Answer:
r = 5
Step-by-step explanation:
Sine trigonometric ratio
[tex]\sf \sin(\theta)=\dfrac{O}{H}[/tex]
where:
- [tex]\theta[/tex] is the angle
- O is the side opposite the angle
- H is the hypotenuse (the side opposite the right angle)
The sides of a triangle are opposite their angles. Therefore, in ΔPQR:
⇒ side r is opposite angle R
⇒ side p is opposite angle P
⇒ side q is opposite angle Q
Using the given ratio of sin Q:
[tex]\sf \implies \sin Q=\dfrac{q}{H}=\dfrac{2}{5}[/tex]
Substitute the given value of q = 6 and solve for H:
[tex]\sf \implies \dfrac{6}{H}=\dfrac{2}{5}[/tex]
[tex]\sf \implies H=6 \cdot\dfrac{5}{2}[/tex]
[tex]\sf \implies H=15[/tex]
Using the given ratio of sin R:
[tex]\implies \sf \sin R=\dfrac{r}{H}=\dfrac{1}{3}[/tex]
Substitute the found value of H and solve for r:
[tex]\implies \sf \dfrac{r}{15}=\dfrac{1}{3}[/tex]
[tex]\implies \sf r=15 \cdot\dfrac{1}{3}[/tex]
[tex]\implies \sf r=5[/tex]
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