1. Given any triangle ABC with sides BC=a, AC=b and AB=c, the following are true :
i) the larger the angle, the larger the side in front of it, and the other way around as well. (Sine Law) Let a=20 in, then the largest angle is angle A.
ii) Given the measures of the sides of a triangle. Then the cosines of any of the angles can be found by the following formula:
a^{2}=b ^{2}+c ^{2}-2bc(cosA)
2.
20^{2}=9 ^{2}+13 ^{2}-2*9*13(cosA) 400=81+169-234(cosA) 150=-234(cosA) cosA=150/-234= -0.641
3. m(A) = Arccos(-0.641)≈130°,
4. Remark: We calculate Arccos with a scientific calculator or computer software unless it is one of the well known values, ex Arccos(0.5)=60°, Arccos(-0.5)=120° etc
I am pretty sure that it is B
Answer:
5:15
Step-by-step explanation:
Split the 30 mins in half. 45+15 = 60 (a new hour) Now, its 5:00. Add 15 and you get the answer.
Answer:
Option B.
Step-by-step explanation:
Let as consider the given equation:
It can be written as
On comparing both sides, we get
Multiply both sides by 5.
Divide both sides by 2.
Therefore, the correct option is B.