SATHEE: Properties of Triangles 1 Question 17

Properties of Triangles 1 Question 17

17. In a $△ A B C, A D$ is the altitude from $A$ . Given $b > c, ∠ C = 23^{\circ}$ and $A D = \frac{a b c}{b^{2} - c^{2}}$ , then $∠ B = \dots$ .

(1994, 2M)

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Answer:

Correct Answer: 17. $113^{\circ}$

Solution:

In $△ A D C, \frac{A D}{b} = \sin 23^{\circ}$

$\begin{aligned} \Rightarrow & A D & = b \sin 23^{\circ} \\ But & A D & = \frac{a b c}{b^{2} - c^{2}} \\ \Rightarrow & \frac{a b c}{b^{2} - c^{2}} & = b \sin 23^{\circ} \\ \Rightarrow & \frac{a}{b^{2} - c^{2}} & = \frac{\sin 23^{\circ}}{c} \end{aligned}$

[given]

Again, in $△ A B C$ ,

$\begin{aligned} \frac{\sin A}{a} & = \frac{\sin 23^{\circ}}{c} \\ \Rightarrow & \frac{\sin A}{a} & = \frac{a}{b^{2} - c^{2}} \\ \Rightarrow & \sin A & = \frac{a^{2}}{b^{2} - c^{2}} \end{aligned}$

$\begin{aligned} \Rightarrow \sin A = \frac{k^{2} \sin^{2} A}{k^{2} \sin^{2} B - k^{2} \sin^{2} C} \\ \Rightarrow \sin A = \frac{\sin^{2} A}{\sin^{2} B - \sin^{2} C} \\ \Rightarrow \sin A = \frac{\sin^{2} A}{\sin (B + C) \sin (B - C)} \\ \Rightarrow \sin A = \frac{\sin^{2} A}{\sin A \cdot \sin (B - C)} \\ \Rightarrow \sin (B - C) = 1 [∵ \sin A \neq 0] \\ \Rightarrow \sin (B - 23^{\circ}) = \sin 90^{\circ} \\ ∴ B - 23^{\circ} = 90^{\circ} \\ ∴ B = 113^{\circ} \end{aligned}$

Properties of Triangles 1 Question 17

17. In a $△ A B C, A D$ is the altitude from $A$ . Given $b > c, ∠ C = 23^{\circ}$ and $A D = \frac{a b c}{b^{2} - c^{2}}$ , then $∠ B = \dots$ .

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Properties of Triangles 1 Question 17

17. In a △ABC,AD is the altitude from A. Given b>c,∠C=23∘ and AD=abcb2−c2, then ∠B=….

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17. In a $△ A B C, A D$ is the altitude from $A$ . Given $b > c, ∠ C = 23^{\circ}$ and $A D = \frac{a b c}{b^{2} - c^{2}}$ , then $∠ B = \dots$ .