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3 years ago

I am having trouble with this question, can anyone help? This is for Geometry.

Mathematics

2 answers:

Bumek [7]3 years ago

8 0

Answer:

by geometry rules

for a triangle

external angle of a triangle is equal to the sum of two internal angle of the triangle ( other than external adjacent angle)

m<1= sum of m<2 and m<3

m<1= m<1+<2

m<1 = 31+72

m<1= 103

hope you are clear now !

marta [7]3 years ago

8 0

Answer:

The correct answer is last option. 103

Step-by-step explanation:

Points to remember

Sum of any two angles of a triangle is equal to the exterior angle of the third angle

<u>To find the measure of <1</u>

It is given that m<2 = 31° and m< 3 = 72°

From the figure we can see that <2 and <3 are two angles of a triangle. And <1 is the exterior angle of third angle of same triangle.

Therefore we can write,

m<2 + m<3 = m<1

m<1 = 31 + 72

= 103°

Therefore the correct answer is last option. 103

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2 years ago

Help! How would I solve this trig identity?

NeTakaya

Using simpler trigonometric identities, the given identity was proven below.

<h3>How to solve the trigonometric identity?</h3>

Remember that:

$sec(x) = \frac{1}{cos(x)} \\\\tan(x) = \frac{sin(x)}{cos(x)}$

Then the identity can be rewritten as:

$sec^4(x) - sen^2(x) = tan^4(x) + tan^2(x)\\\\\frac{1}{cos^4(x)} - \frac{1}{cos^2(x)} = \frac{sin^4(x)}{cos^4(x)} + \frac{sin^2(x)}{cos^2(x)} \\\\$

Now we can multiply both sides by cos⁴(x) to get:

$\frac{1}{cos^4(x)} - \frac{1}{cos^2(x)} = \frac{sin^4(x)}{cos^4(x)} + \frac{sin^2(x)}{cos^2(x)} \\\\\\\\cos^4(x)*(\frac{1}{cos^4(x)} - \frac{1}{cos^2(x)}) = cos^4(x)*( \frac{sin^4(x)}{cos^4(x)} + \frac{sin^2(x)}{cos^2(x)})\\\\1 - cos^2(x) = sin^4(x) + cos^2(x)*sin^2(x)\\\\1 - cos^2(x) = sin^2(x)*sin^2(x) + cos^2(x)*sin^2(x)$