Answer:
9. 36.87°, 53.13°, 90°
10. 22.62°, 67.38°, 90°
Step-by-step explanation:
9. From the diagram given with the question.
[tex]\tan KLJ =\frac{4y}{3y} =\frac{4}{3}[/tex]
⇒ ∠ KLJ = [tex]\tan^{-1} (\frac{4}{3} )= 53.13[/tex] Degree.
Similarly, [tex]\tan KJL =\frac{3y}{4y} =\frac{3}{4}[/tex]
⇒ ∠ KJL = [tex]\tan^{-1} (\frac{3}{4} )= 36.87[/tex] Degree.
Therefore, the angles of the Δ KJL will be 36.87°, 53.13°, and 90°. (Answer)
10. From the diagram given with the question.
Similarly, [tex]\tan SQR =\frac{5}{12}[/tex]
⇒ ∠ SQR = [tex]\tan^{-1} (\frac{5}{12} )= 22.62[/tex] Degree.
Similarly, [tex]\tan QSR =\frac{12}{5}[/tex]
⇒ ∠ QSR = [tex]\tan^{-1} (\frac{12}{5} )= 67.38[/tex] Degree.
Therefore, the angles of the Δ SRQ will be 22.62°, 67.38°, and 90°. (Answer)
