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2 years ago

In a 30-60-90 triangle, the length of the hypotenuse is 17. Find the length of the side opposite the 30 degree angle.

Mathematics

1 answer:

cluponka [151]2 years ago

4 0

Step-by-step explanation:

we know that on a 30 60 90 trianlge the length of the side opposite the 30 degree angle is half of the length of hypotenuse.So in this problem

the length of the side opposite the 30 angle is 17/2

talking about exercise on the photo the answer is

8√3 because it is given that the length of the side opposite the 30 anlge is 8 so hypotenuse is 16 we mentioned why before.

Now lets use pythagorean theorem

x^2+y^2=z^2

z is hypotenuse

8^2+y^2=16^2

64+y^2=256

y^2=256-64

y^2=192

y=8√3

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2 years ago

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2 years ago

Two ships leave a harbor together, traveling on courses that have an angle of 135°40' between them. If they each travel 402 mile

mario62 [17]

Answer:

Therefore they are 734.106 miles apart.

Step-by-step explanation:

Given that ,

Two ships have a harbor together. The angle between two ships is 135°40'. Each of two ships travel 402 miles.

It forms a isosceles triangle whose two sides are 402 miles and one angle is 135°40'. Since it is isosceles triangle then other two angles of the triangle is equal.

Let ∠B= 135°40', and AB = 402 miles , BC = 402 miles

Then the distance between the ships = AC

We know

The sum of all angles = 180°

⇒∠A+∠B+∠C=180°

⇒∠A+135°40'+∠C=180°

⇒2∠A= 180°- 135°40' [ since ∠A=∠C]

⇒2∠A=44°60'

⇒∠A= 22°30'

Again we know that,

$\frac{AB}{sin\angle C}=\frac{BC}{sin \angle A}=\frac{AC}{sin \angle B}$

Taking last two ratio,

$\frac{BC}{sin \angle A}=\frac{AC}{sin \angle B}$

Putting the value of BC , AC ,∠A,∠B

$\frac{402}{sin 22^\circ30'}=\frac{AC}{sin 135^\circ40'}$

$\Rightarrow AC=\frac{402 \times sin135^\circ40'}{sin 22^\circ30'}$

≈734.106 miles

Therefore they are 734.106 miles apart.

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2 years ago