Qu. In a triangle ABC, side AB has length 10cm, side AC has length 5cm, and angle BAC = θ where θ is measured in degrees. The area of triangle ABC is 15cm^2
(a) Find the two possible values of cos θ
Given that BC is the longest side of the triangle,
(b) find the exact length of BC.

To find cosθ, use the formula for the area of a triangle i.e. AREA=1/2 x a x b x sinC.=> For this case: 15= 1/2 x 10 x 5 x sinC to find sinC.=> SinC = 3/5 thus, Arcsin(3/5)=+- 4/5 or +-0.8

To find the exact length of BC, use the cosine rule.=> c(sq)=a(sq)+b(sq)-2abCosC=> c(sq)=10(sq)+5(sq)-2(10)(5)(+-4/5)=> c(sq)= Square root of 205

< 4, -5>

step-by-step explanation:

starting from g. we move 4 right and then 5 down to get to h. that means the vector is < 4, -5>