Question

The interior angles of a triangle are 60,45and 75.the shortest side is 10cm less than the longest side, determine the perimeter of the triangle to the nearest cm.include a diagram please and thank you!!

Answer 1

Answer:

20 cm.

Step-by-step explanation:

We know that the permiter is defined as

$P=L+S+M$

Where $L$ is the longest side, $S$ is the smallest and $M$ is the middle side.

If $L=x$ then, $S=x-20$, where we need to find an expression for $M$ using the law of cosines.

$M^{2} =L^{2}+S^{2}-2 \times L \times S \times cos(60\°)\\M=\sqrt{x^{2}+(x-20)^{2}-x(x-20)}\\M=\sqrt{x^{2} +x^{2}-40x+400-x^{2}+20x}=\sqrt{x^{2}-20x+400}$

Replacing all expression, the perimeter is

$P=x+x-20+\sqrt{x^{2}-20x+400}\\P=2x-20+\sqrt{x^{2}-20x+400}$

Using a calculator, the perimeter is 20 units centimeters.