Answer:
![\displaystyle x \approx 47.44](https://img.qammunity.org/2022/formulas/mathematics/high-school/9lowzv57xuh9ol1zd288yk7ot740og30q4.png)
Explanation:
we are given a right angle triangle
we want to figure out x i.e hypotenuse
in order to do so we can consider trigonometry i.e sin
remember that,
![\displaystyle \sin( \theta) = (opp)/(hypo)](https://img.qammunity.org/2022/formulas/mathematics/high-school/xlz4oe7ixx9augqkywri3msk3q5p6i8tis.png)
given that,theta=21° and opp=17 and hypo=x
Thus substitute:
![\displaystyle \sin( {21}^( \circ) ) = (17)/(x)](https://img.qammunity.org/2022/formulas/mathematics/high-school/qqymzic2p256aw8ac7ajymbu5c8swx24et.png)
cross multiplication:
![\displaystyle x\sin( {21}^( \circ) ) = {17}](https://img.qammunity.org/2022/formulas/mathematics/high-school/8nzbhjqorv5xq38fhjrlsqewgbgb59atxw.png)
divide both sides by sin21°:
![\displaystyle x \frac{\sin( {21}^( \circ) ) }{ \sin( {21}^( \circ)) }= \frac{17}{ \sin( {21}^( \circ) ) }](https://img.qammunity.org/2022/formulas/mathematics/high-school/zfeny46hqap93asa5cs54406u7x9k1w7xo.png)
By using calculator we acquire:
![\displaystyle x \approx 47.44](https://img.qammunity.org/2022/formulas/mathematics/high-school/9lowzv57xuh9ol1zd288yk7ot740og30q4.png)