Question 1208910
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In the diagram to the right, Triangle ABC is isosceles, and Triangle MPQ is equilateral. 
Find the perimeter, in cm, of Triangle MBP.
https://ibb.co/gWnd1MQ
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<pre>
Let x be the side length of the equilateral triangle MPQ.


Consider triangle BMP.


Its angle BPM is  180° - 45° - 60° = 75°.

Its angle B is 45 degrees.


Apply the sine law

    {{{40/sin(BPM)}}} = {{{x/sin(45^o)}}},

or

    {{{40/sin(75^o)}}} = {{{x/sin(45^o)}}}.


From it

    x = {{{40*(sin(45^o)/sin(75^o))}}}.


Next use sin(45^o) = {{{sqrt(2)/2}}},  sin(75^o) = {{{(sqrt(6) + sqrt(2))/4}}}   <<<---=== from your textbook or from the Internet.


Substitute it into the formula for x.  You will get

    x = MP = {{{40*(2*sqrt(2))/(sqrt(6) + sqrt(2))}}} = 29.2820323...


To find the side BP, apply the sine law in this form

    {{{40/sin(BPM)}}} = {{{BP/sin(60^o)}}}.


It gives

    BP = {{{40*(sin(60^o)/sin(75^o))}}} = {{{40*(2*sqrt(3))/(sqrt(6)+sqrt(2))}}} = 35.86301889... cm



Now the perimeter of the triangle  MBP  is  

    P(MBP) = MP + BP + MB = 29.2820323 + 35.86301889 + 40 = 105.1450512.


It is your  <U>ANSWER</U>,  and you can round it 105.145 cm, approximately.
</pre>

Solved.