plzz bhai log q.noii and iii its 50 marks not wrong answer
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solution of ii ..as u can observe that it is a right angle triangle.. therefore sum of other two angles willbe equal to 90 therefore 19x+3+30=90
19x+33=90
19x=90-33
x=57÷19
x=3
iii sum of all angles of triangle =180
7x-4+6x-4+5x+8=180
18x-8+8=180
18x=180
x=10
19x+33=90
19x=90-33
x=57÷19
x=3
iii sum of all angles of triangle =180
7x-4+6x-4+5x+8=180
18x-8+8=180
18x=180
x=10
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2)
In first question
Let the triangle ABC be a right angled triangle
Angle A=90
Angle B=19x+3
Angle C=30
In triangle ABC,
Angle A+B+C= 180 {Sum of measure of all angle of triangle }
90+19x+3+30=180 {substituting the values}
19x+123=180
19x=180 -123
x=57÷19
x=3
Substituting the value of x=3 in Angle B we get;
B=19×3+3o
=57+30
=87
Angle B=87 degree
In first question
Let the triangle ABC be a right angled triangle
Angle A=90
Angle B=19x+3
Angle C=30
In triangle ABC,
Angle A+B+C= 180 {Sum of measure of all angle of triangle }
90+19x+3+30=180 {substituting the values}
19x+123=180
19x=180 -123
x=57÷19
x=3
Substituting the value of x=3 in Angle B we get;
B=19×3+3o
=57+30
=87
Angle B=87 degree
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