Question

If two sides of a triangle are \[12\] and $17$, and the included angle is $60^\circ $, what is the area of the triangle?

Answer 1

Hint: Here, in the given question, we are given that the two sides of a triangle are \[12\] and $17$, and the included angle (angle between two adjacent sides) is $60^\circ $ and we need to find the area of the triangle. If two adjacent sides and the angle between two sides (included angle) is given then the area of the triangle can be found by the formula given below: Let us suppose the sides of a triangle are named as $a$, $b$ and $c$. The angles are named as $A$, $B$ and $C$.

Then area of triangle = $\dfrac{1}{2}ab\sin C = \dfrac{1}{2}bc\sin A = \dfrac{1}{2}ca\sin B$.Thus, the area of a triangle is half the product of the length of two sides and $\sin e$ of the angle between the two sides. Now, as we know the formula, we will find the area of the triangle given in the question using this formula.

Complete step by step answer:
Given, the two sides of a triangle are \[12\] and $17$, and the included angle is $60^\circ $.

$\text{Area} =\dfrac{1}{2}ca\sin B$
On substituting the values, we get
$\text{Area} = \dfrac{1}{2} \times 17 \times 12\sin 60^\circ $
As we know $\sin 60^\circ = \dfrac{{\sqrt 3 }}{2}$. Therefore, we get
$\text{Area} = \dfrac{1}{2} \times 17 \times 12 \times \dfrac{{\sqrt 3 }}{2}$
On canceling out common terms, we get
$\text{Area} = 17 \times 3 \times \sqrt 3 $
As we know $\sqrt 3 = 1.732$. Therefore, we get
$\text{Area} = 17 \times 3 \times 1.732$
On multiplication of terms, we get
$\therefore \text{Area} = 88.332\,uni{t^2}$

Therefore, the area of the triangle is $88.332\,uni{t^2}$.

Note: We should be careful about the unit. Calculation should be done in the same unit. We should take care of the calculations so as to be sure of our final answer. There are other formulas too to find the area of the triangle. For example:
Area of a right-angled triangle = $\dfrac{1}{2} \times base \times height$
Area of an equilateral triangle = \[\dfrac{{\sqrt 3 }}{4} \times sid{e^2}\]
Area of an isosceles triangle = $\dfrac{1}{4} \times b \times \sqrt {4{a^2} - {b^2}} $.....($b$ is the base and $a$ is the measure of one of the equal sides.)
We use them according to the given problem.