Exercise Zone : Trigonometri
Table of Contents
Berikut ini adalah kumpulan soal mengenai Trigonometri. Jika ingin bertanya soal, silahkan gabung ke grup Telegram, Signal, Discord, atau WhatsApp.
Tipe:
No.
Sebuah tangga disandarkan pada tembol rumah dengan membentuk sudut 60° terhadap tanah. Jarak antara ujung tangga dan permukaan tanah adalah $2\sqrt3$ m. Panjang tangga tersebut adalah .....- 4 m
- 4,5 m
- 5 m
- 5,5 m
- 6 m
ALTERNATIF PENYELESAIAN
\begin{aligned}
\sin60\degree&=\dfrac{2\sqrt3}x\\[3.7pt]
\dfrac12\sqrt3&=\dfrac{2\sqrt3}x\\[3.7pt]
x&=\dfrac{2\sqrt3}{\dfrac12\sqrt3}\\
&=\boxed{\boxed{4}}
\end{aligned}
Jadi, panjang tangga tersebut adalah 4 m.
JAWAB: A
JAWAB: A
No.
Jika $x-y=\dfrac12\pi$ maka- $\dfrac{1+\tan y^2}{y}$
- $-\dfrac{1-y^2}{\tan y}$
- $\dfrac{\tan(1-y)}{(1+y)^2}$
- $\dfrac{\tan y}{(1+y)^2}$
- $-\dfrac1{\tan y}$
ALTERNATIF PENYELESAIAN
$x=\dfrac12\pi+y$
\begin{aligned}
\tan x&=\tan\left(\dfrac12\pi+y\right)\\
&=-\cot y\\
&=-\dfrac1{\tan y}
\end{aligned}
Jadi, $\tan x=-\dfrac1{\tan y}$.
JAWAB: E
JAWAB: E
No.
Jika- $\dfrac{a-1}{\sqrt{a^2-1}}$
- $\dfrac{1-a}{\sqrt{a^2-1}}$
- $\dfrac{a-1}{\sqrt{1-a^2}}$
- $\dfrac{1-a}{\sqrt{1-a^2}}$
- $\dfrac{-a-1}{\sqrt{1-a^2}}$
ALTERNATIF PENYELESAIAN
\begin{aligned}
\cot257\degree+\csc257\degree&=\cot\left(270\degree-13\degree\right)+\csc\left(270\degree-13\degree\right)\\
&=\tan13\degree-\sec13\degree\\
&=\dfrac{a}{\sqrt{1-a^2}}-\dfrac1{\sqrt{1-a^2}}\\
&=\boxed{\boxed{\dfrac{a-1}{\sqrt{1-a^2}}}}
\end{aligned}
Jadi, $\cot257\degree+\csc257\degree=\dfrac{a-1}{\sqrt{1-a^2}}$.
JAWAB: C
JAWAB: C
No.
Diketahui $\cos\alpha = \dfrac{a}{2b}$, dengan α sudut lancip dan- $\dfrac{2b}a$
- $\dfrac{\sqrt{a^2-4b^2}}{2a}$
- $\dfrac{\sqrt{4b^2-a^2}}{2a}$
- $\dfrac{\sqrt{a^2-4b^2}}a$
- $\dfrac{\sqrt{4b^2-a^2}}a$
ALTERNATIF PENYELESAIAN
$\cos\alpha = \dfrac{a}{2b}=\dfrac{sa}{mi}$
sa = a , mi = 2b
\begin{aligned}
de&=\sqrt{(2b)^2-a^2}\\
&=\sqrt{4b^2-a^2}
\end{aligned}
\begin{aligned}
\tan\alpha&=\dfrac{de}{sa}\\
&=\boxed{\boxed{\dfrac{\sqrt{4b^2-a^2}}a}}
\end{aligned}
Jadi, $\tan\alpha=\dfrac{\sqrt{4b^2-a^2}}a$.
JAWAB: E
JAWAB: E
No.
Jika θ sudut lancip dan $\cos\theta=\dfrac35$, maka nilai dari $\dfrac{\sin\theta\tan\theta-1}{2\tan^2\theta}$ adalahALTERNATIF PENYELESAIAN
$\sin\theta=\dfrac45$
$\tan\theta=\dfrac43$ \begin{aligned} \dfrac{\sin\theta\tan\theta-1}{2\tan^2\theta}&=\dfrac{\left(\dfrac45\right)\left(\dfrac43\right)-1}{2\left(\dfrac43\right)^2}\\ &=\dfrac{\dfrac{16}{15}-1}{2\left(\dfrac{16}9\right)}\\ &=\dfrac{\dfrac1{15}}{\dfrac{32}9}\\ &=\dfrac1{15}\cdot\dfrac9{32}\\ &=\boxed{\boxed{\dfrac3{160}}} \end{aligned}
$\tan\theta=\dfrac43$ \begin{aligned} \dfrac{\sin\theta\tan\theta-1}{2\tan^2\theta}&=\dfrac{\left(\dfrac45\right)\left(\dfrac43\right)-1}{2\left(\dfrac43\right)^2}\\ &=\dfrac{\dfrac{16}{15}-1}{2\left(\dfrac{16}9\right)}\\ &=\dfrac{\dfrac1{15}}{\dfrac{32}9}\\ &=\dfrac1{15}\cdot\dfrac9{32}\\ &=\boxed{\boxed{\dfrac3{160}}} \end{aligned}
Jadi, nilai dari $\dfrac{\sin\theta\tan\theta-1}{2\tan^2\theta}$ adalah $\dfrac3{160}$.
No.
Nilai dari sin 135° adalah ....- $\dfrac13\sqrt2$
- $\dfrac12\sqrt2$
- $\dfrac14\sqrt2$
- $\sqrt2$
- $-\dfrac12\sqrt2$
ALTERNATIF PENYELESAIAN
\begin{aligned}
\sin135°&=\sin\left(180°-45°\right)\\
&=\sin45°\\
&=\color{blue}\boxed{\boxed{\color{black}\dfrac12\sqrt2}}
\end{aligned}
Jadi, nilai dari sin 135° adalah $\dfrac12\sqrt2$.
JAWAB: B
JAWAB: B
No.
Diketahui $\sin \alpha=\dfrac45$, $0 \lt \alpha \lt \dfrac{\pi}2$ dan $\cos\beta =\dfrac{12}{13}$, $-\dfrac{\pi}2\lt\beta\lt0$. Tentukan nilaiALTERNATIF PENYELESAIAN
α kuadran I dan β kuadran IV
$\cos\alpha=\dfrac35$, $\sin\beta=-\dfrac5{13}$ \begin{aligned} \sin (\alpha+\beta)&=\sin\alpha\cos\beta+\cos\alpha\sin\beta\\ &=\dfrac45\cdot\dfrac{12}{13}+\dfrac35\cdot\left(-\dfrac5{13}\right)\\ &=\dfrac{48}{65}-\dfrac{15}{65}\\ &=\boxed{\boxed{\dfrac{33}{65}}} \end{aligned} \begin{aligned} \sin (\alpha-\beta)&=\sin\alpha\cos\beta-\cos\alpha\sin\beta\\ &=\dfrac45\cdot\dfrac{12}{13}-\dfrac35\cdot\left(-\dfrac5{13}\right)\\ &=\dfrac{48}{65}+\dfrac{15}{65}\\ &=\boxed{\boxed{\dfrac{63}{65}}} \end{aligned}
$\cos\alpha=\dfrac35$, $\sin\beta=-\dfrac5{13}$ \begin{aligned} \sin (\alpha+\beta)&=\sin\alpha\cos\beta+\cos\alpha\sin\beta\\ &=\dfrac45\cdot\dfrac{12}{13}+\dfrac35\cdot\left(-\dfrac5{13}\right)\\ &=\dfrac{48}{65}-\dfrac{15}{65}\\ &=\boxed{\boxed{\dfrac{33}{65}}} \end{aligned} \begin{aligned} \sin (\alpha-\beta)&=\sin\alpha\cos\beta-\cos\alpha\sin\beta\\ &=\dfrac45\cdot\dfrac{12}{13}-\dfrac35\cdot\left(-\dfrac5{13}\right)\\ &=\dfrac{48}{65}+\dfrac{15}{65}\\ &=\boxed{\boxed{\dfrac{63}{65}}} \end{aligned}
Jadi, $\sin (\alpha+\beta)=\dfrac{33}{65}$ dan $\sin (\alpha-\beta)=\dfrac{63}{65}$.
No.
Segitiga ABC siku siku di C jika panjangALTERNATIF PENYELESAIAN
\begin{aligned}\tan60\degree&=\dfrac{BC}{AC}\\\sqrt3&=\dfrac{BC}{20}\\BC&=\boxed{\boxed{20\sqrt3}}\end{aligned}
Jadi, BC = $20\sqrt3$ cm.
No.
Jika sudut θ di kuadran IV dan $\cos\theta=\dfrac1a$, maka- $-\sqrt{a^2-1}$
- $-\sqrt{1-a^2}$
- $\dfrac{-1}{\sqrt{a^2-1}}$
- $\dfrac{-\sqrt{a^2-1}}a$
- $\dfrac{\sqrt{a^2-1}}a$
ALTERNATIF PENYELESAIAN
θ di kuadran IV ⟶ a > 0
$\sin\theta=-\dfrac{\sqrt{a^2-1}}a$ ⟶ sinus bernilai negatif di kuadran IV.
Jadi, $\cot257\degree+\csc257\degree=\dfrac{a-1}{\sqrt{1-a^2}}$.
JAWAB: C
JAWAB: C
No.
Bila- $\dfrac56$
- $\dfrac{25}{36}$
- $\dfrac16\sqrt{11}$
- $\dfrac5{36}$
- $\dfrac1{36}\sqrt{11}$
ALTERNATIF PENYELESAIAN
\begin{aligned}
x&=\sqrt{\left(\sqrt{11}\right)^2+5^2}\\
&=\sqrt{11+25}\\
&=\sqrt{36}\\
&=\boxed{6}
\end{aligned}
\begin{aligned}
\sin a&=\dfrac5x\\
&=\color{blue}\boxed{\boxed{\color{black}\dfrac56}}
\end{aligned}
Jadi, $\sin a=\dfrac56$.
JAWAB: A
JAWAB: A
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