In\( ∆ PQR, P\hat{Q}P = 84^°, |Q\hat{P}R |= 43^°\) and |PQ| = 5cm. Find /QR/ in cm, correct to 1 decimal place.
- 3.4
-
4.3
- 5.9
- 6.2
The correct answer is: B
Explanation
\(\frac{|QR|}{sinP}=\frac{|PQ|}{sinR}\\
\frac{x}{sin43^o}=\frac{5}{sin53^o}\\
x = \frac{5\times sin43^o}{sin53^o}\\
=\frac{5\times 0.6820}{0.7986}\\
=4.3\)