What is the area of a triangle that has 6in high and 7in of the base
1 answer:
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Solve for y
let's put our equation first
Flip the equation
Now follow the rule of
- Brackets
- Of multiplication
- Division
- Addition
- Subtraction
- Parenthesis
- Exponents
- Multiplication
- Division
- Addition
- Subtraction
_________________________
_________________________
add to both sides
Divide both sides by
would equal
cosU:-
As U lies in Q3
- cosU=-24/25
sinV
As V lies in Q3
- sinV=-3/5
So
- sin(V-U)=sinVcosU-cosVsinU=(-3/5)(-24/25)-(-4/5)(-7/25)=72/125-28/125=72-28/125=<u>4</u><u>4</u><u>/</u><u>1</u><u>2</u><u>5</u>
- cos(U-V)=cosUcosV+sinUsinV=(-24/25)(-4/5)+(-7/25)(-3/5)=96/125+21/125=117/125