C-10-Q073mediumsingle_mcq

Moving back $CD=42$ m to point $D$ gives elevation $\angle ADB = 45^\circ$ , so $\tan 45^\circ$ equals—

a $\dfrac{h}{x}$
b $\dfrac{h}{x+42}$
c $\dfrac{x+42}{h}$
d $\dfrac{1}{h}$

ব্যাখ্যা

At the farther point $D$ , the horizontal distance becomes $DB = DC + CB = 42 + x$ , while the height $AB=h$ stays the same. Hence $\tan 45^\circ = \frac{h}{x+42}$ .

এই অধ্যায়ের আরও প্রশ্ন

C-10-Q074Therefore also equals—
C-10-Q075Combining gives
C-10-Q076Hence approximately equals—
C-10-Q077Rationalizing, is equivalent to—
C-10-Q078As the observer moves backwards on level ground, the angle of elevation of a fixed building top—
C-10-Q079The pole stands , breaks without separation, and the broken piece slopes down to at distance m, with . From ,
C-10-Q080Solving for the standing piece, equals—
C-10-Q081Numerically equals—
C-10-Q082From the same triangle, equals—
C-10-Q083So the broken piece equals—

Chapter 10: Distance and Elevation — সব প্রশ্ন দেখুন (160টি) →

Moving back $CD=42$ m to point $D$ gives elevation $\angle ADB = 45^\circ$ , so $\tan 45^\circ$ equals—

\dfrac{h}{x}

\dfrac{h}{x+42}

\dfrac{x+42}{h}

\dfrac{1}{h}

ব্যাখ্যা

At the farther point $D$ , the horizontal distance becomes $DB = DC + CB = 42 + x$ , while the height $AB=h$ stays the same. Hence $\tan 45^\circ = \frac{h}{x+42}$ .

Moving back CD=42CD=42CD=42 m to point DDD gives elevation ∠ADB=45∘\angle ADB = 45^\circ∠ADB=45∘, so tan⁡45∘\tan 45^\circtan45∘ equals—

ব্যাখ্যা

এই অধ্যায়ের আরও প্রশ্ন

Moving back CD=42CD=42CD=42 m to point DDD gives elevation ∠ADB=45∘\angle ADB = 45^\circ∠ADB=45∘, so tan⁡45∘\tan 45^\circtan45∘ equals—

ব্যাখ্যা

এই অধ্যায়ের আরও প্রশ্ন

Moving back $CD=42$ m to point $D$ gives elevation $\angle ADB = 45^\circ$ , so $\tan 45^\circ$ equals—

Moving back $CD=42$ m to point $D$ gives elevation $\angle ADB = 45^\circ$ , so $\tan 45^\circ$ equals—