3) In Δ BDC
|DC|/|BC| = cos C
cos C= 16/17.89
C= cos⁻¹( 16/17.89)=26.57⁰
In triangle ABC:
x=180-(90+26.57)=63.43
x=63.43
2)
AB/BD= tan(70⁰), AB=BD*tan(70⁰)
AB/BC=tan(40⁰), AB=BC*tan(40⁰)
BD*tan(70⁰)=BC*tan(40⁰)
BD=BC-CD=BC - 15
(BC -15)*tan(70⁰)=BC*tan(40⁰)
BC*tan(70⁰) -15*tan(70⁰)= BC*tan(40⁰)
BC*tan(70⁰) - BC*tan(40⁰) = 15*tan(70⁰)
BC(tan(70⁰)-tan(40⁰))= 15*tan(70⁰)
BC = 15*tan(70⁰)/(tan(70⁰)-tan(40⁰)) = 21.60
BC=21.60
1) In a quadrilateral sum of angles =360⁰.
PQS=SQR=50⁰, because SQ bisects PQR.
Using Law of sine in ΔSRQ
SR/sin ∠SQR = SQ/sinR, SQ = SR * sinR/sin SQR = 3*sin30/sin50 =3.26
SQ=3.26 cm
ΔPQS:
cos PQS= PQ/SQ
PQ=SQ*cosPQS =3.26*cos 50⁰=2.09=2.1
PQ=2.1 cm
Answer: 13 1/3
Step 1: Find the LCD
Step 2: Add the fractions
Step 3: Simplify
Answer:
A: (2,2) B: (3,-1) C: (-1,0)
Step-by-step explanation:
Change the coordinates of each point using the given translation. So 2 left 3 up.
