Question

In a $\triangle$ ABC sides b, c, $\angle$C are given, which of the following can determine a unique $\triangle$ ABC

• A. c>b sin C, $\angle$C<$\pi$/2,c>b
• B. c >b sin C, $\angle$C<$\pi$/2, c<b
• C. c> b sin C, $\angle$C>$\pi$/2,c>b
• D. c> b sin C, $\angle$C<$\pi$/2, b = c

Answer 1

Answer: (A)

c>b sin C, $\angle$C<$\pi$/2,c>b

Answer 2

Given $b, c, C$. We are looking for angle B using $\sin B = \frac{b \sin C}{c}$.

Condition (A): $c > b \sin C$, $C < \pi/2$, $c > b$.

Since $c > b \sin C$, there are potentially two values for B: $B_1$ (acute) and $B_2 = \pi - B_1$ (obtuse).

Since $C < \pi/2$, we are in the scenario where the ambiguous case can occur.

However, we are also given $c > b$.

We know that two triangles exist if $C < \pi/2$, $c > b \sin C$, and $c < b$.

One triangle exists if $C < \pi/2$, $c > b \sin C$, and $c \ge b$.

Option (A) gives $C < \pi/2$, $c > b \sin C$, and $c > b$. This fits the condition $c \ge b$.

Thus, under condition (A), there is a unique triangle.