Question

In triangle ABC AB=BC and C = 35 find angle A and B​

Answer 1

Answer:

Yes! Here is your answer!

Step-by-step explanation:

Answer 2

There are a few methods of obtaining right triangle side lengths. Depending on what is given, you can use different relationships or laws to find the missing side:

Given two sides

If you know two other sides of the right triangle, it's the easiest option; all you need to do is apply the Pythagorean theorem:

a² + b² = c²

if leg a is the missing side, then transform the equation to the form when a is on one side, and take a square root:

a = √(c² - b²)

if leg b is unknown, then

b = √(c² - a²)

for hypotenuse c missing, the formula is

c = √(a² + b²)

Given angle and hypotenuse

Right triangle with law of sines formulas. a over sin(α) equals b over sin(β) equals c, because sin(90°) = 1

Apply the law of sines or trigonometry to find the right triangle side lengths:

a = c * sin(α) or a = c * cos(β)

b = c * sin(β) or b = c * cos(α)

Given angle and one leg

Find the missing leg using trigonometric functions:

a = b * tan(α)

b = a * tan(β)

Given area and one leg

As we remember from basic triangle area formula, we can calculate the area by multiplying triangle height and base and dividing the result by two. A right triangle is a special case of a scalene triangle, in which one leg is the height when the second leg is the base, so the equation gets simplified to:

area = a * b / 2

For example, if we know only the right triangle area and the length of the leg a, we can derive the equation for other sides:

b = 2 * area / a

c = √(a² + (2 * area / a)²)

How to find the angle of a right triangle

If you know one angle apart from the right angle, calculation of the third one is a piece of cake:

Givenβ: α = 90 - β

Givenα: β = 90 - α

However, if only two sides of a triangle are given, finding the angles of a right triangle requires applying some basic trigonometric functions:

for α

sin(α) = a / c so α = arcsin(a / c) (inverse sine)

cos(α) = b / c so α = arccos(b / c) (inverse cosine)

tan(α) = a / b so α = arctan(a / b) (inverse tangent)

cot(α) = b / a so α = arccot(b / a) (inverse cotangent)

and for β

sin(β) = b / c so β = arcsin(b / c) (inverse sine)

cos(β) = a / c so β = arccos(a / c) (inverse cosine)

tan(β) = b / a so β = arctan(b / a) (inverse tangent)

cot(β) = a / b so β = arccot(a / b) (inverse cotangent)