In a - ABC- if - A - 60- then the value 1-ac-bc 1-cb-ab -

Given nbsp; ∠ A=60^∘. nbsp; Then ( 1+ac+bc ) ( 1+cb ab ) =(a+b+c)(b+c a)bc =(a+b+c)(a+b+c 2a)bc =2s(2s 2a)bc =4.s(s a)bc =4cos ^2 A2(Using half angle prope ...

You visited us 1 times! Enjoying our articles? Unlock Full Access!

Byju's Answer

Standard XII

Mathematics

Trigonometric Ratios of Half Angles

In a Δ ABC, i...

Question

In a $Δ A B C,$ if $∠ A = 60^{\circ}$ , then the value $(1 + \frac{a}{c} + \frac{b}{c}) (1 + \frac{c}{b} - \frac{a}{b}) =$

3

Right on! Give the BNAT exam to get a 100% scholarship for BYJUS courses

2

No worries! We‘ve got your back. Try BYJU‘S free classes today!

1

No worries! We‘ve got your back. Try BYJU‘S free classes today!

0

No worries! We‘ve got your back. Try BYJU‘S free classes today!

Open in App

Solution

The correct option is A $3$
Given $∠ A = 60^{\circ}$ . Then $(1 + \frac{a}{c} + \frac{b}{c}) (1 + \frac{c}{b} - \frac{a}{b})$
$= \frac{(a + b + c) (b + c - a)}{b c}$
$= \frac{(a + b + c) (a + b + c - 2 a)}{b c}$
$= \frac{2 s (2 s - 2 a)}{b c}$
$= 4. \frac{s (s - a)}{b c}$
$= 4 {cos}^{2} \frac{A}{2}$ (Using half-angle property)
$= 4 {cos}^{2} 30^{0}$ (Since $A = 60^{0}$ )
$= 4 \times \frac{3}{4} = 3$

Suggest Corrections

Similar questions

Q. If in a

Δ A B C, ∠ A = 60^{\circ},

then find then value of

(1 + \frac{a}{c} + \frac{b}{c}) (1 + \frac{c}{b} - \frac{a}{b})

Q. In

Δ A B C

, If

∠ C = 60^{\circ}

and

\frac{1}{a + c} + \frac{1}{(b + c)} = \frac{k}{a + b + c}

, then the value of

k

Q. If in a

Δ ABC,

b = \sqrt{3} + 1, c = \sqrt{3} - 1, ∠ A = 60^{\circ} .

Then the value of

tan (\frac{B - C}{2})

is:

Q. If

0 < a, b, c < 1

and

a + b + c = 2

, then the greatest value of

\frac{a}{1 - a} \times \frac{b}{1 - b} \times \frac{c}{1 - c}

Q. If a, b, c are any three positive numbers, then the least value of

(a + b + c) (\frac{1}{a} + \frac{1}{b} + \frac{1}{c})

Join BYJU'S Learning Program

Explore more

Trigonometric Ratios of Half Angles

Standard XII Mathematics

Join BYJU'S Learning Program

In a ΔABC, if ∠A=60∘, then the value (1+ac+bc)(1+cb−ab)=

The correct option is A 3Given ∠A=60∘. Then (1+ac+bc)(1+cb−ab) =(a+b+c)(b+c−a)bc =(a+b+c)(a+b+c−2a)bc =2s(2s−2a)bc =4.s(s−a)bc =4cos2A2 (Using half-angle property) =4cos2300 (Since A=600) =4×34=3

In a $Δ A B C,$ if $∠ A = 60^{\circ}$ , then the value $(1 + \frac{a}{c} + \frac{b}{c}) (1 + \frac{c}{b} - \frac{a}{b}) =$