Question

Find the measure of one of the larger angles of a parallelogram, having its vertices with coordinates $(2,1),(5,1),(3,5)$ and $(6,5)$.

• A. ${93.4}^o$
• B. ${96.8}^o$
• C. ${104.0}^o$
• D. ${108.3}^o$
• E. ${119.0}^o$

Answer 1

Answer: (C)

Area of parallelogram$=base\times height$

$=4\times CD$

$=4\times \sqrt{{(5-2)}^2+{(1-1)}^2}$

$=12$

Area of parallelogram $=AB\times AD\sin \theta$

$\therefore \sqrt{{(6-3)}^2+{(5-5)}^2}\times \sqrt{{(3-2)}^2+{(5-1)}^2}\sin \theta =12$

$3\times \sqrt{17}\sin \theta =12$

$\sin \theta =\frac{4}{4.123}$

$\sin \theta =0.9701$

$\therefore \theta ={104}^o$

$\because \sin {105}^o=\sin ({90}^o+{15}^o)=\cos {15}^o$

$\cos {15}^o=\cos ({45}^o-{30}^o)$

$=\cos {45}^o\cos {30}^o+\sin {45}^o\sin {30}^o$

$=\frac{\sqrt{3}}{2\sqrt{2}}+\frac{1}{2\sqrt{2}}$

$=0.968$ which is closer to $0.97$

$\therefore \theta ={104}^o$