Question

in ∆ABC,AB=AC,angle A=x,and angleB=x+10° than find value of each angle of ∆ABC​

Answer 1

➤ Correct Question :-

In ∆ABC, AB=AC, ∠A = x, and ∠B = x+30°. Then find value of each angle of ∆ABC.

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★ How to do :-

Here, we are given that in a triangle two of it's sides are equal which says that it's an isosceles triangle. We are given that the value of one angle with a variable. The other angle is 10° greater than the variable x. As it is an isosceles triangle, two of it's angles also remains same as the other angle is also 30° greater than the variable x. We are asked to find the measurement of all the angles in that triangle. So, here we'll use the concept of shifting the numbers from one hand side to the other which changes it's sign. So, let's solve!!

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➤ Solution :-

${\bf \leadsto \underline{\boxed{\bf Sum \: of \: all \: angles \: (triangle) = {180}^{\circ}}}}$

Write the names of all angles.

${\sf \leadsto \angle{A} + \angle{B} + \angle{C} = {180}^{\circ}}$

Substitute the given values.

${\tt \leadsto x + (x + 30) + (x + 30) = 180}$

Add all the variables and constants on LHS.

${\tt \leadsto 3x + 60 = 180}$

Shift the number 60 from LHS to RHS, changing it's sign.

${\tt \leadsto 3x = 180 - 60}$

Subtract the values on RHS.

${\tt \leadsto 3x = 120}$

Shift the number 3 from LHS to RHS, changing it's sign.

${\tt \leadsto x = \dfrac{120}{3}}$

Simplify the fraction on RHS.

${\tt \leadsto x = 40}$

$\bf \leadsto \red{\underline{\boxed{\bf Value \: of \: \angle{A} = {40}^{\circ}}}}$

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$\tt \leadsto x + 30 = 40 + 30$

$\bf \leadsto \red{\underline{\boxed{\bf Value \: of \: \angle{B} = {70}^{\circ}}}}$

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$\tt \leadsto x + 30 = 40 + 30$

$\bf \leadsto \red{\underline{\boxed{\bf Value \: of \: \angle{C} = {70}^{\circ}}}}$