Question and Answers Forum

All Questions      Topic List

Geometry Questions

Previous in All Question      Next in All Question      

Previous in Geometry      Next in Geometry      

Question Number 128193 by Algoritm last updated on 05/Jan/21

Answered by mr W last updated on 05/Jan/21

Commented by Algoritm last updated on 05/Jan/21

?=

?=

Commented by mr W last updated on 05/Jan/21

((AE)/(sin β))=((BE)/(sin (110+β)))=((AB)/(sin 110)) ⇒AE=((sin β)/(sin 70)) ⇒BE=((sin (70−β))/(sin 70)) ((EC)/(sin (45−β)))=((BC)/(sin (45−β+γ)))=((BE)/(sin γ)) ((√2)/(sin (45−β+γ)))=((sin (70−β))/(sin 70 sin γ)) (((√2) sin 70)/(sin (70−β)))=(((√2) cos (β−γ)−(√2) sin (β−γ))/(2 sin γ)) ((2 sin 70)/(sin (70−β)))=((cos (β−γ)−sin (β−γ))/(sin γ)) ((2 sin 70)/(sin (70−β)))=((cos β cos γ+sin β sin γ−sin β cos γ+cos β sin γ)/(sin γ)) ((2 sin 70)/(sin (70−β)))=sin β+cos β+((cos β−sin β)/(tan γ)) ⇒tan γ=((cos β−sin β)/(((2 sin 70)/(sin (70−β)))−sin β−cos β)) ...(i) EC=(((√2) sin (45−β))/(sin (45−β+γ))) α=180−70−(45−γ)=65+γ AD=(1/(tan α))=(1/(tan (65+γ))) DC=1−(1/(tan (65+γ))) ((DC)/(sin 70))=((EC)/(sin α)) (1/(sin 70))(1−(1/(tan (65+γ))))=(((√2) sin (45−β))/(sin (45−β+γ)sin (65+γ))) (1/(sin 70))(sin (65+γ)−cos (65+γ))=(((√2) sin (45−β))/(sin (45−β+γ))) ⇒sin (65+γ)−cos (65+γ)=(((√2) sin 70 sin (45−β))/(sin (45−β+γ))) ...(ii) ⇒β=31.434768°, γ=11.434768° ?=45−β+γ=25°

AEsinβ=BEsin(110+β)=ABsin110⇒AE=sinβsin70⇒BE=sin(70−β)sin70ECsin(45−β)=BCsin(45−β+γ)=BEsinγ2sin(45−β+γ)=sin(70−β)sin70sinγ2sin70sin(70−β)=2cos(β−γ)−2sin(β−γ)2sinγ2sin70sin(70−β)=cos(β−γ)−sin(β−γ)sinγ2sin70sin(70−β)=cosβcosγ+sinβsinγ−sinβcosγ+cosβsinγsinγ2sin70sin(70−β)=sinβ+cosβ+cosβ−sinβtanγ⇒tanγ=cosβ−sinβ2sin70sin(70−β)−sinβ−cosβ...(i)EC=2sin(45−β)sin(45−β+γ)α=180−70−(45−γ)=65+γAD=1tanα=1tan(65+γ)DC=1−1tan(65+γ)DCsin70=ECsinα1sin70(1−1tan(65+γ))=2sin(45−β)sin(45−β+γ)sin(65+γ)1sin70(sin(65+γ)−cos(65+γ))=2sin(45−β)sin(45−β+γ)⇒sin(65+γ)−cos(65+γ)=2sin70sin(45−β)sin(45−β+γ)...(ii)⇒β=31.434768°,γ=11.434768°?=45−β+γ=25°

Commented by mr W last updated on 05/Jan/21

here a check:

hereacheck:

Commented by mr W last updated on 05/Jan/21

Commented by Algoritm last updated on 06/Jan/21

thanks

thanks

Terms of Service

Privacy Policy

Contact: info@tinkutara.com