In a triangle ABC, angle A+B=110
°and angle B+C=132°. Find the measure of each angle of the triangle.
Ayush3008:
A+B+C=180° .............. (1) . IT IS GIVEN THAT A+B=110.PUTTING THE VALUE OF A+B IN EQUATION NO. 1 WE WILL FIND THAT A+B+C=180° 110+C=180° C=180°-110° C=70° . SIMILARLY WE WILL FIND THE VALUE OF B AND A. THANK YOU
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Sum of all angles in a triangle = 180°
⇒ ∠A + ∠B + ∠C = 180 -------- [ 1 ]
Find ∠B:
∠A + ∠B = 110° --------- [ 2 ]
∠B + ∠C = 132° -------- [ 3 ]
[ 2 } + [ 3 ] :
∠A + 2∠B + ∠C = 110 + 132
∠A + 2∠B + ∠C = 242°-------- [ 4 ]
Putting equation [ 4 ] and [ 1 ] together:
∠A + 2∠B + ∠C = 242° --------- [ 4 ]
∠A + ∠B + ∠C = 180 --------- [ 1 ]
[ 4 ] - [ 1 ] :
∠B = 242 - 180
∠B = 62°
Find ∠A:
∠A + ∠B = 110° --------- [ 2 ]
Since ∠B = 62
∠A + 62 = 110
∠A = 110 - 62
∠A = 48°
Find ∠C:
∠B + ∠C = 132° -------- [ 3 ]
∠B = 62
62 + ∠C = 132
∠C = 132 - 62
∠C = 70°
Answer: ∠A = 48°, ∠B = 62° and ∠C = 70°
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