SATHEE: Properties of Triangles 2 Question 1

Properties of Triangles 2 Question 1

1. With the usual notation, in $△ A B C$ , if $∠ A + ∠ B = 120^{\circ}, a = \sqrt{3} + 1$ and $b = \sqrt{3} - 1$ , then the ratio $∠ A : ∠ B$ , is

(2019 Main, 10 Jan II)

(a) $7 : 1$

(b) $3 : 1$

(d) $5 : 3$

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Answer:

Correct Answer: 1. (a)

Solution:

For a $△ A B C$ , it is given that $a = \sqrt{3} + 1$ ,

$b = \sqrt{3} - 1$ and $∠ A + ∠ B = 120^{\circ}$

Clearly, $∠ C = 60^{\circ} [∵ ∠ A + ∠ B + ∠ C = 180^{\circ}]$

Now, by tangent law, we have

$\begin{aligned} \tan \frac{A - B}{2} & = \frac{a - b}{a + b} \cot \frac{C}{2} \\ = \frac{(\sqrt{3} + 1) - (\sqrt{3} - 1)}{(\sqrt{3} + 1) + (\sqrt{3} - 1)} \cot \frac{60^{\circ}}{2} \\ = \frac{2}{2 \sqrt{3}} \cot (30^{\circ}) \\ = \frac{1}{\sqrt{3}} \times \sqrt{3} = 1 \\ \Rightarrow \tan \frac{A - B}{2} & = 1 = \tan 45^{\circ} \\ \Rightarrow \frac{A - B}{2} & = 45^{\circ} \\ \Rightarrow ∠ A - ∠ B & = 90^{\circ} \end{aligned}$

On solving $∠ A - ∠ B = 90^{\circ}$ and $∠ A + ∠ B = 120^{\circ}$ , we get $∠ A = 105^{\circ}$ and $∠ B = 15^{\circ}$

So, $∠ A : ∠ B = 7 : 1$

Properties of Triangles 2 Question 1

1. With the usual notation, in $△ A B C$ , if $∠ A + ∠ B = 120^{\circ}, a = \sqrt{3} + 1$ and $b = \sqrt{3} - 1$ , then the ratio $∠ A : ∠ B$ , is

Answer:

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Properties of Triangles 2 Question 1

1. With the usual notation, in △ABC, if ∠A+∠B=120∘,a=3+1 and b=3−1, then the ratio ∠A:∠B, is

Answer:

Engineering Preparation

Medical Preparation

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Important Resources

Other Resources

Syllabus

Test Platform

Sample Mock Test

Mentorship

NCERT Exercise Video Solution

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1. With the usual notation, in $△ A B C$ , if $∠ A + ∠ B = 120^{\circ}, a = \sqrt{3} + 1$ and $b = \sqrt{3} - 1$ , then the ratio $∠ A : ∠ B$ , is