All categories

3 years ago

Which triangle can be drawn as it is described?

Mathematics

1 answer:

yawa3891 [41]3 years ago

8 0

Answer:

An isosceles triangle with angles measuring 20° and 80°

Step-by-step explanation:

Verify each case

case A) Scalene triangle with angles measuring 110° and 35°

Is not a scalene triangle because the third angle is (180-110°-35°=35°), therefore is an isosceles triangle

case B) An obtuse triangle with sides measuring 5,10 and 15

we know that

Triangle Inequality Theorem, states that the sum of the lengths of any two sides of a triangle is greater than the length of the third side

so in this problem

$5+10> 15$ -----> is not true

therefore

with these measurements can not draw any triangle

case C) An isosceles triangle with angles measuring 20° and 80°

we know that

An isosceles triangle has two equal sides and two equal angles

In this problem the third angle is (180-20°-80°=80°),

therefore

is an isosceles triangle and can be drawn as it is described

case D) An acute triangle with sides measuring 7,4 and 2

we know that

Triangle Inequality Theorem, states that the sum of the lengths of any two sides of a triangle is greater than the length of the third side

so in this problem

$4+2> 7$ -----> is not true

therefore

with these measurements can not draw any triangle

You might be interested in

The product of a number and -7 amounts to seven times the sum of that number and 20. Find the number.

MA_775_DIABLO [31]

Answer: -10

Step-by-step explanation:

If you put it in an equation it would be -7x=7(x+20). You would then -7x to both sides to get -14x=140. Just divide -14 on both sides to isolate x. This equates to -10. If you put -10 back into the equation then it says that 70 is 7 times larger than 10. Which it is.

8 0

3 years ago

A ship leaves port at 10 miles per hour, with a heading of N 35° W. There is a warning buoy located 5 miles directly north of th

Leno4ka [110]

The value of the angle subtended by the distance of the buoy from the

port is given by sine and cosine rule.

The bearing of the buoy from the is approximately 307.35°

Reasons:

Location from which the ship sails = Port

The speed of the ship = 10 mph

Direction of the ship = N35°W

Location of the warning buoy = 5 miles north of the port

Required: The bearing of the warning buoy from the ship after 7.5 hours.

Solution:

The distance travelled by the ship = 7.5 hours × 10 mph = 75 miles

By cosine rule, we have;

a² = b² + c² - 2·b·c·cos(A)

Where;

a = The distance between the ship and the buoy

b = The distance between the ship and the port = 75 miles

c = The distance between the buoy and the port = 5 miles

Angle ∠A = The angle between the ship and the buoy = The bearing of the ship = 35°

Which gives;

a = √(75² + 5² - 2 × 75 × 5 × cos(35°))

By sine rule, we have;

$\displaystyle \frac{a}{sin(A)} = \mathbf{ \frac{b}{sin(B)}}$

Therefore;

$\displaystyle sin(B)= \frac{b \cdot sin(A)}{a}$

Which gives;

$\displaystyle sin(B) = \mathbf{\frac{75 \cdot sin(35^{\circ})}{\sqrt{75^2 + 5^2 - 2 \times 75\times5\times cos(35^{\circ}) } }}$

$\displaystyle B = arcsin\left( \frac{75 \cdot sin(35^{\circ})}{\sqrt{75^2 + 5^2 - 2 \times 75\times5\times cos(35^{\circ}) } }\right) \approx 37.32^{\circ}$

Similarly, we can get;

$\displaystyle B = arcsin\left( \frac{75 \cdot sin(35^{\circ})}{\sqrt{75^2 + 5^2 - 2 \times 75\times5\times cos(35^{\circ}) } }\right) \approx \mathbf{ 142.68^{\circ}}$