Question

1. prove that in any triangle ABC, the area of triangle ABC = 1/2 b.c.SinA

2. Determine the area of triangle PQR in which p = 3.7; q = 5,2 and R = 112°
3. Without using a calculator, determine the possible sizes of angle A if the area of triangle ABC is 6(root 3)m^2; b = 8m and c = 3m

Answer 1

the possible sizes of angle A in triangle ABC are 60° and 120°.

Explanation:

1. To prove that in any triangle ABC, the area of triangle ABC is equal to 1/2 times the product of sides b and c, multiplied by the sine of angle A, we can use the formula for the area of a triangle.

The formula for the area of a triangle is given as:

Area = (1/2) * base * height

In triangle ABC, we can take side b as the base and drop a perpendicular from vertex A to side b, forming the height of the triangle.

Let's call the height of the triangle h. Now, we can express the area of triangle ABC as:

Area = (1/2) * b * h

To find the height of the triangle, we can use the sine of angle A. The sine of an angle is defined as the ratio of the opposite side to the hypotenuse in a right triangle.

In triangle ABC, we have:

sin A = h / c

Rearranging the equation, we get:

h = c * sin A

Now, substituting the value of h in the formula for the area, we have:

Area = (1/2) * b * (c * sin A)

Simplifying the expression, we get:

Area = (1/2) * b * c * sin A

Therefore, we have proven that in any triangle ABC, the area of triangle ABC is equal to 1/2 times the product of sides b and c, multiplied by the sine of angle A.

2. To determine the area of triangle PQR, we can use the formula derived in the previous question:

Area = (1/2) * b * c * sin A

Given p = 3.7, q = 5.2, and R = 112°, we can substitute these values into the formula.

Let's assign p to side b, q to side c, and R to angle A.

b = p = 3.7

c = q = 5.2

A = R = 112°

Now, we can calculate the area using the formula:

Area = (1/2) * p * q * sin R

Substituting the given values, we have:

Area = (1/2) * 3.7 * 5.2 * sin 112°

Calculating sin 112° using trigonometric tables or a calculator, we get:

Area ≈ 8.023 square units (rounded to three decimal places)

Therefore, the area of triangle PQR is approximately 8.023 square units.

3. To determine the possible sizes of angle A, given the area of triangle ABC as 6√3 m², b = 8 m, and c = 3 m, we can use the formula derived in the first question:

Area = (1/2) * b * c * sin A

Given Area = 6√3 m², b = 8 m, and c = 3 m, we can substitute these values into the formula.

Let's assign b = 8, c = 3, and Area = 6√3.

b = 8

c = 3

Area = 6√3

Now, we can calculate the sine of angle A using the formula:

Area = (1/2) * b * c * sin A

Substituting the given values, we have:

6√3 = (1/2) * 8 * 3 * sin A

Simplifying the equation, we get:

6√3 = 12 * sin A

Dividing both sides by 12, we have:

√3/2 = sin A

To determine the possible values of angle A, we need to find the inverse sine (arcsin) of √3/2.

Using trigonometric tables or a calculator, we find that the possible values of A are:

A = 60° or A = 120°

Therefore, without using a calculator, the possible sizes of angle A in triangle ABC are 60° and 120°.

Answer 2

1) See below for proof.

2) 8.9 square units

3) A = 60° or A = 120°

Explanation:

Question 1

Let side b of triangle ABC be its base.

If we draw an altitude from vertex B to side b (shown in blue on the attached diagram), the altitude is the height (h) of triangle ABC.

The sine of an angle in a right triangle is defined as the ratio of the length of the side opposite the angle to the length of the hypotenuse:

`\boxed{\begin{minipage}{9 cm}\underline{Sine trigonometric ratio} \\\\$\sf \sin(\theta)=(O)/(H)$\\\\where:\\ \phantom{ww}$\bullet$ $\theta$ is the angle. \\ \phantom{ww}$\bullet$ $\sf O$ is the side opposite the angle. \\\phantom{ww}$\bullet$ $\sf H$ is the hypotenuse (the side opposite the right angle). \\\end{minipage}}`

The altitude divides the triangle ABC into two right triangles.

For the right triangle that includes angle A, the altitude (h) is the side opposite angle A, and side c is its hypotenuse.

Therefore:

• 
  `\theta = A`
• 
  `\textsf{O}=h`
• 
  `\textsf{H}=c`

Substitute the values into the sine ratio to create an expression for h:

`\sin A=(h)/(c)\implies h=c \cdot \sin A`

Therefore, for triangle ABC:

• 
  `\textsf{base} = b`

• 
  `\textsf{height} = c \cdot \sin A`

We know that the area of a triangle is half the product of its base and height. Therefore, we can substitute the found expressions for the height and the base of the triangle into the equation for the area of a triangle:

`\begin{aligned}\textsf{Area of}\; \triangle ABC&=(1)/(2) \cdot \sf base \cdot height\\\\&=(1)/(2) \cdot b \cdot c \cdot \sin A\end{aligned}`

Therefore, this proves that in any triangle ABC, the area of the triangle ABC is:

`\large\boxed{\textsf{Area of triangle}\;ABC=(1)/(2) \cdot b \cdot c \cdot \sin A}`

`\hrulefill`

Question 2

Given information for triangle PQR:

• Side p (opposite angle P) = 3.7
• Side q (opposite angle Q) = 5.2
• Angle R = 112°

Using the Sine Rule for Area formula:

`\textsf{Area of}\;\triangle PQR=(1)/(2) \cdot p \cdot q \cdot \sin R`

Substitute the given values, and solve for area:

`\begin{aligned}\textsf{Area of}\;\triangle PQR&=(1)/(2) \cdot 3.7 \cdot 5.2 \cdot \sin 112^(\circ)\\\\&=1.85 \cdot 5.2 \cdot \sin 112^(\circ)\\\\&=9.62 \cdot \sin 112^(\circ)\\\\&=8.91950868...\\\\&=8.9\; \sf square\;units\;(nearest\;tenth)\end{aligned}`

Therefore, the area of triangle PQR is 8.9 square units (rounded to the nearest tenth).

`\hrulefill`

Question 3

Given information for triangle ABC:

• Area = 6√3 m²
• Side b (opposite angle B) = 8 m
• Side c (opposite angle C) = 3 m

Using the Sine Rule for Area formula:

`\textsf{Area of}\;\triangle ABC=(1)/(2) \cdot b \cdot c \cdot \sin A`

Substitute the given values into the formula:

`6√(3)=(1)/(2) \cdot 8 \cdot 3 \cdot \sin A`

`6√(3)=4 \cdot 3 \cdot \sin A`

`6√(3)=12\sin A`

Divide both sides by 6:

`√(3)=2\sin A`

Divide both sides by 2:

`\sin A=(√(3))/(2)`

According to the unit circle (attachment 2), when the sine of an angle is √3/2, the angle is 60° or 120°.

Therefore, the possible sizes of angle A are:

• A = 60° or A = 120°