Respuesta :

We shall use the mnemonics SOH-CAH-TOA to find the missing angles.

SOH- means sine ratio is [tex]\frac{Opposite}{Hypotenuse}[/tex]

CAH- means cosine ratio is [tex]\frac{Adjacent}{Hypotenuse}[/tex]

TOA- means tangent ratio is [tex]\frac{Opposite}{Adjacent}[/tex]

11) From the given right-angle triangle,the side length adjacent to [tex]\theta[/tex] is 11 units.

The opposite side  is 8 units.

We use the tangent ratio to obtain:

[tex]\tan \theta=\frac{8}{11}[/tex]

[tex]\theta=\tan ^{-1}(\frac{8}{11})[/tex]

[tex]\theta=36.0\degree[/tex] to the nearest tenth.

12) From the given right-angle triangle,the side length adjacent to [tex]\theta[/tex] is 7 units.

The opposite side  is 13 units.

We use the tangent ratio to obtain:

[tex]\tan \theta=\frac{13}{7}[/tex]

[tex]\theta=\tan ^{-1}(\frac{13}{7})[/tex]

[tex]\theta=61.7\degree[/tex] to the nearest tenth.

13) From the given right-angle triangle,the side length adjacent to [tex]\theta[/tex] is 8 units.

The opposite side  is 11 units.

We use the tangent ratio to obtain:

[tex]\tan \theta=\frac{11}{8}[/tex]

[tex]\theta=\tan ^{-1}(\frac{11}{8})[/tex]

[tex]\theta=54.0\degree[/tex] to the nearest tenth.

14. This time we were given the hypotenuse to be 9.7 units and the opposite side of the right-angle triangle is 7 units.

We use the sine ratio to obtain:

[tex]\sin \theta=\frac{7}{9.7}[/tex]

[tex]\implies \sin \theta=\frac{70}{97}[/tex]

[tex]\implies \sin \theta=\frac{70}{97}[/tex]

[tex]\implies \theta=\sin^{-1}(\frac{70}{97})[/tex]

[tex]\implies \theta=46.2\degree[/tex]

15. For question 15; we the hypotenuse to be 7 units and the adjacent side is 4 units.

We use the cosine ratio to get;

[tex]\cos \theta=\frac{4}{7}[/tex]

[tex]\implies \theta=\cos^{-1}(\frac{4}{7})[/tex]

[tex]\implies \theta=55.2\degree[/tex] to the nearest tenth.

16) From the given right-angle triangle,the side length adjacent to [tex]\theta[/tex] is 13 units.

The opposite side  is 12 units.

We use the tangent ratio to obtain:

[tex]\tan \theta=\frac{12}{13}[/tex]

[tex]\theta=\tan ^{-1}(\frac{12}{13})[/tex]

[tex]\theta=42.7\degree[/tex] to the nearest tenth.

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