A triangle is a polygon with three sides and three angles. Types of triangles are right angled triangle, scalene triangle, equilateral triangle and isosceles triangle.

Given a triangle with angles A, B, C and the corresponding sides opposite to the angles as a, b, c. Sine rule states that for the triangle, the following holds:

$\frac{a}{sin(A)}=\frac{b}{sin(B)}=\frac{c}{sin(C)}$

In triangle JKL, k=4.1 cm, j=3.8 cm and angle J=63°.

Using sine rule, we can find ∠K:

$\frac{k}{sin(K)}=\frac{j}{sin(J)} \\\\\frac{4.1}{sin(K)}=\frac{3.8}{sin(63)} \\\\sin(K)=\frac{4.1*sin(63)}{3.8}\\\\sin(K)=0.9613\\\\K=sin^{-1} (0.9613)\\\\K=74.0^o \\$

6 0

3 years ago

Given: ABCD ~ AEFG Find x. A) 4 B) 8 C) 9 D) 10

gayaneshka [121]

Answer:

Step-by-step explanation:

find CD which is 4

x/4=6/3

x=2×4

x=8

5 0

3 years ago

Which function has a greater rate of change over the interval [0,2]

Illusion [34]

Answer:

g(x) has a greater average rate of change

Step-by-step explanation:

From the given information, the table is:

x | g(x)

-1 7

0 5

1 7

2 13

From this table, we have g(0)=5 and g(2)=13

The average rate of change over [a,b] of g(x) is given by: $\frac{g(b)-g(a)}{b-a}$

This implies that on the [0,2]. the average rate of change is:

$\frac{g(2)-g(0)}{2-0}=\frac{13-5}{2}=\frac{7}{2}=3.5$

Also, we have that: f(0)=-4 and f(2)=-1.

This means that the average rate of change of f(x) on [0,2] is

$\frac{f(2)-f(0)}{2-0}=\frac{-1--4}{2} =\frac{3}{2} =1.5$

Hence g(x) has a greater average rate of change on [0,2]

6 0

3 years ago