Write the standard equation for the circle 30. Center (2,7), r=4. 31. Center (-6,-8), that passes treough (0,0)
1 answer:
You might be interested in
The question is four parts, It is required the angles of <span>parallelogram KLMN
</span><span>
</span>So, It is required ⇒ ∠K , ∠L , ∠M and ∠N
See the attached figure which represents the explanation of the problem
========================
Part (1): Find ∠N:
==============
in the shape FLDN
<span>∵ LF ⊥ KN ⇒⇒⇒ ∴ ∠LFN = 90°</span><span>
</span>∵ <span>LD ⊥ NM ⇒⇒⇒ ∴ ∠LDN = 90°</span><span>
</span>∵ The sum of all angels of FLDN = 360°
∵ <span>∠FLD = 35°</span><span>
</span>∴ ∠FLD + <span>∠LDN + ∠DNF + </span><span>∠LFN = 360°</span><span>
</span>∴ 35° + 90° + <span>∠DNF + 90° = 360°</span><span>
</span>∴ ∠DNF = 360° - ( 90° + 90° + 35°) = 360° - 215<span>° = 145°
</span>
∴ The measure of angle N = 145°
=============================
Part (2): Find ∠M:
==============
∵ KLMN is <span>parallelogram , ∠N = 145°</span>
∴ The angles N and M are supplementary angles ⇒ property of the parallelogram
∴ ∠M + ∠N = 180°
∴ ∠M = 180° - ∠N = 180° - 145° = 35°
∴ The measure of angle M = 35°
================================
Part (3): Find ∠L:
==============
∵ KLMN is parallelogram , ∠M = 35°
∴ The angles N and M are supplementary angles ⇒ property of the parallelogram
∴ ∠M + ∠KLM = 180°
∴ ∠KLM = 180° - ∠M = 180° - 35° = 145°
OR ∠KLM = ∠N = 145° ⇒⇒⇒ property of the parallelogram
∴ The measure of angle KLM = 145°
===================================
Part (4): Find ∠K:
==============∵ KLMN is parallelogram , ∠N = 35°
∴ The angles N and M are supplementary angles ⇒ property of the parallelogram
∴ ∠K + ∠N = 180°
∴ ∠K = 180° - ∠N = 180° - 145° = 35°
OR ∠K = ∠M = 35° ⇒⇒⇒ property of the parallelogram
∴ The measure of angle K = 35°
</span><span>
</span>So, It is required ⇒ ∠K , ∠L , ∠M and ∠N
See the attached figure which represents the explanation of the problem
========================
Part (1): Find ∠N:
==============
in the shape FLDN
<span>∵ LF ⊥ KN ⇒⇒⇒ ∴ ∠LFN = 90°</span><span>
</span>∵ <span>LD ⊥ NM ⇒⇒⇒ ∴ ∠LDN = 90°</span><span>
</span>∵ The sum of all angels of FLDN = 360°
∵ <span>∠FLD = 35°</span><span>
</span>∴ ∠FLD + <span>∠LDN + ∠DNF + </span><span>∠LFN = 360°</span><span>
</span>∴ 35° + 90° + <span>∠DNF + 90° = 360°</span><span>
</span>∴ ∠DNF = 360° - ( 90° + 90° + 35°) = 360° - 215<span>° = 145°
</span>
∴ The measure of angle N = 145°
=============================
Part (2): Find ∠M:
==============
∵ KLMN is <span>parallelogram , ∠N = 145°</span>
∴ The angles N and M are supplementary angles ⇒ property of the parallelogram
∴ ∠M + ∠N = 180°
∴ ∠M = 180° - ∠N = 180° - 145° = 35°
∴ The measure of angle M = 35°
================================
Part (3): Find ∠L:
==============
∵ KLMN is parallelogram , ∠M = 35°
∴ The angles N and M are supplementary angles ⇒ property of the parallelogram
∴ ∠M + ∠KLM = 180°
∴ ∠KLM = 180° - ∠M = 180° - 35° = 145°
OR ∠KLM = ∠N = 145° ⇒⇒⇒ property of the parallelogram
∴ The measure of angle KLM = 145°
===================================
Part (4): Find ∠K:
==============∵ KLMN is parallelogram , ∠N = 35°
∴ The angles N and M are supplementary angles ⇒ property of the parallelogram
∴ ∠K + ∠N = 180°
∴ ∠K = 180° - ∠N = 180° - 145° = 35°
OR ∠K = ∠M = 35° ⇒⇒⇒ property of the parallelogram
∴ The measure of angle K = 35°