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Noma'lum burchakni toping (1-rasm).
Uchburchakning tashqi burchagi
120^(@) bo'lib, unga qo'shni bo'Imagan ichki burchaklari 1:2 nisbatda bo'lsa, uchburchakning burchaklarini toping.
Agar 2 -rasmda
/_ACB=90^(@),CD=BD va
AB=24sm bo"lsa,
CD kesmani toping.

$1$ . Noma'lum burchakni toping ( $1$ -rasm). $\newline$ $2$ . Uchburchakning tashqi burchagi $120^{\circ}$ bo'lib, unga qo'shni bo'Imagan ichki burchaklari $1$ : $2$ nisbatda bo'lsa, uchburchakning burchaklarini toping. $\newline$ $3$ . Agar $2$ -rasmda $\angle A C B=90^{\circ}, C D=B D$ va $A B=24 \mathrm{sm}$ bo

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Evaluate definite integrals using the chain rule

Full solution

Q.

1

. Noma'lum burchakni toping (

1

-rasm).

\newline

2

. Uchburchakning tashqi burchagi

120^{\circ}

bo'lib, unga qo'shni bo'Imagan ichki burchaklari

1

:

2

nisbatda bo'lsa, uchburchakning burchaklarini toping.

\newline

3

. Agar

2

-rasmda

\angle A C B=90^{\circ}, C D=B D

va

A B=24 \mathrm{sm}

bo

New Math Problem: New Math Problem: Find the length of segment CD in the second figure where triangle ACB is a right triangle with ACB being $90$ degrees, CD is equal to BD, and AB is $24$ cm. $\newline$ Since triangle ACB is a right triangle, we can use the Pythagorean theorem to find the length of BC. $\newline$ Let $x$ be the length of CD (and also BD since CD $=$ BD). $\newline$ Then, $AC^2 + BC^2 = AB^2$ $\newline$ Since AB is the hypotenuse and is $24$ cm, we have: $\newline$ $AC^2 + BC^2 = 24^2$ $\newline$ $AC^2 + (2x)^2 = 576$ $\newline$ We know that ACB is a right angle, so AC is also the height of the triangle, and it drops down to the midpoint of AB, dividing AB into two segments of equal length, $12$ cm each. $\newline$ So, $AC = 12$ cm. $\newline$ Now we plug AC into the equation: $\newline$ $24$ $0$ $\newline$ $24$ $1$ $\newline$ $24$ $2$ $\newline$ $24$ $3$ $\newline$ $24$ $4$ $\newline$ $24$ $5$ $\newline$ $24$ $6$ $\newline$ $24$ $7$

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