$math['eq']} 角度值：{$bo}{$math['o']}"); array_push($step, "求三角形的{$math['ang']}C："); array_push($step, "{$math['eq']} arccos{$math['lsqb']}{$math['lpar']}$a{$math['sup2']}{$math['add']}$b{$math['sup2']}{$math['sub']}$c{$math['sup2']}{$math['rpar']}{$math['div']}{$math['lpar']}2{$math['mul']}$a{$math['mul']}$b{$math['rpar']}{$math['rsqb']}"); $aa = bcmul($a, $a); $bb = bcmul($b, $b); $cc = bcmul($c, $c); array_push($step, "{$math['eq']} arccos{$math['lsqb']}{$math['lpar']}$aa{$math['add']}$bb{$math['sub']}$cc{$math['rpar']}{$math['div']}{$math['lpar']}2{$math['mul']}$a{$math['mul']}$b{$math['rpar']}{$math['rsqb']}"); $aabb = bcadd($aa, $bb); $ab = bcmul($a, $b); array_push($step, "{$math['eq']} arccos{$math['lsqb']}{$math['lpar']}$aabb{$math['sub']}$cc{$math['rpar']}{$math['div']}{$math['lpar']}2{$math['mul']}$ab{$math['rpar']}{$math['rsqb']}"); $aabbcc = bcsub($aabb, $cc); $ab2 = bcmul($ab ,2); array_push($step, "{$math['eq']} arccos{$math['lpar']}$aabbcc{$math['div']}$ab2{$math['rpar']}"); $aabbccab2 = $aabbcc/$ab2; array_push($step, "{$math['eq']} arccos{$math['lpar']}$aabbccab2{$math['rpar']}"); $acosaabbccab2= acos($aabbccab2); array_push($step, "{$math['eq']} 弧度值：{$acosaabbccab2}"); $co = rad2deg($acosaabbccab2); array_push($step, "{$math['eq']} 角度值：{$co}{$math['o']}"); //算出結果 $ends = array(); $aoo = round($ao,1); $boo = round($bo,1); $coo = round($co,1); array_push($ends, "三角形的{$math['ang']}A：{$aoo}{$math['o']}"); array_push($ends, "三角形的{$math['ang']}B：{$boo}{$math['o']}"); array_push($ends, "三角形的{$math['ang']}C：{$coo}{$math['o']}"); //公式表示 $home = array(); array_push($home, "三角形的{$math['ang']}A：{$math['ang']}A{$math['eq']}arccos{$math['lsqb']}{$math['lpar']}c{$math['sup2']}{$math['add']}b{$math['sup2']}{$math['sub']}a{$math['sup2']}{$math['rpar']}{$math['sol']}2bc{$math['rsqb']}"); array_push($home, "三角形的{$math['ang']}B：{$math['ang']}B{$math['eq']}arccos{$math['lsqb']}{$math['lpar']}a{$math['sup2']}{$math['add']}c{$math['sup2']}{$math['sub']}b{$math['sup2']}{$math['rpar']}{$math['sol']}2ac{$math['rsqb']}"); array_push($home, "三角形的{$math['ang']}C：{$math['ang']}C{$math['eq']}arccos{$math['lsqb']}{$math['lpar']}a{$math['sup2']}{$math['add']}b{$math['sup2']}{$math['sub']}c{$math['sup2']}{$math['rpar']}{$math['sol']}2ab{$math['rsqb']}"); //公式說明 $info = array(); array_push($info, "三角函式的餘弦定理，可求任意三角形"); array_push($info, "已知任意三角形的兩邊及兩邊所夾的角，求第三邊："); array_push($info, "a{$math['sup2']}{$math['eq']}b{$math['sup2']}{$math['add']}c{$math['sup2']}{$math['sub']}2bc{$math['mul']}cos{$math['ang']}A"); array_push($info, "b{$math['sup2']}{$math['eq']}a{$math['sup2']}{$math['add']}c{$math['sup2']}{$math['sub']}2ac{$math['mul']}cos{$math['ang']}B"); array_push($info, "c{$math['sup2']}{$math['eq']}a{$math['sup2']}{$math['add']}b{$math['sup2']}{$math['sub']}2ab{$math['mul']}cos{$math['ang']}C"); array_push($info, "已知三角形的兩邊及與第三邊所夾的兩角，求第三邊："); array_push($info, "a{$math['eq']}b{$math['mul']}cos{$math['ang']}C{$math['add']}c{$math['mul']}cos{$math['ang']}B"); array_push($info, "b{$math['eq']}c{$math['mul']}cos{$math['ang']}A{$math['add']}a{$math['mul']}cos{$math['ang']}C"); array_push($info, "c{$math['eq']}a{$math['mul']}cos{$math['ang']}B{$math['add']}b{$math['mul']}cos{$math['ang']}A"); array_push($info, "已知三角形的三邊，求角度數："); array_push($info, "cos{$math['ang']}A{$math['eq']}{$math['lpar']}c{$math['sup2']}{$math['add']}b{$math['sup2']}{$math['sub']}a{$math['sup2']}{$math['rpar']}{$math['sol']}2bc"); array_push($info, "cos{$math['ang']}B{$math['eq']}{$math['lpar']}a{$math['sup2']}{$math['add']}c{$math['sup2']}{$math['sub']}b{$math['sup2']}{$math['rpar']}{$math['sol']}2ac"); array_push($info, "cos{$math['ang']}C{$math['eq']}{$math['lpar']}a{$math['sup2']}{$math['add']}b{$math['sup2']}{$math['sub']}c{$math['sup2']}{$math['rpar']}{$math['sol']}2ab"); array_push($info, ""); array_push($info, ""); array_push($info, ""); array_push($info, ""); array_push($info, ""); know($know); ends($ends); home($home); step($step); info($info); } ?> <?php include ROOT.'/assets/php/foot.php'; ?>
結果：
☁參考上圖[2022-11-30] ☀求三角形各角度已知三邊 30 40 50 a邊長：30 b邊長：40 a邊長：50 ♠算出結果 三角形的∠A：36.9° 三角形的∠B：53.1° 三角形的∠C：90° ♦公式表示 三角形的∠A：∠A=arccos[(c²+b²−a²)/2bc] 三角形的∠B：∠B=arccos[(a²+c²−b²)/2ac] 三角形的∠C：∠C=arccos[(a²+b²−c²)/2ab] ♣計算步驟 求三角形的∠A： = arccos[(50²+40²−30²)÷(2×40×50)] = arccos[(2500.00+1600.00−900.00)÷(2×40×50)] = arccos[(4100.00−900.00)÷(2×2000.00)] = arccos(3200.00÷4000.00) = arccos(0.8) = 弧度值：0.64350110879328 = 角度值：36.869897645844° 求三角形的∠B： = arccos[(30²+50²−30²)÷(2×30×50)] = arccos[(900.00+2500.00−900.00)÷(2×30×50)] = arccos[(3400.00−900.00)÷(2×1500.00)] = arccos(1800.00÷3000.00) = arccos(0.6) = 弧度值：0.92729521800161 = 角度值：53.130102354156° 求三角形的∠C： = arccos[(30²+40²−50²)÷(2×30×40)] = arccos[(900.00+1600.00−2500.00)÷(2×30×40)] = arccos[(2500.00−2500.00)÷(2×1200.00)] = arccos(0.00÷2400.00) = arccos(0) = 弧度值：1.5707963267949 = 角度值：90° ♥公式解釋 三角函式的餘弦定理，可求任意三角形 已知任意三角形的兩邊及兩邊所夾的角，求第三邊： a²=b²+c²−2bc×cos∠A b²=a²+c²−2ac×cos∠B c²=a²+b²−2ab×cos∠C 已知三角形的兩邊及與第三邊所夾的兩角，求第三邊： a=b×cos∠C+c×cos∠B b=c×cos∠A+a×cos∠C c=a×cos∠B+b×cos∠A 已知三角形的三邊，求角度數： cos∠A=(c²+b²−a²)/2bc cos∠B=(a²+c²−b²)/2ac cos∠C=(a²+b²−c²)/2ab