Answered

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8. Triangle ENT is a right triangle with angle N as its angle. Given the measure of angle T=16° and the length of NT=11.2 dm, which trigonometric ratio can be used to solve for the length of NE?
A. Secant B. Tangent C. Sine D. Gosine

9. What is the length of UV in item No. 8?
A. 40.63 dm B. 11.65 dm C. 39.06 dm D. 3.21 dm​

Sagot :

Rktmobilezidn

Question:

8. Triangle ENT is a right triangle with angle N as its angle. Given the measure of angle T=16° and the length of NT=11.2 dm, which trigonometric ratio can be used to solve for the length of NE?

A. Secant B. Tangent C. Sine D. Gosine

9. What is the length of UV in item No. 8?

A. 40.63 dm B. 11.65 dm C. 39.06 dm D. 3.21 dm​

Answer:

Based on the given figure,

Reference Angle = Angle A = 16°

Opposite Side = Length of NE = ?

Hypotenuse Side = ET = not given

Adjacent Side = Length of NT = 11.2 dm

In order to solve the Length NE (Opposite) we need to use a trigonometric ratio that uses both the Adjacent and Opposite Side.

Based on the choices, only Tangent ratio is possible to solve the length of NE,

[tex]\large {\sf \tan\theta = \frac{opposite}{adjacent}[/tex]

The other choices all uses Hypotenuse as one of its ratio and since the length of Hypotenuse is not given, we cannot solve the length NE by using these trigonometric ratios.

[tex]\large {\sf \sec\theta = \frac{hypotenuse}{adjacent}}\\\\\large {\sf \sin\theta = \frac{opposite}{hypotenuse}}\\\\\large {\sf \cos\theta = \frac{adjacent}{hypotenuse}}[/tex]

Hence the correct answer is letter B. Tangent

9. What is the length of UV in item No. 8?

What is UV? Our triangle is ENT ?

In any case let's solve for the missing lengths of the Opposite side (NE) and Hypotenuse (ET)

solve for NE (opposite side),

[tex]\large {\sf \tan\theta = \frac{opposite}{adjacent}}\\\\\large {\sf tan16^\circ = \frac{NE}{11.2}}\\\\\large {\sf NE = (11.2)(tan16^\circ)}\\\\\large {\sf NE = 3.21 \;dm}\\\\[/tex]

solve for ET (hypotenuse),

[tex]\large {\sf \cos\theta = \frac{adjacent}{hypotenuse}}\\\\\large {\sf cos16^\circ = \frac{11.2}{ET}}\\\\\large {\sf ET = \frac{11.2}{cos16^\circ }}\\\\\large {\sf ET = 11.65 \;dm}[/tex]

* as always, double check my answers for errors or carelessness.

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#No to plagiarism

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