Question

Question 9 and 10 in picture, help please

Answer 1

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.

$\tan KLJ =\frac{4y}{3y} =\frac{4}{3}$

⇒ ∠ KLJ = $\tan^{-1} (\frac{4}{3} )= 53.13$ Degree.

Similarly, $\tan KJL =\frac{3y}{4y} =\frac{3}{4}$  

⇒ ∠ KJL = $\tan^{-1} (\frac{3}{4} )= 36.87$ 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,  $\tan SQR =\frac{5}{12}$

⇒ ∠ SQR = $\tan^{-1} (\frac{5}{12} )= 22.62$ Degree.

Similarly, $\tan QSR =\frac{12}{5}$

⇒ ∠ QSR = $\tan^{-1} (\frac{12}{5} )= 67.38$ Degree.

Therefore, the angles of the Δ SRQ will be 22.62°, 67.38°, and 90°. (Answer)