In ΔABC; with usual notations, (bsinB-csinC)/(sin⁡B-C) = ……

Correct answer: (C) a. We have, (bsinB-csinC)/(sin⁡B-C) =(ksinBsinB-ksinCsinC)/(sin⁡B-c) (Using Sine rule) =(ksin^2B-sin^2C)/(sin⁡B-c) =(ksinB+Csin⁡B-C)/(sin⁡B-C) =ksinB+C =ksin180^o-A =ksinA=a

Q99 / 150 · MHT CET 2019 (Shift 2) · Mathematics

In $Δ A B C;$ with usual notations, $\frac{b s i n B - c s i n C}{\sin (B - C)}$ = ……

A
$b$
B
$a + b + c$
Correct answer
C
$a$
D
$c$

Solution

We have,

\frac{b s i n B - c s i n C}{\sin (B - C)}

= \frac{k s i n B s i n B - k s i n C s i n C}{\sin (B - c)}

(Using Sine rule)

= \frac{k {s i n}^{2} B - {s i n}^{2} C}{\sin (B - c)}

= \frac{k s i n (B + C) \sin (B - C)}{\sin (B - C)}

= k s i n (B + C)

= k s i n (180^{o} - A)

= k s i n A = a

Hence, the correct answer is option: 'C'.

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In ΔABC; with usual notations, bsinB-csinCsin⁡B-C = ……

Solution

Relevant Questions

In $Δ A B C;$ with usual notations, $\frac{b s i n B - c s i n C}{\sin (B - C)}$ = ……