TRIGNOMETRY-solve these.

QUESTIONS

 The area of a right triangle is 50. One of its angles is 45^o. Find the lengths of the sides and hypotenuse of the triangle.

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In a right triangle ABC, tan(A) = 3/4. Find sin(A) and cos(A).

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In a right triangle ABC with angle A equal to 90^o, find angle B and C so that sin(B) = cos(B).

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A rectangle has dimensions 10 cm by 5 cm. Determine the measures of the angles at the point where the diagonals intersect.

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The lengths of side AB and side BC of a scalene triangle ABC are 12 cm and 8 cm respectively. The size of angle C is 59^o. Find the length of side AC.

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From the top of a 200 meters high building, the angle of depression to the bottom of a second building is 20 degrees. From the same point, the angle of elevation to the top of the second building is 10 degrees. Calculate the height of the second building.

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Karla is riding vertically in a hot air balloon, directly over a point P on the ground. Karla spots a parked car on the ground at an angle of depression of 30^o. The balloon rises 50 meters. Now the angle of depression to the car is 35 degrees. How far is the car from point P?

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If the shadow of a building increases by 10 meters when the angle of elevation of the sun rays decreases from 70^o to 60^o, what is the height of the building?  

    ANSWERS

The triangle is right and the size one of its angles is 45^o; the third angle has a size 45^o and therefore the triangle is right and isosceles. Let x be the length of one of the sides and H be the length of the hypotenuse.

Area = (1/2)x² = 50 , solve for x: x = 10

We now use Pythagoras  to find H: x² + x² = H²

Solve for H: H = 10 sqrt(2)

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Let a be the length of the side opposite angle A, b the length of the side adjacent to angle A and h be the length of the hypotenuse.

tan(A) = opposite side / adjacent side = a/b = 3/4

We can say that: a = 3k and b = 4k , where k is a coefficient of proportionality. Let us find h.

Pythagoras theorem: h² = (3k)² + (5k)²

Solve for h: h = 5k

sin(A) = a / h = 3k / 5k = 3/5 and cos(A) = 4k / 5k = 4/5

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Let b be the length of the side opposite angle B and c the length of the side opposite angle C and h the length of the hypotenuse.

sin(B) = b/h and cos(B) = c/h

sin(B) = cos(B) means b/h = c/h which gives c = b

The two sides are equal in length means that the triangle is isosceles and angles B and C are equal in size of 45^o.

The diagram below shows the rectangle with the diagonals and half one of the angles with size x.

tan(x) = 5/2.5 = 2 , x = arctan(2)

larger angle made by diagonals 2x = 2 arctan(2) = 127^o (3 significant digits)

Smaller angle made by diagonals 180 - 2x = 53^o.

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Let x be the length of side AC. Use the cosine law

12² = 8² + x² - 2*8*x*cos(59^o)

Solve the quadratic equation for x: x = 14.0 and x = -5.7

x cannot be negative and therefore the solution is x = 14.0 (rounded to one decimal place).