# Question Video: Using Sine and Tangent Ratios to Solve Problems Mathematics • 10th Grade

Find the value of 3 tan 𝐶 + sin 𝐵, given that the line segment 𝐴𝐷 ⊥ line segment 𝐵𝐶, 𝐴𝐶 = 35 cm, 𝐷𝐶 = 28 cm, and 𝐴𝐵 = 29 cm.

03:53

### Video Transcript

Find the value of three tan 𝐶 plus sin 𝐵, given that the line segment 𝐴𝐷 is perpendicular to the line segment 𝐵𝐶, 𝐴𝐶 is equal to 35 centimeters, 𝐷𝐶 is equal to 28 centimeters, and 𝐴𝐵 is equal to 29 centimeters.

In order to find the value of the expression, we will need to use our knowledge of the trigonometric ratios in right triangles. These state that the sin of angle 𝜃 is equal to the opposite over the hypotenuse. The cos of angle 𝜃 is equal to the adjacent over the hypotenuse. And the tan of angle 𝜃 is equal to the opposite over the adjacent. One way of remembering these ratios is using the acronym SOHCAHTOA.

In this question, we are interested in the angles at 𝐶 and 𝐵. Let’s begin by considering the left triangle 𝐴𝐶𝐷, which contains angle 𝐶. In this triangle, the hypotenuse 𝐴𝐶 has length 35 centimeters and the adjacent has length 28 centimeters. As we need to find the tan of angle 𝐶, we will firstly need to calculate the length of the opposite. We can do this using the Pythagorean theorem such that 𝐴𝐷 squared plus 28 squared is equal to 35 squared. We can subtract 28 squared from both sides of this equation. This means that 𝐴𝐷 squared is equal to 441. We can then square root both sides and take the positive answer as 𝐴𝐷 is a length. The side length 𝐴𝐷 is equal to 21 centimeters. Recalling that tan 𝜃 is the opposite over the adjacent, the tan of angle 𝐶 is equal to 21 over 28. We can divide the numerator and denominator by seven. tan 𝐶 is therefore equal to three-quarters.

Our next step is to calculate the sin of angle 𝐵. We know that the sin of angle 𝜃 is equal to the opposite over the hypotenuse. And in triangle 𝐴𝐵𝐷, the side opposite angle 𝐵 is 21 centimeters and the hypotenuse has length 29 centimeters. The sin of angle 𝐵 is therefore equal to 21 over 29. We can now calculate three multiplied by tan 𝐶 plus sin 𝐵. This is equal to three multiplied by three-quarters plus 21 over 29. Three multiplied by three-quarters is nine-quarters. So we need to add this to 21 over 29. Obtaining a common denominator by multiplying the numerator and denominator of the first fraction by 29 and the second fraction by four, we have 261 over 116 plus 84 over 116. This is equal to 345 over 116. This is the value of three tan 𝐶 plus sin 𝐵.