Question

Tai
(2) in triangle abc angle a + angle b is equal to 125 degree and angle b + angle c is equal to 150 degree find all the angles of a triangle ABC
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Answer 1

$\sf\large\underline{Given:}$

• $\rm{\angle A+\angle B=125\degree-----(i)}$
• $\rm{\angle B+\angle C=150\degree-----(ii)}$

$\sf\large\underline{To\: Find:}$

• $\rm{All\:the\:angles\:of\:\triangle\:ABC}$

$\sf\large\underline{Solution:}$

• As we know that, sum of of all angles of Triangle is equal to 180°

$\rm{\implies As\:per\:the\:sum\:of\:angles\:of\:triangle\: properties}$

$\rm{\implies \angle A+\angle B+\angle C=180\degree}$

$\rm{\implies (\angle A+\angle B)+\angle C=180\degree}$

• $\sf{putting\: value\: from\:eq\:1st}$

$\rm{\implies 125+\angle C=180\degree}$

$\rm{\implies \angle C=180-125}$

$\rm\red{\implies \angle C=55\degree}$

• $\sf{putting\:the\: value\:of\:\angle C=55\:in\:eq\:ii}$

$\rm{\implies \angle B+\angle C=150\degree}$

$\rm{\implies \angle B+55=150}$

$\rm{\implies \angle B=150-55}$

$\rm\purple{\implies \angle B=95\degree}$

• $\sf{Than\: putting\:the\: value\:of\:\angle B=95\:in\:eq\:1st}$

$\rm{\implies \angle A+\angle B=125\degree}$

$\rm{\implies \angle A+95=125}$

$\rm{\implies \angle A=125-95}$

$\rm\green{\implies \angle A=30\degree}$

Hence,

$\rm{\implies The\: angles\:of\:\triangle\:are\:30\degree,\:95\degree,\:55\degree}$

$\sf\large\underline{Verification:}$

$\rm{\implies (\angle A+\angle B)+\angle C=180\degree}$

$\rm{\implies 30+95+55=180}$

$\rm{\implies 180\degree=180\degree}$

$\sf\large\underline{Hence,\: Verified:}$

Answer 2

Given that ,

In a triangle Δ ABC

$\sf \angle A + \angle B = 125 \: - - - \: (i)\\ \\ \sf \angle B + \angle C = 150 \: - - - \: (ii)$

As we know that ,

The sum of all angles of triangle is 180

Thus ,

$\Rightarrow \sf \angle A + \angle B + \angle C = 180 \\ \\ \Rightarrow \sf 125 + \angle C = 180 \: \: \{from \: eq \: (i) \}\\ \\ \Rightarrow \sf \angle C = 180 - 125 \\ \\ \Rightarrow \sf \angle C = 55$

Put the value of $\sf \angle C = 55$ in eq (ii) , we get

$\Rightarrow \sf \angle B + 55 = 150 \\ \\\Rightarrow \sf \angle B = 150 - 55 \\ \\\Rightarrow \sf \angle B = 95$

Put the value of $\sf \angle B = 95$ in eq (i) , we get

$\Rightarrow \sf \angle A + 95 = 125 \\ \\\Rightarrow \sf \angle A = 125 - 95 \\ \\\Rightarrow \sf \angle A =30$

$\therefore \bold{ \sf \underline{The \: angles \: of \: triangle \: ΔABC \: are \: 30 , 95 \: and \: 55}}$