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How do I solve this?

Answer 1

Mason Adams

#/_MPL=80^o#

Join line #MN# and now examine #DeltaMNP#.

As #m hat(ML)=175^o# and #m hat(ON)=15^o#

#/_MPL=/_MNL-/_PMN#,

as exterior angle is equal to sum of interior opposite angles.

Now #/_MNL=1/2xx(m hat(ML))=1/2xx175^o# and

#/_PMN=1/2xx(m hat(ON))=1/2xx15^o#

Hence #/_MPL=/_MNL-/_PMN#

= #1/2xx175^o-1/2xx15^o#

= #1/2xx(175^o-15^o)#

= #80^o#

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Answer 2

Autumn Armstrong

#angle MPL = (angle ML)/2-(angle ON)/2 = 80^o#

#angle OMN = (angle ON)/2# #angle MNL = (angle ML)/2# #angle MPL + pi-angle MNL+angle OMN=pi# Solving for #angle MPL = (angle ML)/2-(angle ON)/2 = 80^o #

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Answer 3

Allison Allen

To provide an accurate answer, I'll need more information about what you're trying to solve. Could you please specify the problem or question you need help with?

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Answer from HIX Tutor

When evaluating a one-sided limit, you need to be careful when a quantity is approaching zero since its sign is different depending on which way it is approaching zero from. Let us look at some examples.