The bases of a trapezoid are 425 feet and 250 feet respectively. The angles that the sides make with the longer base are 64 degree and 75 degree. What are the sides of the trapezoid?

Question

The bases of a trapezoid are 425 feet and 250 feet respectively. The angles that the sides make with the longer base are 64 degree and 75 degree. What are the sides of the trapezoid?

1 Answer

Impact of this question

1586 views around the world

You can reuse this answer
Creative Commons License

Answer 1 · 2018-04-25T05:27:58

$A D = 257.3$ $AD=257.3$ feet and $B C = 239.5$ $BC=239.5$ feet

Explanation:

The trapezium $A B C D$ $ABCD$ is shown in the diagram below, where we have also drawn $D L$ $DL$ and $C M$ $CM$ perpendicular to $A B$ $AB$ as shown. Further, height of trapezium is $h$ $h$

enter image source here

Now as $\frac{A L}{h} = {cot 64}^{\circ}$ $(AL)/h=cot64^@$ and $\frac{B M}{h} = {cot 75}^{\circ}$ $(BM)/h=cot75^@$

i.e. $A L + B M = h ({cot 64}^{\circ} + {cot 75}^{\circ})$ $AL+BM=h(cot64^@+cot75^@)$

But $A L + B M = 425 - 250 = 175$ $AL+BM=425-250=175$

Hence, $h = \frac{175}{{cot 64}^{\circ} + {cot 75}^{\circ}}$ $h=175/(cot64^@+cot75^@)$

= $\frac{175}{0.4887 + 0.2679}$ $175/(0.4887+0.2679)$

= $\frac{175}{0.7566} = 231.3$ $175/0.7566=231.3$ feet

and as $\frac{h}{A D} = {sin 64}^{\circ}$ $h/(AD)=sin64^@$ and $\frac{h}{B C} = {sin 75}^{\circ}$ $h/(BC)=sin75^@$

i.e. $A D = \frac{h}{{sin 64}^{\circ} =} \frac{231.3}{0.8988} = 257.3$ $AD=h/sin64^@=231.3/0.8988=257.3$ feet and

$B C = \frac{h}{{sin 75}^{\circ} =} \frac{231.3}{0.9659} = 239.5$ $BC=h/sin75^@=231.3/0.9659=239.5$ feet.

Science

Math

Humanities

... and beyond

The bases of a trapezoid are 425 feet and 250 feet respectively. The angles that the sides make with the longer base are 64 degree and 75 degree. What are the sides of the trapezoid?

1 Answer

Explanation:

Related questions

Impact of this question