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dangina [55]
3 years ago
15

Question 9 and 10 in picture, help please

Mathematics
1 answer:
ss7ja [257]3 years ago
6 0

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)

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