Asked by siddharth
construct a triangle PQR in which angle Q = 60 degree , angle R = 75 degree and the sum of three sides is 12 cm
Answers
Answered by
Reiny
We will need to find the sides
angle P = 180 - 60 - 75 = 45°
let PQ=x , QR = y , PR = z
x/sin75 = y/sin45 = z/sin60
let's assign an arbitrary value of y = 1 to our smallest side
using x/sin75 = 1/sin45
we get x = .1.366025
using 1/sin45 = z/sin60
z = 1.224745
so sum of those sides
= 1 + 1.366025+1.224745
= 3.59077
but we want this sum to be 12, so each of our sides must be multiplied by a factor of 12/3.59077 = 3.3419
giving me sides of
x = 4.565122
y = 3.3419
z = 4.09298
Now that you have the sides, round them off to suitable values, and construct your triangle
angle P = 180 - 60 - 75 = 45°
let PQ=x , QR = y , PR = z
x/sin75 = y/sin45 = z/sin60
let's assign an arbitrary value of y = 1 to our smallest side
using x/sin75 = 1/sin45
we get x = .1.366025
using 1/sin45 = z/sin60
z = 1.224745
so sum of those sides
= 1 + 1.366025+1.224745
= 3.59077
but we want this sum to be 12, so each of our sides must be multiplied by a factor of 12/3.59077 = 3.3419
giving me sides of
x = 4.565122
y = 3.3419
z = 4.09298
Now that you have the sides, round them off to suitable values, and construct your triangle
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