Question

Trigonometry

Objective: Use trigonometry functions to find the area of triangles.
In ΔABC, AB=21, AC=16, and m< A=67*. Find the area of ΔABC, to the nearest tenth of a square unit.

Answer 1

The area of the triangle ABC is 154.6 square units

Step-by-step explanation:

Given that the measurements of the sides of the triangles are
`AB=21` ,
`AC=16` and
`m\angle A=67^(\circ)`

We need to determine the area of the triangle ABC

Area of triangle ABC:

The area of the triangle can be determined using the formula,

`\text {Area}=(1)/(2) b c \sin A`

Substituting the values, we get,

`\text {Area}=(1)/(2)(21)(16) \sin 67^(\circ)`

Simplifying the terms, we get,

`\text {Area}=(1)/(2)(21)(16) (0.92)`

Multiplying the values, we have,

`\text {Area}=(309.12)/(2)`

Dividing, we get,

`\text {Area}=154.56`

Rounding off to the nearest tenth, we get,

`Area = 154.6`

Thus, the area of the triangle ABC is 154.6 square units