Find the area of a triangle ABC
If AC=9CM, AB=8CM, CB=10cm
Give your answer to one decimal place

Area = 34.2 cm²

Step-by-step explanation:

The general formula of the area of the triangle is half the product of two sides multiplied by the sine the angle between them.

So, for the given triangle ABC

AC = 9 cm, AB = 8 cm, CB = 10 cm

Area = 0.5 AC * AB * sin A or 0.5 BC * BA * sin B or = 0.5 CA * CB * sin C

Using the first form, so we need the measure of angle A

Using cosine low: cos A = (b² + c² - a²)/(2bc)

Where: a = BC = 10 , b = AC = 9  and c = AB = 8

So. cos A = (9² + 8² - 10²)/(2 * 9 * 8 ) = 0.3125

∠A = cos⁻¹0.3125 = 71.79°

So, Area = 0.5 AC * AB * sin A = 0.5 * 9 * 8 * sin 71.79° = 34.2 cm²

So, Area = 34.2 cm² to the nearest one decimal place

3

step-by-step explanation:

i can't solve unless i know what the line is

step-by-step explanation:

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