Solution of Triangles Questions with Solutions

6 questions3 easy2 medium1 hard

Solving a triangle means finding all its unknown sides and angles from the given information. Every triangle has six elements (three sides and three angles), and in general three suitable elements, including at least one side, are enough to find the rest. This topic covers the standard cases: three sides given (use the cosine rule), two sides and the included angle (cosine rule, then sine rule), two angles and a side (sine rule), and two sides with a non-included angle, the ambiguous case where zero, one or two triangles may be possible. Exam questions test choosing the right rule, handling the ambiguous case correctly, and using the angle sum A + B + C = 180 degrees to find the remaining angle. Practise the solution of triangles questions below, each with a step-by-step solution.

All Questions Easy Medium Hard

Q1.
If $B C = a$ , $A C = b$ and $A B = c$ are the sides of a triangle $A B C$ , and $∠ C \neq = \frac{π}{2}$ , then which one of the following is not correct?
A. $\frac{a}{s in A} = \frac{b}{s in B} = \frac{c}{s in C}$ B. $\frac{a - b}{a + b} = \frac{s in A - s in B}{s in A + s in B}$ C. $\frac{a - b}{a + b} = co t (\frac{A + B}{2}) t an (\frac{A - B}{2})$ D. $\frac{a - b}{a + b} = \frac{t an ( \frac{A - B}{2} )}{t an ( \frac{C}{2} )}$
mediumAsked in NIMCET 2026View solution ›
Q2.
In $△ A B C$ , $B = 45^{\circ}$ , $C = 105^{\circ}$ and $c = 2$ . Then
A. $A = 30^{\circ}$ , $a = 3 - 1$ , $b = 2 (3 - 1)$ B. $A = 30^{\circ}$ , $a = 3 + 1$ , $b = 2 (3 - 1)$ C. $A = 30^{\circ}$ , $a = 1 - 3$ , $b = 2 (3 + 1)$ D. $A = 30^{\circ}$ , $a = 3 - 1$ , $b = 2 (3 + 1)$
mediumAsked in NIMCET 2011View solution ›