Question #164862

Include a diagram for each question.


Solve triangle β–³ π‘‹π‘Œπ‘, given βˆ π‘‹ = 90Β°, π‘₯ = 21π‘π‘š, π‘§ = 15π‘π‘š


1
Expert's answer
2021-03-02T05:12:20-0500


By the Pythagorean Theorem


x2=y2+z2x^2=y^2+z^2

y=x2βˆ’z2y=\sqrt{x^2-z^2}

Given x=21 cm,z=15 cmx=21\ cm, z=15\ cm


y=212βˆ’152=66(cm)y=\sqrt{21^2-15^2}=6\sqrt{6}(cm)

sin⁑∠Z=zx=1521=57\sin \angle Z=\dfrac{z}{x}=\dfrac{15}{21}=\dfrac{5}{7}

∠Z=sinβ‘βˆ’1(57)β‰ˆ45.6Β°\angle Z=\sin^{-1}(\dfrac{5}{7})\approx45.6\degree

∠Y=90Β°βˆ’sinβ‘βˆ’1(57)β‰ˆ44.4Β°\angle Y=90\degree-\sin^{-1}(\dfrac{5}{7})\approx 44.4\degree

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