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Please help! i’m giving 20 points. THANK YOU!

Please help! i’m giving 20 points. THANK YOU!-example-1
User Eng Mghase
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1 Answer

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8 votes

Answer:

A. Distance KL = 7.36 m

B. Height ML = 4.25 m

C. The maximum height is 6.96 m

Explanation:

A. Determination of distance KL

Hypothenus = KM = 8.5 m

Angle θ = 30°

Adjacent = KL =?

Cos θ = Adjacent / Hypothenus

Cos 30 = KL / 8.5

0.866 = KL / 8.5

Cross multiply

KL = 0.866 × 8.5

KL = 7.36 m

Thus, the distance is 7.36 m

B. Determination of height (ML)

Hypothenus = KM = 8.5 m

Angle θ = 30°

Opposite = ML =?

Sine θ = Opposite /Hypothenus

Sine 30 = ML /8.5

0.5 = ML / 8.5

Cross multiply

ML = 0.5 × 8.5

ML = 4.25 m

Thus, the height is 4.25 m

C. Determination of the maximum height.

Hypothenus = KM = 8.5 m

Angle θ = 55°

Opposite = ML =?

Sine θ = Opposite /Hypothenus

Sine 55 = ML /8.5

0.8192 = ML / 8.5

Cross multiply

ML = 0.8192 × 8.5

ML = 6.96 m

Thus, the maximum height is 6.96 m

User Morgane
by
2.4k points