Answer to Question #164862 in Trigonometry for Jankarlos

Question

Include a diagram for each question.

Solve triangle △ 𝑋𝑌𝑍, given ∠𝑋 = 90°, 𝑥 = 21𝑐𝑚, 𝑧 = 15𝑐𝑚

Answer 1

By the Pythagorean Theorem

$$x^2=y^2+z^2$$

$$y=\sqrt{x^2-z^2}$$

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

$$y=\sqrt{21^2-15^2}=6\sqrt{6}(cm)$$

$$\sin \angle Z=\dfrac{z}{x}=\dfrac{15}{21}=\dfrac{5}{7}$$

$$\angle Z=\sin^{-1}(\dfrac{5}{7})\approx45.6\degree$$

$$\angle Y=90\degree-\sin^{-1}(\dfrac{5}{7})\approx 44.4\degree$$