Wie löst man das $D r e i e c k H J K$ $Dreieck HJK$ mit $h = 18, j = 10, k = 23$ $h = 18, j = 10, k = 23$ ?

Antworten:

$ˆ H = {48.47}^{\circ}, ˆ J = {24.57}^{\circ}, ˆ K = {106.96}^{\circ}$ $color(brown)(hat H = 48.47^@, hat J = 24.57^@, hat K = 106.96^@$

$Area of triangle A_{t} = 86.086 sq units$ $color(indigo)("Area of triangle " A_t = )color(indigo)(86.086 " sq units"$

Erläuterung:

$Given : h = 18, j = 10, k = 23, To solve the triangle$ $"Given : " h = 18, j = 10, k = 23, " To solve the triangle"$

Anwenden der Kosinusregel,

$cos K = \frac{h^{2} + j^{2} - k^{2}}{2 \cdot h \cdot j}$ $cos K = (h^2 + j^2 - k^2) / (2 * h * j)$

$cos K = \frac{18^{2} + 10^{2} - 23^{2}}{2 \cdot 18 \cdot 10} = - 0.2917$ $cos K = (18^2 + 10^2 - 23^2) / (2 * 18 * 10) = -0.2917$

$ˆ K = {cos}^{- 1} - 0.2917 = {106.96}^{\circ}$ $hat K = cos ^(-1) -0.2917 = 106.96^@$

Ebenso $cos H = \frac{23^{2} + 10^{2} - 18^{2}}{2 \cdot 23 \cdot 10} = 0.663$ $cos H = (23^2 + 10^2 - 18^2) / (2 * 23 * 10) = 0.663$

$ˆ H = {cos}^{- 1} 0.663 = {48.47}^{\circ}$ $hat H = cos ^(-1) 0.663= 48.47^@$

Ebenso $cos J = \frac{23^{2} + 18^{2} - 10^{2}}{2 \cdot 23 \cdot 18} = 0.9094$ $cos J = (23^2 + 18^2 - 10^2) / (2 * 23 * 18) = 0.9094$

$ˆ J = {cos}^{- 1} 0.9094 = {24.57}^{\circ}$ $hat J = cos ^(-1) 0.9094 = 24.57^@$

$Area of triangle A_{t} = (\frac{1}{2}) h j sin K = (\frac{1}{2}) \cdot 18 \cdot 10 \cdot sin ({106.96}^{\circ}) = 86.086 sq units$ $color(indigo)("Area of triangle " A_t = )(1/2) h j sin K = (1/2) * 18 * 10 * sin (106.96^@) = color(indigo)(86.086 " sq units"$

Wie löst man das Dreieck HJK DreieckHJK Dreieck HJK mit h = 18, j = 10, k = 23 h=18,j=10,k=23 h = 18, j = 10, k = 23 ?

Antworten:

Erläuterung:

Wie löst man das $D r e i e c k H J K$ $Dreieck HJK$ mit $h = 18, j = 10, k = 23$ $h = 18, j = 10, k = 23$ ?