Q68 / 160 · TS EAMCET 2024 (10 May Shift 1) · Mathematics
In triangle ABC , if $a=4, b=3, c=2$ thèn $2(a-b \cos C)(a-c \sec B)=$
- A0
- B1
- C2
- Correct answerD3
Solution
$a=4, b=3, c=2$, in $\triangle A B C$
Using $\cos$ formula, $\begin{aligned} & \cos C=\frac{a^2+b^2-c^2}{2 a b}=\frac{16+9-4}{2 \times 4 \times 3}=\frac{21}{24} \\ & \cos B=\frac{a^2+c^2-b^2}{2 a c}=\frac{16+4-9}{2 \times 4 \times 2}=\frac{11}{16} \Rightarrow \sec B=\frac{16}{11} \\ & 2(a-b \cos C)(a-c \sec B)=2\left(4-3 \times \frac{21}{24}\right)\left(4-2 \times \frac{16}{11}\right) \\ & =2\left(4-\frac{21}{8}\right)\left(4-\frac{32}{11}\right)=3 . \end{aligned}$
Using $\cos$ formula, $\begin{aligned} & \cos C=\frac{a^2+b^2-c^2}{2 a b}=\frac{16+9-4}{2 \times 4 \times 3}=\frac{21}{24} \\ & \cos B=\frac{a^2+c^2-b^2}{2 a c}=\frac{16+4-9}{2 \times 4 \times 2}=\frac{11}{16} \Rightarrow \sec B=\frac{16}{11} \\ & 2(a-b \cos C)(a-c \sec B)=2\left(4-3 \times \frac{21}{24}\right)\left(4-2 \times \frac{16}{11}\right) \\ & =2\left(4-\frac{21}{8}\right)\left(4-\frac{32}{11}\right)=3 . \end{aligned}$
Hence, the correct answer is option: 'D'.