Leaving Cert Mathematics Trigonometry: A stand is being used to prop up

Question

A stand is being used to prop up a portable solar panel. It consists of a support that is hinged to the panel near the top, and an adjustable strap joining the panel to the support near the bottom. By adjusting the length of the strap, the angle between the panel and the ground can be changed. The dimensions are as follows:

$|AB| = 30 cm$
$|AD| = |CB| = 5 cm$
$|CF| = 22 cm$
$|EF| = 4 cm$

We want to find out how long the strap has to be in order to make the angle $\alpha$ between the panel and the ground equal to $60^{\circ}$.

(a) Two diagrams are given below – one showing triangle $CAF$ and the other showing triangle $CDE$. Use the measurements given in the question, as well as the diagram, to find the lengths of two sides in each triangle.

(b) Taking $\alpha = 60^{\circ}$, as shown, use the triangle $CAF$ to find $\angle LCA$, correct to one decimal place.

The sine rule tells us that in a triangle, if the angle $\alpha$ is opposite a side of length $a$ and the angle $\beta$ is opposite a side of length $b$, then:

$$ \frac{\sin(\alpha)}{a} = \frac{\sin(\beta)}{b} $$

In our case, we have:

$$ \sin(\angle LCA) = \frac{25}{22} \cdot \sin(60^{\circ}) $$

Since $\sin(60^{\circ}) = \frac{\sqrt{3}}{2}$, this means that $\angle LCA = \sin^{-1}(\frac{25}{22} \cdot \frac{\sqrt{3}}{2})$ which is $79.8^{\circ}$ correct to one decimal place.

(c) Hence find $\angle ACF$, correct to one decimal place.

Since all three angles in a triangle have to sum to $180^{\circ}$, we know that $180 - 60 - 79.8 = \angle ACF$, and so $\angle ACF = 40.2^{\circ}$ correct to one decimal place.

(d) Use triangle $CDE$ to find $DE$, the length of the strap, correct to one decimal place.

Firstly, we note that since $D$ lies on the line $AC$ and $E$ lies on the line $CF$, the $\angle ACF$ and $\angle LDC$ must be the same. We now know an angle and two sides of a triangle and wish to find the third side. The cosine rule tells us that if $c$ is the side opposite the angle $\gamma$ and $a$ and $b$ are the other two sides, then:

$$ c^{2} = a^{2} + b^{2} - 2ab\cdot\cos(\gamma) $$

In our case, we have:

$$ |DE| = \sqrt{|CD|^2 + |DE|^2 - 2(|CD|)(|DE|)\cos(\angle LDC)} $$

$$ |DE| = \sqrt{(20)^{2} + (18)^{2} - (20)(22.0)(\cos(40.2^{\circ}))} $$

Thus, $|DE| = \sqrt{173.2568} = 13.2 	ext{ cm correct to one decimal place.}$

A stand is being used to prop up a portable solar panel - Leaving Cert Mathematics - Question 8 - 2013

Worked Solution & Example Answer:A stand is being used to prop up a portable solar panel - Leaving Cert Mathematics - Question 8 - 2013

Two diagrams are given below – one showing triangle CAF and the other showing triangle CDE. Use the measurements given in the question, as well as the diagram, to find the lengths of two sides in each triangle.

Taking α = 60°, as shown, use the triangle CAF to find LCA, correct to one decimal place.

Hence find ACF, correct to one decimal place.

Use triangle CDE to find DE, the length of the strap, correct to one decimal place.

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