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.
An engineer standing at a point $P$ wishes to determine the width of a new rectangular pond. She finds the distance to the western-most point $A$ of the pond from $P$ to be 60 m, while the distance to the northern-most point $B$ of the pond from $P$ is 80 m. If the angle between the two lines of sight at $P$ is $60^{\circ}$ , then the width $A B$ (in metres) of the pond, where $A B$ is not parallel to the North–South line, is:
A. $2013$ B. $1320$ C. $1013$ D. $1310$
easyAsked in NIMCET 2026View solution ›
Q3.
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 ›
Q4.
In triangle ABC, $a = 3$ , $b = 5$ and $c = 7$ . The angle C is
A. $60^{\circ}$ B. $90^{\circ}$ C. $120^{\circ}$ D. $150^{\circ}$
easyView solution ›
Q5.
In triangle ABC, $A = 30^{\circ}$ , $B = 45^{\circ}$ and $a = 4$ . Then $b$ equals
A. $22$ B. $42$ C. $43$ D. $8$
easyView solution ›
Q6.
A harbour lies in a direction $60^{\circ}$ South of West from a fort and at a distance 30 km from it. A ship sets out from the harbour at noon and sails due East at 10 km an hour. The time at which the ship will be 70 km from the fort is
A. 7 PMB. 8 PMC. 5 PMD. 10 PM
hardAsked in NIMCET 2015View solution ›