Circle O is shown. Chords K M and J L intersect at point A. The measure of arc K J is 170 degrees. The measure of arc L M is 80
2 answers:
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Answer:
Perimeter = (2 + √3)·a
Step-by-step explanation:
Given: ΔABC is equilateral and AB = a
The diagram is given below :
AM is a median , PB ⊥ AB , PM = b
Now, by using properties of equilateral triangle, median is perpendicular bisector and each angle is of 60°.
We get, ∠AMB = 90°. So, by linear pair ∠AMB + ∠PMB = 180° ⇒ ∠PMB = 90°. Also, ∠ABC = 60° and ∠ABP = 90° (given) So, ∠PBM = 30°
Since, AM is perpendicular bisector of BC. So,
Now in ΔAMB , By using Pythagoras theorem
Now, in ΔBMP :
Perimeter of ABM = AB + PB + PM + AM
Hence, Perimeter of ΔABP = (2 + √3)·a units
t N.
In the figure shown below
Answer:
A horizontal line segment M K intersects with line segment J L at their midpoint N.
∠J N M =(5x+2)°
∠ LN M=3( x+ 14)°
So, ∠J N M + ∠ LN M =180°[ These two angles form linear pair.Angles forming linear pair are supplementary.]
⇒5 x+ 2+ 3 (x+ 14) =180 [ By Substitution]
⇒ 5 x+2 +3 x+42°= 180°
⇒ 8 x=180°-44°
⇒8 x= 136°
⇒x= 136°÷8
⇒x=17°
So, ∠J N M =5×17 +2=87°
∠ LN M= 3×(17 +14)=3×31=93
∠J N M =∠K N L [Vertically opposite angles]
∠K N L=87°
