Question

journal
draw and label right triangle pqr with ∠q = 90°. then, describe the sine ratio in terms of ∠r and ∠p.
remember
the sine of an acute angle in a right triangle is the ratio of the length of the side that is opposite the angle to the length of the hypotenuse.
the cosecant of an acute angle is the inverse of the sine of the same angle.
you can use the inverse sine of x, or sin⁻¹x, to determine the measure of an acute angle whose sine is x.
practice
1 use the sine ratio, the cosecant ratio, or the inverse sine to solve for x. round each answer to the nearest tenth.
a
b
c

Explanation:

Part (a)

Step1: Identify the sides

In right triangle \( FMY \), \( \angle M = 90^\circ \), \( \angle F = 57^\circ \), opposite side to \( \angle F \) is \( MY = 11 \) ft, hypotenuse is \( FY = x \).

Step2: Apply sine ratio

Sine of \( \angle F \) is \( \sin(57^\circ)=\frac{\text{opposite}}{\text{hypotenuse}}=\frac{11}{x} \)

Step3: Solve for \( x \)

\( x = \frac{11}{\sin(57^\circ)} \)
\( \sin(57^\circ)\approx0.8387 \)
\( x\approx\frac{11}{0.8387}\approx13.1 \)

Part (b)

Step1: Identify the sides

In right triangle \( PSC \), \( \angle S = 90^\circ \), hypotenuse \( PC = 25 \) yd, opposite side to \( \angle C \) (which is \( x \)) is \( PS = 17 \) yd.

Step2: Apply sine ratio

\( \sin(x)=\frac{\text{opposite}}{\text{hypotenuse}}=\frac{17}{25} = 0.68 \)

Step3: Find \( x \) using inverse sine

\( x=\sin^{-1}(0.68) \)
\( x\approx42.8^\circ \)

Part (c)

Step1: Identify the sides

In the right triangle, opposite side to \( x \) is \( 2\sqrt{2} \) m, hypotenuse is \( 15 \) m.

Step2: Apply sine ratio

\( \sin(x)=\frac{2\sqrt{2}}{15} \)
\( 2\sqrt{2}\approx2.828 \)
\( \sin(x)\approx\frac{2.828}{15}\approx0.1885 \)

Step3: Find \( x \) using inverse sine

\( x = \sin^{-1}(0.1885) \)
\( x\approx10.9^\circ \)

Answer:

s:
a) \( x\approx\boldsymbol{13.1} \) ft
b) \( x\approx\boldsymbol{42.8^\circ} \)
c) \( x\approx\boldsymbol{10.9^\circ} \)