plz solve in detail....do not short cut...
we will mark brainliest...
Answers
On the face of it, the problem reduces to solving a system of 4 linear equations in 4 variables:
a+b=160
b+c=130
x+a=140
x+c=110
which can be solved in a number of ways to get the solution (all numbers in degrees).
Note that a and b are the other two angles of the right-most inner-triangle in the above figure and c is the angle opposite to x.
However, on further inspection the above system turns out to be singular and its inverse does not exist. So, the above can be reduced to three simultaneous equations in 4 variables:
x+a=140
x+c=110
b−x=20
Since, none of the angles is negative, the solution is bounded, i.e x∈[0,110], but nothing more can be said using the above alone.
However, we can also use extra information from the triangles via the sine-rule. Denoting by l the side from angle x to the top-left vertex of the largest triangle and by b the left edge of the largest triangle, we get:
bsin40=lsin80
and
bsinx=lsin(c+50)
Substituting c=110−x (from our simultaneous system) in the above and after some elementary algebra/trigonometry, we get a neat equation for x:
sin(20+x)sinx=sin80sin40
This can be solved by brute force, which will involve expansion of sin(20+x), re-arranging and taking the tan−1 of the RHS terms. Or, better still, we can write the above as:
2sinx⋅sin(70+30)=sin(70+x)+sin(x−30)
which leads to the obvious answer of x=30.
here is ur answer
(a)
x°+70°=180° [LP]
➡️x°=110°
a°+80°=180°[LP]
➡️a°=100°
z°+70°=180°[LP]
➡️z°=110°
y°+60°=180°
➡️y°=120°
a°+y°+z°+x°+b°=540°[angle sum property of pentagon]
➡️so b°=100°
(b)
z°+150°=180°[LP]
➡️z°=30°
x°+100°=180°[LP]
➡️x°=80°
y°+140°=180°[LP]
➡️y°=40°
t°+145°=180°[LP]
➡️t°=35°
➡️q°=115° ( by angle sum property of pentagon)
q°+r°=180°[LP]
115°+r°=180°
➡️r°=65°
I hope this will help you
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