Trigonometry Problems and Questions with Solutions  Grade 10
Grade 10 trigonometry problems and questions with answers and solutions are presented.

Find x and H in the right triangle below.

Find the lengths of all sides of the right triangle below if its area is 400.

BH is perpendicular to AC. Find x the length of BC.

ABC is a right triangle with a right angle at A. Find x the length of DC.

In the figure below AB and CD are perpendicular to BC and the size of angle ACB is 31^{o}. Find the length of segment BD.

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.

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

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

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

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.

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.

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?
 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?
Solutions to the Above Problems

x = 10 / tan(51^{o}) = 8.1 (2 significant digits)
H = 10 / sin(51^{o}) = 13 (2 significant digits)

Area = (1/2)(2x)(x) = 400
Solve for x: x = 20 , 2x = 40
Pythagora's theorem: (2x)^{2} + (x)^{2} = H^{2}
H = x sqrt(5) = 20 sqrt(5)

BH perpendicular to AC means that triangles ABH and HBC are right triangles. Hence
tan(39^{o}) = 11 / AH or AH = 11 / tan(39^{o})
HC = 19  AH = 19  11 / tan(39^{o})
Pythagora's theorem applied to right triangle HBC: 11^{2} + HC^{2} = x^{2}
solve for x and substitute HC: x = sqrt [ 11^{2} + (19  11 / tan(39^{o}) )^{2} ]
= 12.3 (rounded to 3 significant digits)

Since angle A is right, both triangles ABC and ABD are right and therefore we can apply Pythagora's theorem.
14^{2} = 10^{2} + AD^{2} , 16^{2} = 10^{2} + AC^{2}
Also x = AC  AD
= sqrt( 16^{2}  10^{2} )  sqrt( 14^{2}  10^{2} ) = 2.69 (rounded to 3 significant digits)

Use right triangle ABC to write: tan(31^{o}) = 6 / BC , solve: BC = 6 / tan(31^{o})
Use Pythagora's theorem in the right triangle BCD to write:
9^{2} + BC^{2} = BD^{2}
Solve above for BD and substitute BC: BD = sqrt [ 9 + ( 6 / tan(31^{o}) )^{2} ]
= 13.4 (rounded to 3 significant digits)

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)x^{2} = 50 , solve for x: x = 10
We now use Pythagora to find H: x^{2} + x^{2} = H^{2}
Solve for H: H = 10 sqrt(2)

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.
Pythagora's theorem: h^{2} = (3k)^{2} + (5k)^{2}
Solve for h: h = 5k
sin(A) = a / h = 3k / 5k = 3/5 and cos(A) = 4k / 5k = 4/5

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}.

Let x be the length of side AC. Use the cosine law
12^{2} = 8^{2} + x^{2}  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).

The diagram below show the two buildings and the angles of depression and elevation.
tan(20^{o}) = 200 / L
L = 200 / tan(20^{o})
tan(10^{o}) = H2 / L
H2 = L * tan(10^{o})
= 200 * tan(10^{o}) / tan(20^{o})
Height of second building = 200 + 200 * tan(10^{o}) / tan(20^{o})
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