Triangle ABC has are 30cm^2, AC=9 cm and BC= 7cm. Calculate the measureof angle ACB and the larget possible length of AB.
Area = (1/2)(AC)(BC)sinØ , where Ø is angle ACB
30 =(1/2)(9)(7)sinØ
sinØ = 60/63 = .95238
Ø = 72.25° or Ø = 107.75°
The largest value of AB occurs when Ø = 107.75
by cosine law
AB^2 = 7^2 + 9^2 - 2(7)(9)cos 107.75
= 168.4187
AB = √168.4187 = 12.98
thanks :)
how did yu get 107.75?
the second angle is in quadrant II
the sine is positive in first ad second quadrants.
so when you take sine-inverse, you get two answers, even though your calculator only gives you the angle in the first quadrant.
so the second angle is 180-72.25 = 107.75
pls. explain
oh yeah thaNKS.:)
To solve for the measure of angle ACB, we can use the formula for the area of a triangle:
Area = (1/2) * base * height
Given that the area of triangle ABC is 30cm^2, and AC is 9cm, we can substitute these values into the formula to find the height of the triangle:
30 = (1/2) * 7 * height
Simplifying the equation:
60 = 7 * height
Dividing both sides by 7:
height = 60 / 7 ≈ 8.57 cm
Now, we have the base (BC = 7 cm) and the height (8.57 cm) of triangle ABC. To find the measure of angle ACB, we can use the tangent ratio:
tan(ACB) = opposite / adjacent
In this case, the opposite side is the height (8.57 cm), and the adjacent side is BC (7 cm). Substituting these values into the equation:
tan(ACB) = 8.57 / 7
Using a calculator, we can find the inverse tangent of this ratio to solve for ACB:
ACB = arctan(8.57 / 7) ≈ 51.4 degrees
To find the largest possible length of AB, we can use the Triangle Inequality Theorem, which states that the sum of any two sides of a triangle must be greater than the third side.
In this case, we have AC = 9 cm, BC = 7 cm, and we want to find the largest possible length of AB. The sum of AC and BC is 16 cm, so AB must be greater than 16 cm for it to be a valid triangle.
Therefore, the largest possible length of AB is any value greater than 16 cm.