Papers of Surds and Indices

Question:

$I f \sin θ = \frac{3}{7} a n d 0 ° \leq θ \leq 90 ° . f i n d \cos θ i n s u r d f o r m . F i n d t h e a r e a o f t h e t r i a n g l e b e l o w l e a v i n g y o u r a n s w e r i n s u r d f o r m .$

Answer:

$\sin θ = \frac{3}{7}$

$\cos θ = \frac{\sqrt{40}}{7} A = \frac{1}{2} a b \sin C = \frac{1}{2} \times 3 \times 6 \times \sin 120 = 9 \times \frac{\sqrt{3}}{2} = \frac{9 \sqrt{3}}{2} u n i t s^{2}$