Question

In ΔQRS, the measure of ∠S=90°, the measure of ∠Q=71°, and RS = 2.5 feet. Find the length of SQ to the nearest tenth of a foot

Answer 1

Answer is in the picture

Answer 2

Answer: 0.9

tan71=

x

2.5

​

x\tan 71=2.5

xtan71=2.5

\frac{x\tan 71}{\tan 71}=\frac{2.5}{\tan 71}

tan71

xtan71

​

=

tan71

2.5

​

x=\frac{2.5}{\tan 71}=0.8608\approx 0.9\text{ feet}

x=

tan71

2.5

​

=0.8608≈0.9 feet