Question

Can you help me to solve this
Using special right triangles and for the other using trigonometry

Answer 1

Answer:

$\text{Triangle 1:}\\\\x=30\sqrt{2},\\\\\text{Triangle 2:}\\x\approx 36.18,\\?\approx12.37$

Step-by-step explanation:

*Notes (clarified by the person who asked this question):

-The triangle on the right has a right angle (angle that appears to be a right angle is a right angle)

-The bottom side of the right triangle is marked with a question mark (?)

Triangle 1 (triangle on left):

Special triangles:

In all 45-45-90 triangles, the ratio of the sides is $x:x:x\sqrt{2}$, where $x\sqrt{2}$ is the hypotenuse of the triangle. Since one of the legs is marked as $30$, the hypotenuse must be $\boxed{30\sqrt{2}}$

It's also possible to use a variety of trigonometry to solve this problem. Basic trig for right triangles is applicable and may be the simplest:

$\cos 45^{\circ}=\frac{30}{x},\\x=\frac{30}{\cos 45^{\circ}}=\frac{30}{\frac{\sqrt{2}}{2}}=30\cdot \frac{2}{\sqrt{2}}=\frac{60}{\sqrt{2}}=\frac{60\cdot\sqrt{2}}{\sqrt{2}\cdot \sqrt{2}}=\frac{60\sqrt{2}}{2}=\boxed{30\sqrt{2}}$

Triangle 2 (triangle on right):

We can use basic trig for right triangles to set up the following equations:

$\cos 20^{\circ}=\frac{34}{x},\\x=\frac{34}{\cos 20^{\circ}}\approx \boxed{36.18}$,

$\tan 20^{\circ}=\frac{?}{34},\\?=34\tan 20^{\circ}\approx \boxed{12.37}$

We can verify these answers using the Pythagorean theorem. The Pythagorean theorem states that in all right triangles, the following must be true:

$a^2+b^2=c^2$, where $c$ is the hypotenuse of the triangle and $a$ and $b$ are two legs of the triangle.

Verify $34^2+(34\tan20^{\circ})^2=\left(\frac{34}{\cos20^{\circ}}\right)^2\:\checkmark$