1. ## sine rule

(Give non-exact answers to 3 significant figures.)

In a triangle, the largest side has length 2 cm and one of the other sides has length √2 cm. Given that the area of the triangle is 1cm2, show that the triangle is right-angled and isosceles.

Am not sure how to state off, can anyone just start me off . thanks.

2. Hi

Using cross product properties we know that the area of the triangle is half the product of 2 lengths of the triangle multiplied by the absolute value of sine of the angle formed by the two sides

$A = \frac12\:||\overrightarrow{AB}\wedge\overrightarro w{AC}|| = \frac12\:AB\:AC\:|sin(BAC)|$

With this formula you can compute the angle BAC, the length of the other side and conclude

3. $1 = \frac{1}{2} \times 2 \times \sqrt{2} \times sin\theta$

$\theta = 45$

where do I go from here? how do I show it is a right angled triangle and isosceles.?

4. Hello, Tweety

In a triangle, the largest side has length 2 cm and one of the other sides has length √2 cm.
Given that the area of the triangle is 1 cm², show that the triangle is right-angled and isosceles.
Code:
            *
*  *
_  *     *
√2 *        *
*           *
* θ            *
*  *  *  *  *  *  *
2

Formula: . $A \;=\;\frac{1}{2}\,bc\sin A$

The area is one-half the product of two sides and the sine of the included angle.

We have: . $\frac{1}{2}\left(\sqrt{2}\right)(2)\sin\theta \:=\:1 \quad\Rightarrow\quad \sin\theta \:=\:\frac{1}{\sqrt{2}} \quad\Rightarrow\quad \theta \:=\:45^o$

Hence, the triangle looks like this:

The right half is identical to the left half.
Code:
                  *
* | *
_   *   |   *    _
√2 *     |1    * √2
*       |       *
* 45°     |     45° *
*  *  *  *  *  *  *  *  *
1           1

Therefore, the triangle is an isosceles right triangle.