Question

The diagram shows a 9cm x 7cm rectangle-based pyramid. all the diagonal sides - TA, TB, TC, AND TD- are length 12cm. M is the midpoint of the rectangular base.
Work out angle TAC, to 1 decimal place.

Answer 1

Answer:

∠TAC is approximately equal to 61.6°

Step-by-step explanation:

The given parameters for the pyramid are;

The dimension for the rectangular base are; Length = 9 cm, width = 7 cm

The length of the diagonal sides, TA, TB, TC, and TD = 12 cm each

The midpoint of the rectangular base = Point M

The diagonal AC = AM + MC

AM = MC as given M is the midpoint of the rectangular base

∴ AC = AM + MC = 2·AM

By Pythagoras' theorem, AC = √(9² + 7²) = √130

AC = √130 cm

∴ AM = AC/2 = (√130)/2 cm

Alternatively, AM = √((9/2 cm)² + (7/2 cm)²) = √(32.5) cm

∠TAC = ∠TAM

By trigonometric ratios, we have;

$cos (\theta) = \dfrac{Length \ of \ adjacent \ side \ to \ angle }{Length \ of \ hypotenuse\ side \ to \ angle}$

$\therefore cos (\angle TAM) = cos (\angle TAC) = \dfrac{\left (\dfrac{\sqrt{130} }{2} \right )}{12} = \dfrac{\sqrt{130} }{2 \times 12} = \dfrac{\sqrt{130} }{24}$

$\angle TAC = arccos \left ( \dfrac{\sqrt{130} }{24} \right ) \approx 61.6 ^{\circ} \ to 1 \ decimal \ place$

Answer 2

The measure of angle TAC is 61.6 degrees

The given parameters are:

$\mathbf{AB = 7}$

$\mathbf{BC = 9}$

Start by calculating length AC, using the following Pythagoras theorem.

$\mathbf{AC = \sqrt{AB^2 + BC^2}}$

So, we have:

$\mathbf{AC = \sqrt{7^2 + 9^2}}$

$\mathbf{AC = \sqrt{49 + 81}}$

$\mathbf{AC = \sqrt{130}}$

$\mathbf{AC = 11.40}$

Next, we calculate side length AM

$\mathbf{AM = \frac{AC}{2}}$

So, we have:

$\mathbf{AM = \frac{11.40}{2}}$

$\mathbf{AM = 5.70}$

The measure of angle TAC is calculated using the following cosine ratio

$\mathbf{cos(A)= \frac{AM}{TA}}$

So, we have:

$\mathbf{cos(A)= \frac{5.70}{12}}$

$\mathbf{cos(A)= 0.475}$

Take arc cos of both sides

$\mathbf{A= 61.6}$

Hence, the measure of angle TAC is 61.6 degrees

Read more about pyramids at:

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