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nordsb [41]
3 years ago
10

PLEASE HELP ME WITH THESE ALL WORK IS APPRECIATED!! THANK YOU:)❤️

Mathematics
1 answer:
Vsevolod [243]3 years ago
3 0

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


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