Question

if one of the angle of the basic is 15 degree less than angle of the vertex of an isosceles triangle then find the angle at each base​

Answer 1

$Let \: vertex \: angle \:of \\ isosceles \: triangle = x$

/* According to the problem given*/

$Base \: angle = x - 15 \: ---(1)$

/* We know that,

$\blue {( Base \:angles \: are \: equal \: in }$

$\blue { isosceles \: triangle . }$

$\underline { \orange { By \:Angle \:sum \: property : }}$

$\pink { Sum \:of \: angles = 180\degree }$

$\implies x + ( x - 15 ) + ( x - 15 ) = 180\degree$

$\implies 3x - 30 = 180$

$\implies 3x = 180 + 30$

$\implies 3x = 210$

$\implies x = \frac{210}{3}$

$\implies x = 70$

/* Substitute x = 70 in equation (1), we get */

$Each \: base \:angle = x - 15 \\= 70 - 15 \\= 55\degree$

Therefore.,

$\red { Each \: base \:angle} \green { = 55\degree}$

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Answer 2

answer:

Let the base angles be x vertical angle=x+15

x+x+x+15= 180 ( angle sum property)

3x+15=180

3x=180-15

3x= 165

x=165/3

x-55

Therefore the angles are 55°, 55° ( base angles) and 70°( vertical angle)

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