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3 years ago

How to find the lateral area of the right triangular prism

Mathematics

1 answer:

german3 years ago

6 0

Lateral area : Perimeter of base * height
LA = ( a+b+c )* H

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Please help:In our amusement park that we built this week, the ride Diablo's

Vedmedyk [2.9K]

ANSWER ; 10003 yup cause 1000+3 =

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3 years ago

Read 2 more answers

drawbridge at the entrance to an ancient castle is raised and lowered by a pair of chains. The figure represents the drawbridge

jasenka [17]

Answer:

4.0 meters, ∠C = 39°, ∠A = 51°

Step-by-step explanation:

Firstly, our diagram shows us that the given triangle is actually a right triangle. So we can use the <em>Pythagorean Theorem</em> to solve for the height of the chain:

$a^{2} +b^{2} =c^{2}$

$(3.3)^{2} +b^{2} =(5.2)^{2}$

$b^{2} =(5.2)^{2}-(3.3)^{2}$

$b =\sqrt{(5.2)^{2}-(3.3)^{2}}$

$b=4.0187...$

$b=4.0 m$

Now, we can use the <em>Law of Cosines</em> to figure out one of the acute angles:

$c^{2} =a^{2} +b^{2} -2ab(cos(C))$

$(3.3)^{2} =(4.0)^{2} +(5.2)^{2} -2(4.0)(5.2)(cos(C))$

$cos(C)=\frac{(3.3)^{2}-(4.0)^{2} -(5.2)^{2}}{-2(4.0)(5.2)}$

$C=cos^{-1}( \frac{(3.3)^{2}-(4.0)^{2} -(5.2)^{2}}{-2(4.0)(5.2)})$

$C=39.3915...$

∠C = 39°

And since we know that all angles in a triangle add up to 180°:

∠A + ∠B + ∠C = 180

∠A + 90 + 39 = 180

∠A = 180 - 90 - 39

∠A = 51°

However, you should always review any answers on the Internet and make sure they are correct! Check my work to see if I made any mistakes!

7 0

3 years ago

Two lighthouses are located 60 miles from one another on a north-south line. If a boat is spotted S 21o E from the northern ligh

jasenka [17]

Answer:

(A)The northern lighthouse is 8.2 miles closer than the southern lighthouse.

Step-by-step explanation:

The triangle attached represents the given problem.

First, let us determine the distance of the Boat from each of the lighthouse.

In Triangle ABC,

∠A+∠B+∠C=180 degrees

21+∠B+16=180

∠B=180-37=143 degrees.

Using Law of Sines

$\frac{a}{Sin A}=\frac{b}{Sin B}\\\frac{a}{Sin 21^0}=\frac{60}{Sin 143^0} \\\text{Cross Multiply}\\a*sin143=60*sin21\\a=60*sin21\div sin143\\a=35.73 miles$

Similarly

$\frac{c}{Sin C}=\frac{b}{Sin B}\\\frac{c}{Sin 16^0}=\frac{60}{Sin 143^0} \\\text{Cross Multiply}\\c*sin143=60*sin16\\c=60*sin16\div sin143\\c=27.48 miles$