## Free Mathematics Tutorials

Trigonometry problems and questions with solutions - grade 10.

Grade 10 trigonometry problems and questions with answers and solutions are presented.

- The area of a right triangle is 50. One of its angles is 45°. 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°, 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°. 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°. 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° to 60°, what is the height of the building?

## Solutions to the Above Problems

- x = 10 / tan(51°) = 8.1 (2 significant digits) H = 10 / sin(51°) = 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 √(5) = 20 √(5)
- BH perpendicular to AC means that triangles ABH and HBC are right triangles. Hence tan(39°) = 11 / AH or AH = 11 / tan(39°) HC = 19 - AH = 19 - 11 / tan(39°) Pythagora's theorem applied to right triangle HBC: 11 2 + HC 2 = x 2 solve for x and substitute HC: x = √ [ 11 2 + (19 - 11 / tan(39°) ) 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 = √( 16 2 - 10 2 ) - √( 14 2 - 10 2 ) = 2.69 (rounded to 3 significant digits)
- Use right triangle ABC to write: tan(31°) = 6 / BC , solve: BC = 6 / tan(31°) 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 = √ [ 9 + ( 6 / tan(31°) ) 2 ] = 13.4 (rounded to 3 significant digits)
- The triangle is right and the size one of its angles is 45°; the third angle has a size 45° 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 √(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°.
- Let x be the length of side AC. Use the cosine law 12 2 = 8 2 + x 2 - 2 · 8 · x · cos(59°) 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).

## More References and links on Trigonometry

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## High school geometry

Let's focus on angle B \goldD B B start color #e07d10, B, end color #e07d10 since that is the angle that is explicitly given in the diagram.

A right triangle A B C. Angle A C B is a right angle. Angle A B C is fifty degrees and is highlighted. Side A C is unknown. Side A B is six units.

Note that we are given the length of the hypotenuse \purpleC{\text{hypotenuse}} hypotenuse start color #aa87ff, start text, h, y, p, o, t, e, n, u, s, e, end text, end color #aa87ff , and we are asked to find the length of the side opposite \blueD{\text{opposite}} opposite start color #11accd, start text, o, p, p, o, s, i, t, e, end text, end color #11accd angle B \goldD B B start color #e07d10, B, end color #e07d10 . The trigonometric ratio that contains both of those sides is the sine . [I'd like to review the trig ratios.]

sin ( B ) = opposite hypotenuse Define sine. sin ( 5 0 ∘ ) = A C 6 Substitute. 6 sin ( 5 0 ∘ ) = A C Multiply both sides by 6. 4.60 ≈ A C Evaluate with a calculator. \begin{aligned}\sin( \goldD{ B}) &= \dfrac{ \blueD{\text{ opposite}} \text{ } }{\purpleC{\text{ hypotenuse} }} ~~~~~~~~\small{\gray{\text{Define sine.}}}\\\\ \sin (\goldD{50^\circ})&= \dfrac{\blueD{AC}}{\purpleC6}~~~~~~~~~~~~~~~~~~~~~~~\small{\gray{\text{Substitute.}}} \\\\\\\\ 6\sin ({50^\circ})&= {{AC}} ~~~~~~~~~~~~~~~~~~~~~~~~~\small{\gray{\text{Multiply both sides by }6.}}\\\\\\\\ 4.60&\approx AC~~~~~~~~~~~~~~~~~~~~~~~~~\small{\gray{\text{Evaluate with a calculator.}}} \end{aligned} sin ( B ) sin ( 5 0 ∘ ) 6 sin ( 5 0 ∘ ) 4 . 6 0 = hypotenuse opposite Define sine. = 6 A C Substitute. = A C Multiply both sides by 6 . ≈ A C Evaluate with a calculator.

## Now let's try some practice problems.

- Your answer should be
- an integer, like 6 6 6 6
- a simplified proper fraction, like 3 / 5 3/5 3 / 5 3, slash, 5
- a simplified improper fraction, like 7 / 4 7/4 7 / 4 7, slash, 4
- a mixed number, like 1 3 / 4 1\ 3/4 1 3 / 4 1, space, 3, slash, 4
- an exact decimal, like 0.75 0.75 0 . 7 5 0, point, 75
- a multiple of pi, like 12 pi 12\ \text{pi} 1 2 pi 12, space, start text, p, i, end text or 2 / 3 pi 2/3\ \text{pi} 2 / 3 pi 2, slash, 3, space, start text, p, i, end text

## Challenge problem

- (Choice A) sin ( 2 8 ∘ ) = 20 z \sin (28^\circ)=\dfrac{20}{z} sin ( 2 8 ∘ ) = z 2 0 sine, left parenthesis, 28, degrees, right parenthesis, equals, start fraction, 20, divided by, z, end fraction A sin ( 2 8 ∘ ) = 20 z \sin (28^\circ)=\dfrac{20}{z} sin ( 2 8 ∘ ) = z 2 0 sine, left parenthesis, 28, degrees, right parenthesis, equals, start fraction, 20, divided by, z, end fraction
- (Choice B) cos ( 2 8 ∘ ) = 20 z \cos (28^\circ)=\dfrac{20}{z} cos ( 2 8 ∘ ) = z 2 0 cosine, left parenthesis, 28, degrees, right parenthesis, equals, start fraction, 20, divided by, z, end fraction B cos ( 2 8 ∘ ) = 20 z \cos (28^\circ)=\dfrac{20}{z} cos ( 2 8 ∘ ) = z 2 0 cosine, left parenthesis, 28, degrees, right parenthesis, equals, start fraction, 20, divided by, z, end fraction
- (Choice C) cos ( 6 2 ∘ ) = 20 z \cos (62^\circ)=\dfrac{20}{z} cos ( 6 2 ∘ ) = z 2 0 cosine, left parenthesis, 62, degrees, right parenthesis, equals, start fraction, 20, divided by, z, end fraction C cos ( 6 2 ∘ ) = 20 z \cos (62^\circ)=\dfrac{20}{z} cos ( 6 2 ∘ ) = z 2 0 cosine, left parenthesis, 62, degrees, right parenthesis, equals, start fraction, 20, divided by, z, end fraction
- (Choice D) tan ( 6 2 ∘ ) = 20 z \tan (62^\circ)=\dfrac{20}{z} tan ( 6 2 ∘ ) = z 2 0 tangent, left parenthesis, 62, degrees, right parenthesis, equals, start fraction, 20, divided by, z, end fraction D tan ( 6 2 ∘ ) = 20 z \tan (62^\circ)=\dfrac{20}{z} tan ( 6 2 ∘ ) = z 2 0 tangent, left parenthesis, 62, degrees, right parenthesis, equals, start fraction, 20, divided by, z, end fraction

## Want to join the conversation?

- Unit circle
- Derive the relationship of right-triangle trigonometry and circles
- Relate degree and radian measures for angles

As we vary the angle θ, the opposite and adjacent sides of the triangle also vary.

By the Pythagorean theorem, we know the following relationship among r, x 1 , and y 1 .

a. 60° b. 270° c. -90° d. 120°

Practice Problem: Convert each of the following angle measures from radians to degrees.

Application to Trigonometric Functions

sin θ = distance of point P from the x -axis

cos θ = distance of point P from the y -axis

We can use the basic facts of angles to redraw this situation in a more familiar form.

Our job is to then calculate y. We can use trigonometry as follows.

Thus, the mark is about 17.07 inches above the ground.

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## Trigonometry : Solving Word Problems with Trigonometry

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Since you want to find the height of the platform, you will need to use tangent.

## Example Question #2 : Solving Word Problems With Trigonometry

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Therefore the shadow cast by the building is 150 meters long.

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Therefore the change in height between Angelina's starting and ending points is 1480 meters.

## Example Question #9 : Solving Word Problems With Trigonometry

## Example Question #10 : Solving Word Problems With Trigonometry

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