Question

Question #ffc84

Answer 1

At some point measurement has to be made. The calculations below
Gives the `"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`. 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

`(z_t+o_t+z_t)/(a+b+c) = z_t/c =y_t/a=o_t/b`

`=>11/(2+1+sqrt(3))=z_t/2 = y_t/sqrt(3)=o_t/1`

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`=> "base "->z_t=(2xx11)/(3+sqrt(3)) ~~4.649` to 3 decimal places

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`=> o_t= 11/(3+sqrt(3))~~2.325` to 3 decimal places

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`=>y_t=(11xxsqrt(3))/(3+sqrt(3)) ~~4.026` to 3 decimal places

Answer 2

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 `PQ=11 cm` is first drawn.
• With the help of a pencil compass and a ruler then we draw `/_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 `/_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`