In ΔQRS, the measure of ∠S=90°, the measure of ∠R=31°, and QR = 46 feet. Find the length of SQ to the nearest tenth of a foot.

Step-by-step explanation:

we know that

In the right triangle QRS

----> by SOH (opposite side divided by the hypotenuse)

substitute the given values

solve for SQ

< coa = 150 ( given)

as ba and bc are tangents

so oc and oa is perpendicular to cb and ab respectively because in circle normal passes through centre

so in quadrilateral oabc

< coa + < oab + < ocb + < abc = 360

150 +90+90 + < abc = 360

330 + < abc = 360

< abc = 360 -330 = 30

so < abc = 30

⭐dhruv⭐