# Question #ffc84

Apr 22, 2016

At some point measurement has to be made. The calculations below
Gives the $\text{base } \to {z}_{t} = \frac{2 \times 11}{3 + \sqrt{3}} \approx 4.649$ to 3 decimal places.

#### Explanation:

The sum of the internal angles on a triangle is ${180}^{o}$. As the two given angles sum to ${90}^{o}$ the remaining angle is also ${90}^{o}$. So we have a right triangle. Not only that, the angle given match those found in 1/2 of an equilateral triangle. So we have: By the properties of similar triangles and ratio of proportionality we can now determine the length of the sides

By ratio

$\frac{{z}_{t} + {o}_{t} + {z}_{t}}{a + b + c} = {z}_{t} / c = {y}_{t} / a = {o}_{t} / b$

$\implies \frac{11}{2 + 1 + \sqrt{3}} = {z}_{t} / 2 = {y}_{t} / \sqrt{3} = {o}_{t} / 1$

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$\implies \text{base } \to {z}_{t} = \frac{2 \times 11}{3 + \sqrt{3}} \approx 4.649$ to 3 decimal places

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$\implies {o}_{t} = \frac{11}{3 + \sqrt{3}} \approx 2.325$ to 3 decimal places

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$\implies {y}_{t} = \frac{11 \times \sqrt{3}}{3 + \sqrt{3}} \approx 4.026$ to 3 decimal places

May 6, 2016

I discuss here a common method
It is not restricted to right angled triangle only.It is applicable when the given angles can be drawn with the help of compass and ruler.

#### Explanation: Construction

• A straight line $P Q = 11 c m$ is first drawn.
• With the help of a pencil compass and a ruler then we draw $\angle Y P Q = {15}^{\circ} \text{and} \angle Y Q P = {30}^{\circ}$.As a result PY and QY intersect at Y
• Again with the help of pencil compass and a ruler we draw $\angle P Y O = {15}^{\circ} \text{and} \angle Q Y Z = {30}^{\circ}$. As a result PO intersects PQ at O and YZ intersects PQ atZ
• $\Delta Y O Z$ is the required triangle drawn.

Proof

• In $\Delta Y P O , \angle O Y P = {15}^{\circ} = \angle O P Y$ by construction, So $\Delta Y P O$ is an isosceles triangle whose $Y O = P O$
• In $\Delta Z Y Q , \angle Z Y Q = {30}^{\circ} = \angle Z Q Y$ by construction, So $\Delta Y P O$ is an isosceles triangle whose $Y Z = Z Q$
• Now in $\Delta Y O Z , \angle Y O Z = \angle O Y P + \angle O P Y = {30}^{\circ} , \text{since"/_YOZ" is the exterior angle of} \Delta P A B$,
• Also in $\Delta Y O Z , \angle Y Z O = \angle Z Y Q + \angle Y Q Z = {60}^{\circ} , \text{since"/_YZO" is the exterior angle of} \Delta Q Y Z$,
• Hence in $\Delta Y O Z , \angle Y O Z = {30}^{\circ} , \angle Y Z O = {60}^{\circ} \text{and Perimeter=} Y O + O Z + Z Y = P O + O Z + Z Q = P Q = 11 c m$