angle A=40 AngleB and E =x Angle C=50 and Angle d =80 whats the angle of b and e
Answers
Answer:
It has been given that,
It has been given that,⇒ ∠A=∠B−40
It has been given that,⇒ ∠A=∠B−40 o
It has been given that,⇒ ∠A=∠B−40 o ....(1)
It has been given that,⇒ ∠A=∠B−40 o ....(1)∠A=∠C−50
It has been given that,⇒ ∠A=∠B−40 o ....(1)∠A=∠C−50 o
It has been given that,⇒ ∠A=∠B−40 o ....(1)∠A=∠C−50 o ....(2)
It has been given that,⇒ ∠A=∠B−40 o ....(1)∠A=∠C−50 o ....(2)Now, in triangle ABC,
It has been given that,⇒ ∠A=∠B−40 o ....(1)∠A=∠C−50 o ....(2)Now, in triangle ABC,∠A+∠B+∠C=180
It has been given that,⇒ ∠A=∠B−40 o ....(1)∠A=∠C−50 o ....(2)Now, in triangle ABC,∠A+∠B+∠C=180 o
It has been given that,⇒ ∠A=∠B−40 o ....(1)∠A=∠C−50 o ....(2)Now, in triangle ABC,∠A+∠B+∠C=180 o
It has been given that,⇒ ∠A=∠B−40 o ....(1)∠A=∠C−50 o ....(2)Now, in triangle ABC,∠A+∠B+∠C=180 o ⇒ ∠A+∠A+40
It has been given that,⇒ ∠A=∠B−40 o ....(1)∠A=∠C−50 o ....(2)Now, in triangle ABC,∠A+∠B+∠C=180 o ⇒ ∠A+∠A+40 o
It has been given that,⇒ ∠A=∠B−40 o ....(1)∠A=∠C−50 o ....(2)Now, in triangle ABC,∠A+∠B+∠C=180 o ⇒ ∠A+∠A+40 o +∠A+50
It has been given that,⇒ ∠A=∠B−40 o ....(1)∠A=∠C−50 o ....(2)Now, in triangle ABC,∠A+∠B+∠C=180 o ⇒ ∠A+∠A+40 o +∠A+50 o
It has been given that,⇒ ∠A=∠B−40 o ....(1)∠A=∠C−50 o ....(2)Now, in triangle ABC,∠A+∠B+∠C=180 o ⇒ ∠A+∠A+40 o +∠A+50 o =180
It has been given that,⇒ ∠A=∠B−40 o ....(1)∠A=∠C−50 o ....(2)Now, in triangle ABC,∠A+∠B+∠C=180 o ⇒ ∠A+∠A+40 o +∠A+50 o =180 o
It has been given that,⇒ ∠A=∠B−40 o ....(1)∠A=∠C−50 o ....(2)Now, in triangle ABC,∠A+∠B+∠C=180 o ⇒ ∠A+∠A+40 o +∠A+50 o =180 o
It has been given that,⇒ ∠A=∠B−40 o ....(1)∠A=∠C−50 o ....(2)Now, in triangle ABC,∠A+∠B+∠C=180 o ⇒ ∠A+∠A+40 o +∠A+50 o =180 o ⇒ 3∠A=180
It has been given that,⇒ ∠A=∠B−40 o ....(1)∠A=∠C−50 o ....(2)Now, in triangle ABC,∠A+∠B+∠C=180 o ⇒ ∠A+∠A+40 o +∠A+50 o =180 o ⇒ 3∠A=180 o
It has been given that,⇒ ∠A=∠B−40 o ....(1)∠A=∠C−50 o ....(2)Now, in triangle ABC,∠A+∠B+∠C=180 o ⇒ ∠A+∠A+40 o +∠A+50 o =180 o ⇒ 3∠A=180 o −50
It has been given that,⇒ ∠A=∠B−40 o ....(1)∠A=∠C−50 o ....(2)Now, in triangle ABC,∠A+∠B+∠C=180 o ⇒ ∠A+∠A+40 o +∠A+50 o =180 o ⇒ 3∠A=180 o −50 o
It has been given that,⇒ ∠A=∠B−40 o ....(1)∠A=∠C−50 o ....(2)Now, in triangle ABC,∠A+∠B+∠C=180 o ⇒ ∠A+∠A+40 o +∠A+50 o =180 o ⇒ 3∠A=180 o −50 o −40
It has been given that,⇒ ∠A=∠B−40 o ....(1)∠A=∠C−50 o ....(2)Now, in triangle ABC,∠A+∠B+∠C=180 o ⇒ ∠A+∠A+40 o +∠A+50 o =180 o ⇒ 3∠A=180 o −50 o −40 o
It has been given that,⇒ ∠A=∠B−40 o ....(1)∠A=∠C−50 o ....(2)Now, in triangle ABC,∠A+∠B+∠C=180 o ⇒ ∠A+∠A+40 o +∠A+50 o =180 o ⇒ 3∠A=180 o −50 o −40 o
It has been given that,⇒ ∠A=∠B−40 o ....(1)∠A=∠C−50 o ....(2)Now, in triangle ABC,∠A+∠B+∠C=180 o ⇒ ∠A+∠A+40 o +∠A+50 o =180 o ⇒ 3∠A=180 o −50 o −40 o ⇒ 3∠A=90
It has been given that,⇒ ∠A=∠B−40 o ....(1)∠A=∠C−50 o ....(2)Now, in triangle ABC,∠A+∠B+∠C=180 o ⇒ ∠A+∠A+40 o +∠A+50 o =180 o ⇒ 3∠A=180 o −50 o −40 o ⇒ 3∠A=90 o
It has been given that,⇒ ∠A=∠B−40 o ....(1)∠A=∠C−50 o ....(2)Now, in triangle ABC,∠A+∠B+∠C=180 o ⇒ ∠A+∠A+40 o +∠A+50 o =180 o ⇒ 3∠A=180 o −50 o −40 o ⇒ 3∠A=90 o
It has been given that,⇒ ∠A=∠B−40 o ....(1)∠A=∠C−50 o ....(2)Now, in triangle ABC,∠A+∠B+∠C=180 o ⇒ ∠A+∠A+40 o +∠A+50 o =180 o ⇒ 3∠A=180 o −50 o −40 o ⇒ 3∠A=90 o ⇒ ∠A=30
It has been given that,⇒ ∠A=∠B−40 o ....(1)∠A=∠C−50 o ....(2)Now, in triangle ABC,∠A+∠B+∠C=180 o ⇒ ∠A+∠A+40 o +∠A+50 o =180 o ⇒ 3∠A=180 o −50 o −40 o ⇒ 3∠A=90 o ⇒ ∠A=30 o
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Answer:
The reason why we only see ~118 elements is because we only see the ones stable enough to observe. Anything common in nature would need to have a half-life comparable to the age of the earth (or be produced as a decay product of something else that does).