QUESTION IMAGE
Question
use $\triangle xyz$, as needed, to complete the given parts.
part 1: let $h$ be the height of $\triangle xyz$.
find the length of the corresponding base.
$\square$ m
part 2: the measure of angle $y$ is $43.69^{\circ}$.
fill in the blank to complete the sine ratio.
$\sin(43.69^{\circ}) = \frac{h}{\square}$
part 3: fill in the blanks in the expression for the area of $\triangle xyz$.
area of $\triangle xyz = \frac{1}{2}\cdot\square\cdot\square\cdot\sin(^{\circ})$ $\text{m}^2$
part 4: write the area of $\triangle abc$ in terms of $a$, $c$, and $b$.
area of $\triangle abc = \square$
Step1: Identify base for height h
The height \( h \) is perpendicular to side \( XY \), so the corresponding base is \( XY \), which has length 13 m.
Step2: Identify hypotenuse for sine ratio
For \( \angle Y \), the height \( h \) is opposite the angle, and the hypotenuse of the right triangle formed is \( YZ \) (length 9 m). So \( \sin(43.69^\circ) = \frac{h}{9} \).
Step3: Apply SAS area formula
The area formula using two sides and included angle is \( \frac{1}{2}ab\sin(C) \). For \( \triangle XYZ \), use sides \( XY=13 \), \( YZ=10 \), and included \( \angle Y=43.69^\circ \). So the expression is \( \frac{1}{2} \cdot 13 \cdot 10 \cdot \sin(43.69^\circ) \).
Step4: Generalize SAS area formula
For \( \triangle ABC \), sides adjacent to \( \angle B \) are \( a \) and \( c \). The area is \( \frac{1}{2}ac\sin(B) \).
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