We need to solve problems 15, 16, 17 and 18 from the image. Problem 15: Find the value of $\sin L$ rounded to the nearest hundredth, if necessary, for the given triangle. Problem 16: Find the value of $\cos R$ rounded to the nearest hundredth, if necessary, for the given triangle. Problem 17: Solve for $x$. Round to the nearest tenth, if necessary. Problem 18: Solve for $x$. Round to the nearest tenth, if necessary.

GeometryTrigonometryRight TrianglesSineCosineTriangle PropertiesRounding
2025/4/1

1. Problem Description

We need to solve problems 15, 16, 17 and 18 from the image.
Problem 15: Find the value of sinL\sin L rounded to the nearest hundredth, if necessary, for the given triangle.
Problem 16: Find the value of cosR\cos R rounded to the nearest hundredth, if necessary, for the given triangle.
Problem 17: Solve for xx. Round to the nearest tenth, if necessary.
Problem 18: Solve for xx. Round to the nearest tenth, if necessary.

2. Solution Steps

Problem 15:
In a right triangle, sinθ=oppositehypotenuse\sin \theta = \frac{\text{opposite}}{\text{hypotenuse}}.
In the given triangle, the side opposite to angle LL is 2, and the hypotenuse is

7. Therefore, $\sin L = \frac{2}{7}$.

sinL0.2857\sin L \approx 0.2857
Rounding to the nearest hundredth, sinL0.29\sin L \approx 0.29.
Problem 16:
In a right triangle, cosθ=adjacenthypotenuse\cos \theta = \frac{\text{adjacent}}{\text{hypotenuse}}.
In the given triangle, the side adjacent to angle RR is 14, and the hypotenuse is
2

9. Therefore, $\cos R = \frac{14}{29}$.

cosR0.4827\cos R \approx 0.4827
Rounding to the nearest hundredth, cosR0.48\cos R \approx 0.48.
Problem 17:
We are given a triangle LMNLMN with LM=xLM = x, MN=45MN = 45, and angle LNM=64LNM = 64^\circ. Angle LMN=90LMN = 90^\circ.
Using the sine function:
sin(64)=x45\sin(64^\circ) = \frac{x}{45}
x=45sin(64)x = 45 \sin(64^\circ)
x45×0.8988x \approx 45 \times 0.8988
x40.446x \approx 40.446
Rounding to the nearest tenth, x40.4x \approx 40.4.
Problem 18:
We are given a triangle KLMKLM with LM=72LM = 72, angle KML=22KML = 22^\circ, and angle KLM=90KLM = 90^\circ. KM=xKM = x.
Using the cosine function:
cos(22)=x72\cos(22^\circ) = \frac{x}{72}
x=72cos(22)x = 72 \cos(22^\circ)
x72×0.9272x \approx 72 \times 0.9272
x66.758x \approx 66.758
Rounding to the nearest tenth, x66.8x \approx 66.8.

3. Final Answer

Problem 15: sinL0.29\sin L \approx 0.29
Problem 16: cosR0.48\cos R \approx 0.48
Problem 17: x40.4x \approx 40.4
Problem 18: x66.8x \approx 66.8

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