Question

$\fcolorbox{red}{gold}\ \textless \ br /\ \textgreater \ {PLEASE SOLVE THIS QUESTION}$
SOLVE [[ANY ONE]]
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Answer 1

Answer:

Let the measure of ∠B be ‘x°’ and that of ∠C be ‘y°’. According to the first condition, m∠A = m∠B + m∠C ∴ m∠A = x° + y° In AABC, m∠A + m∠B + m∠C = 180° …[Sum of the measures of the angles of a triangle is 180°] ∴ x + y + x + y = 180, ∴ 2x + 2y = 180 Dividing both sides by 2, x + y = 90 …(i) According to the second condition, the ratio of the measures of ∠B and ∠C is 4 : 5. ∴ x/y = 4/5 ∴ 5x = 4y ∴ 5x – 4y = 0 …….(ii) Multiplying equation (i) by 4, 4x + 4y = 360 …(iii)

x = 40

Substituting x = 40 in equation (i), x + y = 90 ∴ 40 + y = 90 ∴ y = 90 – 40 ∴ y = 50 ∴ m∠A = x° + y° = 40° + 50° = 90° ∴ The measures of ∠A, ∠B and ∠C are 90° , 40° , and 50°

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Answer 2

Answer:

Let the measure of ∠B be ‘x°’ and that of ∠C be ‘y°’.

According to the first condition, m∠A = m∠B + m∠C ∴ m∠A = x° + y°

In AABC, m∠A + m∠B + m∠C = 180°

…[Sum of the measures of the angles of a triangle is 180°] ∴ x + y + x + y = 180, ∴ 2x + 2y = 180 Dividing both sides by 2, x + y = 90

…(i) According to the second condition, the ratio of the measures of ∠B and ∠C is 4 : 5. ∴ x/y = 4/5 ∴ 5x = 4y ∴ 5x – 4y = 0 ……

.(ii) Multiplying equation (i) by 4, 4x + 4y = 360 …(iii)x = 40

Substituting x = 40 in equation

(i), x + y = 90 ∴ 40 + y = 90 ∴ y = 90 – 40 ∴ y = 50 ∴ m∠A = x° + y° = 40° + 50° = 90° ∴ The measures of ∠A, ∠B and ∠C are 90° , 40° , and 50°