Question #ffc84

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Question #ffc84

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Answer 1 · 2016-04-22T14:10:14

At some point measurement has to be made. The calculations below
Gives the $base \to z_{t} = \frac{2 \times 11}{3 + \sqrt{3}} \approx 4.649$ $"base "->z_t=(2xx11)/(3+sqrt(3)) ~~4.649$ to 3 decimal places.

Explanation:

The sum of the internal angles on a triangle is $180^{o}$ $180^o$ . As the two given angles sum to $90^{o}$ $90^o$ the remaining angle is also $90^{o}$ $90^o$ . 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:

TonyB

By the properties of similar triangles and ratio of proportionality we can now determine the length of the sides

By ratio

$\frac{z_{t} + o_{t} + z_{t}}{a + b + c} = \frac{z_{t}}{c} = \frac{y_{t}}{a} = \frac{o_{t}}{b}$ $(z_t+o_t+z_t)/(a+b+c) = z_t/c =y_t/a=o_t/b$

$\Rightarrow \frac{11}{2 + 1 + \sqrt{3}} = \frac{z_{t}}{2} = \frac{y_{t}}{\sqrt{3}} = \frac{o_{t}}{1}$ $=>11/(2+1+sqrt(3))=z_t/2 = y_t/sqrt(3)=o_t/1$

'~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~

$\Rightarrow base \to z_{t} = \frac{2 \times 11}{3 + \sqrt{3}} \approx 4.649$ $=> "base "->z_t=(2xx11)/(3+sqrt(3)) ~~4.649$ to 3 decimal places

'~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~

$\Rightarrow o_{t} = \frac{11}{3 + \sqrt{3}} \approx 2.325$ $=> o_t= 11/(3+sqrt(3))~~2.325$ to 3 decimal places

'~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~

$\Rightarrow y_{t} = \frac{11 \times \sqrt{3}}{3 + \sqrt{3}} \approx 4.026$ $=>y_t=(11xxsqrt(3))/(3+sqrt(3)) ~~4.026$ to 3 decimal places

Answer 2 · 2016-05-06T12:24:16

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:

selfsrawn

Construction

A straight line $P Q = 11 c m$ $PQ=11 cm$ is first drawn.
With the help of a pencil compass and a ruler then we draw $∠ Y P Q = 15^{\circ} and ∠ Y Q P = 30^{\circ}$ $/_YPQ=15^@ "and"/_YQP = 30^@$ .As a result PY and QY intersect at Y
Again with the help of pencil compass and a ruler we draw $∠ P Y O = 15^{\circ} and ∠ Q Y Z = 30^{\circ}$ $/_PYO=15^@ "and"/_QYZ = 30^@$ . As a result PO intersects PQ at O and YZ intersects PQ atZ
$DeltaYOZ$ is the required triangle drawn.

Proof

In $DeltaYPO,/_OYP=15^@=/_OPY$ by construction, So $DeltaYPO$ is an isosceles triangle whose $YO =PO$
In $DeltaZYQ,/_ZYQ=30^@=/_ZQY$ by construction, So $DeltaYPO$ is an isosceles triangle whose $YZ =ZQ$
Now in $DeltaYOZ,/_YOZ=/_OYP+/_OPY=30^@,"since"/_YOZ" is the exterior angle of" Delta PAB$ ,
Also in $DeltaYOZ,/_YZO=/_ZYQ+/_YQZ=60^@,"since"/_YZO" is the exterior angle of" Delta QYZ$ ,
Hence in $DeltaYOZ,/_YOZ=30^@,/_YZO=60^@ "and Perimeter=" YO+OZ+ZY = PO+OZ+ZQ=PQ =11 cm$

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Question #ffc84

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