Question

If base angle of an isosceles triangle is 15° more than its Vertical angle find all the angles of the triangle .​

Answer 1

Given:

• Base angle of an isosceles triangle is 15°.

To Find:

• All angles of triangle

Solution:

☯ Let vertical angle of the isosceles triangle "x".

• Each base angle = x + 15°

As we know:

★ Sum of angle of a triangle is 180°.

According to the question:

→ x + 15° + x + 15° + x° = 180° [Sum of angles of a triangle]

→ 3x + 30° = 180°

→ 3x = 180° - 30°

→ 3x = 150°

→ x = 150°/3

→ x = 50°

Hence,

• Vertical angle = 50°
• Each base angle = 50° + 15° = 65°

Answer 2

$\large\underline\bold{ANSWER \red{\huge{\checkmark}}}$

$\pink{\overbrace{ \underbrace{ \red{\mid\star\mid \:\: \purple{ vertical\:angle(a)= \orange{ 50^{ \circ}} base\:angle(b,c)=\orange{ 65^{ \circ}} } } \:\: \red{\mid\star\mid}}}}$

$\large{ \purple{ \underbrace{ \pink{ \underline{\mathcal{explanation\:in\:details:-}}} }}}$

$\large\underline\bold{GIVEN,}$

$\dashrightarrow base\:angle\:of\:triangle(isosceles )is\:\: \pink{ 15^{\circ}}$

$\large\underline\bold{TO\:FIND,}$

$\green{ \circ\: all \:the\:angles\:of\:the\:triangle.}$

$\bf{ \mathcal{ \underline{\orange{We\:Know,}}}} \\ \\ \rm{ \boxed{ \underline{ \red{ sum\:of\:angle\:of\:a\:triangle\:is\:180^{ \circ} }}}}$

$\large\underline\bold{SOLUTION,}$

$\natural \: \text{ as given in the question , } \\ \circ\:\text{ base angle of an isosceles triangle is 15° more }$

$\therefore let\:vertical\:angle\:be \: x^{\circ} \\ \\ \therefore \:base\:angle\:formed\:will\:be\: \red{ \rightarrow \: x+15 }$

$\circ\: an\: isosceles\: triangle \: has \:two \:equal \\ \:sides\:and\:two\:equal\:angels. \\ \\ therefore \:according\:to\:the\: statement\: \\ \\\red{ \diamondsuit} \: \: two \:angles\:of\:same\:angles\:will\:be\: \red{ x+15} \\ \\ \leadsto a= x +15 \\ \leadsto b= x+15 \\ \leadsto c= x$

$\\ \\ \clubsuit \: \: \blue{ by\:angle\:sum\:property\:of\:a\:triangle } \\ \\ \mid\implies x+15+x+15+x= 180^{ \circ } \\ \\ \mid\implies x+x+x+15+15=180^{ \circ} \\ \\ \mid\implies 3x+30=180^{ \circ} \\ \\ \mid\implies 3x=180-30 \\ \\ \mid\implies 3x= 150^{ \circ } \\ \\ \mid\implies x= \dfrac{150}{ 3} \\ \\ \mid\implies x= \cancel\dfrac{ 150}{3} \\ \\ \mid\implies x=50^{ \circ} \\ \\ \underline{ vertical\:angle\:of\:triangle\:is\:50^{ \circ} } \\ \\ \rm{ \boxed{ \red{ \overline{\underline{ \mid\heartsuit\mid \:\: x= 50^{ \circ} \mid\heartsuit \mid }}}}} \\ \\ \:base\:angle\:formed\:is\: \red{ \rightarrow \: x+15 } \\ \\ \mid\leadsto50+15 = \rm{ \underline{ \green{\bold{65^{ \circ}}}}}$