# Tutor profile: Jas A.

## Questions

### Subject: Pre-Calculus

Given a vector <-3,-4>, find its unit vector.

A Unit Vector maintains direction of the original vector, with a magnitude of 1. Begin by solving for the magnitude of the vector given: We are given the vector <-3,-4> magnitude = $$ \sqrt{(-3)^{2}+(-4)^{2}} $$ =$$ \sqrt{9+16} $$ =$$ \sqrt{25} $$ =5 To find the unit vector, make the magnitude of the given vector 1. To do this, we divide the magnitude by itself $$ \frac{5}{5} $$=1 Now we must also divide each of the components of the given vector by 5. Answer: <$$ \frac{-3}{5} $$,$$ \frac{-4}{5} $$>

### Subject: Trigonometry

Given a triangle ABC, where angle A=62*, angle B=31*, and side length c=4cm, complete the triangle. Summarize your answer in one place at the end. Round your answers to the nearest hundredth.

We want to complete the triangle, meaning we want to solve for all missing lengths and angles. Begin by finding the last angle of the triangle: 180*-62*-31*= 87*. Angle C=87*. Using the Law of Sines, set up your proportion to find the remaining side lengths. We will solve for side length a first: $$ \frac{4cm}{sin(87*)} $$ = $$ \frac{a}{sin(62*)} $$ Multiply both sides by sin(62*) to get a by itself: a = $$ \frac{4sin(62*)}{sin(87*)} $$ a = 3.54 cm Repeat these steps using the Law of Sines to solve for side length b. Angle A=62* Angle B=31* Angle C=87* side a=3.54 cm side b=2.06 cm side c=4 cm

### Subject: Geometry

Given a right triangle ABC with side AB=24ft and hypotenuse, side AC=74ft, Find the length of the missing side BC.

BC=70ft

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