[tex]\huge\mathcal{\fcolorbox{purple} {lavenderblush}{{Question}}}[/tex]
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In a ∆ABC, ∠B<∠C and the values of B and C satisfy the equation 2 tanx-k(1+tan x)-0, where (0<k<1). Then, the measure of ∠A is
• [tex]\frac{\pi}{3}[/tex]
• [tex]\frac{2\pi}{3}[/tex]
• [tex]\frac{\pi}{2}[/tex]
• [tex]\frac{3\pi}{4}[/tex]
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Thank You
(✯ᴗ✯)
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▬▬▬▬▬▬▬▬▬▬▬▬▬▬▬▬▬▬▬▬▬▬▬▬
No place for ☆SPAMMERS☆
▬▬▬▬▬▬▬▬▬▬▬▬▬▬▬▬▬▬▬▬▬▬▬▬
In a ∆ABC, ∠B<∠C and the values of B and C satisfy the equation 2 tanx-k(1+tan x)-0, where (0<k<1). Then, the measure of ∠A is
• [tex]\frac{\pi}{3}[/tex]
• [tex]\frac{2\pi}{3}[/tex]
• [tex]\frac{\pi}{2}[/tex]
• [tex]\frac{3\pi}{4}[/tex]
▬▬▬▬▬▬▬▬▬▬▬▬▬▬▬▬▬▬▬▬▬▬▬▬
Thank You
(✯ᴗ✯)
▬▬▬▬▬▬▬▬▬▬▬▬▬▬▬▬▬▬▬▬▬▬▬▬
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Verified answer
[tex] \huge \star \rm ANSWER\huge \star[/tex]
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Class 11
>>Maths
>>Trigonometric Functions
>>Trigonometric Equations
>>In a triangle ABC, B > C and the value
Question
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In a triangle ABC, ∠B>∠C and the values of B & C satisfy the equation (2tanx−k)(1+tan
2
x)=0 where ( 0 > k > 1). Then the measure of angle A is
Hard
Updated on : 2022-09-05
Solution
verified
Verified by Toppr
Correct option is C)
Given equation,
2tanx−k(1+tan
2
x)=0
ktan
2
x−2tanx+k=0
tanC and tanB are the roots of the equation.
Thus, tanC+tanB=
k
2
(Sum of the roots)
tanCtanB=
k
k
=1 (Product of the roots)
Now, tan(B+C)=
1−tanAtanB
tanB+tanC
tan(180−A)=
1−1
k
2
(Angle sum property)
tanA=∞
tanA=tan90
A=90
∘