Given RQ = 20 inches and PR = 25 inches what is the m∠Q ?
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Answer:
73.2°
Step-by-step explanation:
Use Law of Sines to solve:
(Sin 50)/20 = (Sin B)/25
Solve for Sin B
[25(Sin 50)]/20 = Sin B
Use Sin^-1 x to solve (sine inverse)
Sin^-1 ( [25(Sin 50)]/20 ) = B
B = 73.24685774
Answer:
Step-by-step explanation:
Use the sine law:
[tex]\dfrac{RQ}{\sin(\angle P)}=\dfrac{PR}{\sin(\angle Q)}[/tex]
We have
[tex]RQ=20\ in\\\\m\angle P=50^o\to\sin50^o\approx0.766\\\\PR=25\ in[/tex]
Substitute:
[tex]\dfrac{20}{0.766}=\dfrac{25}{\sin(\angle Q)}[/tex] cross multiply
[tex]20\sin(\angle Q)=(25)(0.766)[/tex]
[tex]20\sin(\angle Q)=19.15[/tex] divide both sides by 20
[tex]\sin(\angle Q)=0.9575\to m\angle Q\approx73^o[/tex]