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

Antworten:

#color(brown)(hat H = 48.47^@, hat J = 24.57^@, hat K = 106.96^@#

#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"#

https://www.in.pinterest.com/pin/500040364846620813/

Anwenden der Kosinusregel,

#cos K = (h^2 + j^2 - k^2) / (2 * h * j)#

#cos K = (18^2 + 10^2 - 23^2) / (2 * 18 * 10) = -0.2917#

#hat K = cos ^(-1) -0.2917 = 106.96^@#

Ebenso #cos H = (23^2 + 10^2 - 18^2) / (2 * 23 * 10) = 0.663#

#hat H = cos ^(-1) 0.663= 48.47^@#

Ebenso #cos J = (23^2 + 18^2 - 10^2) / (2 * 23 * 18) = 0.9094#

#hat J = cos ^(-1) 0.9094 = 24.57^@#

#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"#