A=1/2(leg)(leg)sin(angle between)
EG: The area of Triangle PQR is 15. If p is 5 and q is 10, find all possible measures of R.
A=15
P=5
Q=10
15=1/2(5)(10)sinR
15=25sinR
sinR=3/5
R=sin-1(3/5)
R≈36.870°,143.130°
Right Triangles
- A=1/2bh
- SOHCAHTOA
- sinΘ=opposite leg/hypotenuse
- cosΘ=adjacent leg/hypotenuse
- tanΘ=opposite leg/adjacent leg
Law of Sines
sinA/a - sinB/b = sinC/c
- used when you know pairs in a non-right triangle
- you are setting up proportions
Law of Cosines
(opposite leg)²=(adjacent leg)² + (other leg)² - 2(adjacent leg)(adjacent leg)cos°
EG: x, 5, 6 angle = 35°
x²=5²+6²-2(5)(6)cos35°
x=√(5²+6²-2(5)(6)cos35°)
x≈3.443
Area of Inscribed Shapes
A=nr²sinΘcosΘ
I don't understand how you know when to use Law of Sines or Law of Cosines. I especially don't understand how to plug them in (for some reason). Also, getting the Area of Inscribed Shapes eludes me.
you use law of sines when you have a pair..meaning if you a have an angle and the adj. side
ReplyDelete& you use law of cosines when you can't use law of sines...
i don't the area of inscribed shapes so sorry..
law of sines is used when there is an angle/side pair in a non right triangle. The formula for that is sin(angle A)/side a=sin(angle B)/side b=sin(angle C)/side c. This means that in triangle ABC, A, B, and C are angles and a, b, and c are the opposite sides.
ReplyDeletelaw of cosines is used when you have 2 legs with an angle inbetween and the formula is: (opp leg)^2=(adj leg)^2 +(adj leg)^2 -2(adj leg)(adj leg)cos(angle).
to find the area of an inscribed shape you use the formula
ReplyDeletenr^2sin(THETA)cos(theta)
N= # of sides of shape
R= radius given
THETA= 360/# of sides
once you solve for all three variables just plug them into the formula and solve for the plugged in formula
You use the Law of Sines is used if you have the measures of angle and the adjacent side measure
ReplyDeleteYou use the Law of Cosines when you have two legs and an angle in between the two leg measures. I'm not too sure about the area of an inscribed shape though.