Answer:
579o
Step-by-step explanation:
61+ 62=123o
180-123= 57
J = 57o
A cylinder has two bases and a height that (unless oblique) runs right along a side, while a cone has one base and whose height begins at the center of its base and goes straight up to the peak.
tan A is 4/3
use sohcahtoa
Angle a is 53.130 if you need it
Answer:
ΔGJH ≅ ΔEKF
HL: GH and EF
SAS: FK and JH (or GH and EF)
ASA: ∠JGH and ∠FEK (or ∠EFK and ∠JHG)
ΔGFJ ≅ ΔEKH
SSS: KH and FJ
SAS: ∠KEH and ∠FGJ
Step-by-step explanation:
List whatever angles/sides need to be congruent for the two triangles to be congruent.
Prove ΔGJH ≅ ΔEKF using....
- HL (Hypotenuse + Leg)
We already have two legs that are congruent (EK and GJ), so we just need the hypotenuses (GH and EF) to be equal.
- SAS (Side + Angle + Side)
1 pair of sides (EK and JG) are equal, and m∠EKF = m∠GJH. So we need one more side. You can either use FK and JH or GH and EF.
- ASA (Angle + Side + Angle)
1 pair of angles (∠EKF and ∠GJH) are already given as equal, and 1 pair of sides (EK and GJ) are equal. We just need one more pair of angles. So either ∠JGH and ∠FEK or ∠EFK and ∠JHG.
Prove ΔGFJ ≅ ΔEKH using...
- SSS (Side + Side + Side)
Two pairs of sides (EK + GJ and EH + FG) are equal, so KH and FJ need to be equal.
- SAS (Side + Angle + Side)
FG + EH and KE + GJ are equal. We need to use the angle in between them to use SAS, so ∠KEH and ∠FGJ need to be equal.
