Angle CDB= α α = 180°-100°-(2*10°) = 60° h = 5√3 sinα = 5√3.√3/2 = 7,5 mts. Similarly of right triangles: x/5√3 = 5√3 /7,5 x = 10 mts (Solved √ ) If the figure were drawn into scale, this will be seen clearly !!!
I solved with trigonometry and sines law in this way: On triangle BCD BC/ sin 60 = BD/sin 100 BC = BD*sin 60 /sin 100 (1) On triangle ABC AC/ sin 160 = BC/sin 10 BC = AC*sin 10/ sin 160 (2) Knowing that: sin 100 = sin (90 + 10) = sin(90 - 10) = cos 10 and sin 160 = sin(180 - 20) = sin 20 we can campare (1) with (2) (BD*√ 3/2)/cos 10 = 10√ 3*sin 10/sin 20 BD/2*sin 20 = 10*sin 10*cos 10 BD/2*sin 20 = 5*sin 20 BD = 10
Angle CDB= α
α = 180°-100°-(2*10°) = 60°
h = 5√3 sinα = 5√3.√3/2 = 7,5 mts.
Similarly of right triangles:
x/5√3 = 5√3 /7,5
x = 10 mts (Solved √ )
If the figure were drawn into scale, this will be seen clearly !!!
I solved with trigonometry and sines law in this way:
On triangle BCD
BC/ sin 60 = BD/sin 100
BC = BD*sin 60 /sin 100 (1)
On triangle ABC
AC/ sin 160 = BC/sin 10
BC = AC*sin 10/ sin 160 (2)
Knowing that:
sin 100 = sin (90 + 10) = sin(90 - 10) = cos 10
and
sin 160 = sin(180 - 20) = sin 20
we can campare (1) with (2)
(BD*√ 3/2)/cos 10 = 10√ 3*sin 10/sin 20
BD/2*sin 20 = 10*sin 10*cos 10
BD/2*sin 20 = 5*sin 20
BD = 10
(10)^2 (3m)^2={100+9m^2}=109m^2 {10°A+10°C+130°B}={150°ACB+30°D}=180°ACBD/109m^2=1.7..8 1.7.2^3 1.3^4.1^1^1 3^2^2.1 3^1^2.1 3^2 (ACBD ➖ 3ACBD+2).
BD=(sin100)*(sin10*)10√3/(sin60)(sin160) m
El ejercicio es Sin calculadora