Exercise Zone : Trigonometri

Berikut ini adalah kumpulan soal mengenai Trigonometri. Jika ingin bertanya soal, silahkan gabung ke grup Telegram, Signal, Discord, atau WhatsApp.

Tipe:

Standar SBMPTN HOTS

No.

Sebuah tangga disandarkan pada tembol rumah dengan membentuk sudut 60\degree terhadap tanah. Jarak antara ujung tangga dan permukaan tanah adalah 2\sqrt3 m. Panjang tangga tersebut adalah .....

1. 4 m
2. 4{,}5 m
3. 5 m

4. 5{,}5 m
5. 6 m

Grup Telegram

$$\begin{array}{rl}\sin 60{}^{\circ }& ={\displaystyle \frac{2\sqrt{3}}{x}}\\ {\displaystyle \frac{1}{2}}\sqrt{3}& ={\displaystyle \frac{2\sqrt{3}}{x}}\\ x& ={\displaystyle \frac{2\sqrt{3}}{{\displaystyle \frac{1}{2}}\sqrt{3}}}\\ & =\boxed{{\displaystyle \boxed{{\displaystyle 4}}}}\end{array}$$

Jadi, panjang tangga tersebut adalah 4 m. JAWAB: A

No.

Jika x-y=\dfrac12\pi maka \tan x adalah....

1. \dfrac{1+\tan y^2}{y}
2. -\dfrac{1-y^2}{\tan y}
3. \dfrac{\tan(1-y)}{(1+y)^2}

4. \dfrac{\tan y}{(1+y)^2}
5. -\dfrac1{\tan y}

x=\dfrac12\pi+y

$$\begin{array}{rl}\tan x& =\tan ({\displaystyle \frac{1}{2}}\pi +y)\\ & =-\cot y\\ & =-{\displaystyle \frac{1}{\tan y}}\end{array}$$

Jadi, \tan x=-\dfrac1{\tan y}. JAWABB: E

No.

Jika \sin13\degree=a, maka nilai {\cot257\degree+\csc257\degree=}

1. \dfrac{a-1}{\sqrt{a^2-1}}
2. \dfrac{1-a}{\sqrt{a^2-1}}
3. \dfrac{a-1}{\sqrt{1-a^2}}

4. \dfrac{1-a}{\sqrt{1-a^2}}
5. \dfrac{-a-1}{\sqrt{1-a^2}}

$$\begin{array}{rl}\cot 257{}^{\circ }+\csc 257{}^{\circ }& =\cot (270{}^{\circ }-13{}^{\circ })+\csc (270{}^{\circ }-13{}^{\circ })\\ & =\tan 13{}^{\circ }-\sec 13{}^{\circ }\\ & ={\displaystyle \frac{a}{\sqrt{1-a^2}}}-{\displaystyle \frac{1}{\sqrt{1-a^2}}}\\ & =\boxed{{\displaystyle \boxed{{\displaystyle {\displaystyle \frac{a-1}{\sqrt{1-a^2}}}}}}}\end{array}$$

Jadi, \cot257\degree+\csc257\degree=\dfrac{a-1}{\sqrt{1-a^2}}. JAWAB: C

No.

Diketahui \cos\alpha = \dfrac{a}{2b}, dengan \alpha sudut lancip dan b\neq0. Nilai dari \tan\alpha=

1. \dfrac{2b}a
2. \dfrac{\sqrt{a^2-4b^2}}{2a}
3. \dfrac{\sqrt{4b^2-a^2}}{2a}

4. \dfrac{\sqrt{a^2-4b^2}}a
5. \dfrac{\sqrt{4b^2-a^2}}a

\cos\alpha = \dfrac{a}{2b}=\dfrac{sa}{mi}
sa=a, mi=2b

$$\begin{array}{rl}de& =\sqrt{(2b{)}^2-a^2}\\ & =\sqrt{4b^2-a^2}\end{array}$$

$$\begin{array}{rl}\tan \alpha & ={\displaystyle \frac{de}{sa}}\\ & =\boxed{{\displaystyle \boxed{{\displaystyle {\displaystyle \frac{\sqrt{4b^2-a^2}}{a}}}}}}\end{array}$$

Jadi, \tan\alpha=\dfrac{\sqrt{4b^2-a^2}}a. JAWAB: E

No.

Jika \theta sudut lancip dan {\cos\theta=\dfrac35}, maka nilai dari \dfrac{\sin\theta\tan\theta-1}{2\tan^2\theta} adalah

Grup WhatsApp

\sin\theta=\dfrac45

\tan\theta=\dfrac43

$\begin{array}{rl}{\displaystyle \frac{\sin \theta \tan \theta -1}{2{\tan }^2\theta }}& ={\displaystyle \frac{\left({\displaystyle \frac{4}{5}}\right)\left({\displaystyle \frac{4}{3}}\right)-1}{2{\left({\displaystyle \frac{4}{3}}\right)}^2}}\\ & ={\displaystyle \frac{{\displaystyle \frac{16}{15}}-1}{2\left({\displaystyle \frac{16}{9}}\right)}}\\ & ={\displaystyle \frac{{\displaystyle \frac{1}{15}}}{{\displaystyle \frac{32}{9}}}}\\ & ={\displaystyle \frac{1}{15}}\cdot {\displaystyle \frac{9}{32}}\\ & =\boxed{{\displaystyle \boxed{{\displaystyle {\displaystyle \frac{3}{160}}}}}}\end{array}$

Jadi, \dfrac{\sin\theta\tan\theta-1}{2\tan^2\theta={\dfrac3{160}}.

No.

6\cos x\sin4x= ...

Grup Whatsapp

$\begin{array}{rl}6\cos x\sin 4x& =3(2\sin 4x\cos x)\\ & =3(\sin (4x+x)+\sin (4x-x))\\ & =3(\sin 5x+\sin 3x)\\ & =\boxed{{\displaystyle \boxed{{\displaystyle 3\sin 5x+3\sin 3x}}}}\end{array}$

Jadi, 6\cos x\sin4x= 3\sin5x+3\sin3x.

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 nilai \sin (\alpha+\beta) dan \sin (\alpha-\beta).

Grup Telegram

\alpha kuadran I dan \beta kuadran II

\cos\alpha=\dfrac35, \sin\beta=-\dfrac5{13}

$\begin{array}{rl}\sin (\alpha +\beta )& =\sin \alpha \cos \beta +\cos \alpha \sin \beta \\ & ={\displaystyle \frac{4}{5}}\cdot {\displaystyle \frac{12}{13}}+{\displaystyle \frac{3}{5}}\cdot (-{\displaystyle \frac{5}{13}})\\ & ={\displaystyle \frac{48}{65}}-{\displaystyle \frac{15}{65}}\\ & =\boxed{{\displaystyle \boxed{{\displaystyle {\displaystyle \frac{33}{65}}}}}}\end{array}$

$\begin{array}{rl}\sin (\alpha -\beta )& =\sin \alpha \cos \beta -\cos \alpha \sin \beta \\ & ={\displaystyle \frac{4}{5}}\cdot {\displaystyle \frac{12}{13}}-{\displaystyle \frac{3}{5}}\cdot (-{\displaystyle \frac{5}{13}})\\ & ={\displaystyle \frac{48}{65}}+{\displaystyle \frac{15}{65}}\\ & =\boxed{{\displaystyle \boxed{{\displaystyle {\displaystyle \frac{63}{65}}}}}}\end{array}$

Jadi, \sin (\alpha+\beta)=\dfrac{33}{65} dan \sin (\alpha-\beta)=\dfrac{63}{65}.

No.

Segitiga ABC siku siku di C jika panjang {AC = 20} cm dan besar sudut A = 60\degree tentukan panjang BC!​

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$$\begin{array}{rl}\tan 60{}^{\circ }& ={\displaystyle \frac{BC}{AC}}\\ \sqrt{3}& ={\displaystyle \frac{BC}{20}}\\ BC& =\boxed{{\displaystyle \boxed{{\displaystyle 20\sqrt{3}}}}}\end{array}$$

Jadi, BC=20\sqrt3.

No.

Jika \sin13\degree=a, maka nilai \cot257\degree+\csc257\degree=

1. \dfrac{a-1}{\sqrt{a^2-1}}
2. \dfrac{1-a}{\sqrt{a^2-1}}
3. \dfrac{a-1}{\sqrt{1-a^2}}

4. \dfrac{1-a}{\sqrt{1-a^2}}
5. \dfrac{-a-1}{\sqrt{1-a^2}}

$$\begin{array}{rl}\cot 257{}^{\circ }+\csc 257{}^{\circ }& =\cot (270{}^{\circ }-13{}^{\circ })+\csc (270{}^{\circ }-13{}^{\circ })\\ & =\tan 13{}^{\circ }-\sec 13{}^{\circ }\\ & ={\displaystyle \frac{a}{\sqrt{1-a^2}}}-{\displaystyle \frac{1}{\sqrt{1-a^2}}}\\ & =\boxed{{\displaystyle \boxed{{\displaystyle {\displaystyle \frac{a-1}{\sqrt{1-a^2}}}}}}}\end{array}$$

Jadi, \cot257\degree+\csc257\degree=\dfrac{a-1}{\sqrt{1-a^2}}. JAWAB: C

No.

Diketahui \cos\alpha = \dfrac{a}{2b}, dengan \alpha sudut lancip dan b\neq0. Nilai dari \tan\alpha=

1. \dfrac{2b}a
2. \dfrac{\sqrt{a^2-4b^2}}{2a}
3. \dfrac{\sqrt{4b^2-a^2}}{2a}

4. \dfrac{\sqrt{a^2-4b^2}}a
5. \dfrac{\sqrt{4b^2-a^2}}a

\cos\alpha = \dfrac{a}{2b}=\dfrac{sa}{mi}
sa=a, mi=2b

$$\begin{array}{rl}de& =\sqrt{(2b{)}^2-a^2}\\ & =\sqrt{4b^2-a^2}\end{array}$$

$$\begin{array}{rl}\tan \alpha & ={\displaystyle \frac{de}{sa}}\\ & =\boxed{{\displaystyle \boxed{{\displaystyle {\displaystyle \frac{\sqrt{4b^2-a^2}}{a}}}}}}\end{array}$$

Jadi, \tan\alpha=\dfrac{\sqrt{4b^2-a^2}}a. JAWAB: E