In trigonometry, we write tan (60\u00b0) mathematically, and its exact value in fraction form is \u221a3. Therefore, we write it in the following form in trigonometry:<\/span><\/b><\/p>\n tan (60\u00b0) = tan \u03c0\/3 = \u221a3<\/b><\/p>\n <\/p>\n The exact value of tan(\u03c0\/3) is 1\/\u221a3 equal to 1.7320508075\u2026 in decimal form.<\/span><\/p>\n It is reciprocal of cot 60 degrees.<\/span><\/p>\n The approximate value of the tangent of angle 60 degrees is equal to 1.7321.<\/span><\/b><\/p>\n tan (60\u00b0) = 1.7320508075\u2026 \u2248 1.7321<\/b><\/p>\n <\/p>\n The exact value of tan (\u03c0\/3) can be derived using three methods explained below.<\/span><\/p>\n <\/span><\/p>\n We can derive the exact value of tan (60\u00b0) by considering an equilateral triangle ABC.<\/span><\/p>\n <\/span><\/p>\n <\/span><\/p>\n We know that each angle in an equilateral triangle is 60\u00b0.<\/span><\/p>\n So, \u2220A = \u2220B = \u2220C = 60\u00b0<\/span><\/span><\/p>\n Now, draw a perpendicular line AD from point A to side BC.<\/span><\/p>\n Now we have two right-angled triangles ABD and ADC.<\/span><\/p>\n Here, \u2220 ADB = \u2220ADC = 90\u00b0 and,<\/span><\/p>\n \u2220 ABD = \u2220ACD = 60\u00b0<\/span><\/p>\n AD = AD<\/span><\/p>\n According to AAS Congruency<\/span><\/p>\n \u0394 ABD \u2245 \u0394 ACD<\/span><\/p>\n From this, we conclude that<\/span><\/p>\n BD = DC\u00a0 \u00a0 (since they are corresponding parts of congruent triangles)<\/span><\/p>\n Take the value of AB = BC = 2a<\/span><\/p>\n Then, BD =\u00a0 \u00bd (BC)<\/span><\/p>\n \u00a0= \u00bd (2a) = a<\/span><\/p>\n <\/span><\/p>\n Now use Pythagoras theorem in the <\/span>\u25b3ABD<\/span><\/p>\n \\mathrm{AB}^{2}=\\mathrm{AD}^{2}-\\mathrm{BD}^{2} <\/span> <\/p>\n Now in right-angled triangle ADB,<\/span><\/p>\n tan (60\u00b0) = (opposite side to the \u2220 ABD) \/ (adjacent side to the \u2220 ABD)<\/span><\/p>\n tan (60\u00b0) = AD\/BD<\/span><\/p>\n tan (60\u00b0) = a\u221a3\/a\u00a0<\/span><\/p>\n \u00a0 \u00a0 \u00a0 \u00a0 \u00a0 \u00a0 \u00a0 \u00a0 = \u221a3<\/span><\/p>\n Therefore,<\/span> tan (60\u00b0) =\u221a3<\/b><\/p>\n <\/b><\/p>\n You can also find the value of the tangent of angle 60\u00b0 practically by constructing a right-angled triangle with a 60\u00b0 angle by geometrical tools.<\/span><\/p>\n Draw a straight horizontal line from Point H and then construct an angle of 60\u00b0 using the protractor.<\/span><\/p>\n <\/span><\/p>\n <\/span><\/p>\n Set the compass to any length by a ruler. Here, the compass is set to 4.4 cm. Now, draw an arc on the 60\u00b0 angle line from point H, and it intersects the line at point I.<\/span><\/p>\n <\/span><\/p>\n <\/span><\/p>\n Finally, draw a perpendicular line on the horizontal line from point I, and it intersects the horizontal line at point J perpendicularly. Thus, a right-angled triangle \u2206HIJ is formed.<\/span><\/p>\n <\/span><\/p>\n <\/span><\/p>\n Now,\u00a0 calculate the value of the tangent of 60 degrees and for this, measure the length of the adjacent side (HJ) with a ruler. You will observe that the length of the opposite side (IJ) is 3.8 cm, and the length of the adjacent side (HJ) is 2.2 cm in this example.<\/span><\/p>\n <\/span><\/p>\n <\/p>\n Now, find the ratio of lengths of the opposite side to the adjacent side and get the value of the tangent of angle 60\u00b0.<\/span><\/p>\n Here,<\/span><\/p>\n tan (60\u00b0) = (opposite side) \/ (adjacent side)<\/span><\/p>\n tan (60\u00b0) = IJ\/HJ = (3.8)\/(2.2)<\/span><\/p>\n So, tan (60\u00b0) = 1.727272\u2026 \u2248 1.7321<\/span><\/p>\n <\/p>\n We can prove the value of tan (60\u00b0) with a trigonometric approach.<\/span><\/p>\n we know,<\/span><\/p>\n sin 60\u00b0 = \u221a3\/2,<\/span><\/p>\n cos 60\u00b0 = 1\/2<\/span><\/p>\n Also, by trigonometric identities,<\/span><\/p>\n sin x\/cos x = tan x<\/b><\/p>\n Put x = 60\u00b0<\/span><\/p>\n tan (60\u00b0) = sin (60\u00b0)\/cos (60\u00b0)<\/span><\/p>\n Put the values of sin 60\u00b0 and cos 60\u00b0<\/span><\/p>\n tan (60\u00b0) = (\u221a3\/2)\/(1\/2)<\/span><\/p>\n tan (60\u00b0) = \u221a3<\/span><\/p>\n Hence, we proved the value of tan (60\u00b0) using different approaches.<\/span><\/p>\n <\/p>\n <\/p>\n 1. Evaluate: tan 60\u00b0 + sin 60\u00b0<\/strong><\/p>\n Solution:<\/strong><\/p>\n We know that tan (60\u00b0) = \u221a3 and sin (60\u00b0) = \u221a3\/2 <\/p>\n 2. Evaluate 2 tan 60\u00b0 \u2013 2 cos 30\u00b0<\/strong><\/p>\n Solution:<\/strong> [\/et_pb_text][\/et_pb_column][et_pb_column type=”2_5″ module_id=”stickysideR” admin_label=”Column R” _builder_version=”4.10.4″ _module_preset=”default” background_color=”#fdefe0″ custom_padding=”25px|25px|25px|25px|true|true” sticky_position=”top” sticky_offset_top=”-280px” sticky_limit_top=”row” sticky_limit_bottom=”row” sticky_position_tablet=”none” sticky_position_phone=”none” sticky_position_last_edited=”on|desktop” sticky_limit_bottom_tablet=”” sticky_limit_bottom_phone=”” sticky_limit_bottom_last_edited=”on|phone” border_radii=”on|15px|15px|15px|15px” box_shadow_style=”preset3″ global_colors_info=”{}”][et_pb_image src=”https:\/\/eistudymaterial.s3.amazonaws.com\/1080×1080.png” alt=”Free Trial banner” title_text=”Mindspark Free Trial Banner” url=”https:\/\/mindspark.in\/free-trial” align=”center” module_class=”adsimg” _builder_version=”4.11.1″ _module_preset=”default” custom_padding=”||||false|false” border_radii=”on|10px|10px|10px|10px” global_colors_info=”{}” transform_styles__hover_enabled=”on|hover” transform_scale__hover_enabled=”on|hover” transform_translate__hover_enabled=”on|desktop” transform_rotate__hover_enabled=”on|desktop” transform_skew__hover_enabled=”on|desktop” transform_origin__hover_enabled=”on|desktop” transform_scale__hover=”102%|102%”][\/et_pb_image][et_pb_text admin_label=”Explore Other Topics [\/et_pb_text][et_pb_text _builder_version=”4.10.7″ _module_preset=”default” text_line_height=”2.2em” link_font_size=”16px” custom_margin=”||0px||false|false” custom_padding=”10px|15px|10px|28px|true|false” locked=”off” global_colors_info=”{}”]<\/p>\n [\/et_pb_text][et_pb_text admin_label=”Related Concepts [\/et_pb_text][et_pb_text _builder_version=”4.13.1″ _module_preset=”default” text_line_height=”2.2em” link_font_size=”16px” custom_margin=”||0px||false|false” custom_padding=”10px|15px|10px|28px|true|false” locked=”off” global_colors_info=”{}”]<\/p>\nValue of Tan 60\u00b0<\/b><\/h2>\n
Proof<\/b><\/h2>\n
\n
Theoretical Method<\/b><\/h3>\n<\/li>\n<\/ul>\n
![\"\"](\"https:\/\/eistudymaterial.s3.amazonaws.com\/Tan-60-Degrees-01-300x278.png\")
<\/span><\/p>\n
\\mathrm{AD}^{2}=\\mathrm{AB}^{2}-\\mathrm{BD}^{2} <\/span>
\\mathrm{AD}^{2}=(2 \\mathrm{a})^{2}-\\mathrm{a}^{2} <\/span>
\\mathrm{AD}^{2}=4 \\mathrm{a}^{2}-\\mathrm{a}^{2} <\/span>
\\mathrm{AD}^{2}=3 \\mathrm{a}^{2} <\/span>
\\text { So, } \\mathrm{AD}=\\mathrm{a} \\sqrt{3}<\/span><\/span><\/p>\n\n
Practical Method
<\/b><\/b><\/h3>\n<\/li>\n<\/ul>\n![\"\"](\"https:\/\/eistudymaterial.s3.amazonaws.com\/Tan-60-Degrees-02-300x258.png\")
<\/span><\/p>\n![\"\"](\"https:\/\/eistudymaterial.s3.amazonaws.com\/Tan-60-Degrees-03-300x285.png\")
<\/span><\/p>\n![\"\"](\"https:\/\/eistudymaterial.s3.amazonaws.com\/Tan-60-Degrees-04-300x296.png\")
<\/span><\/p>\n![\"\"](\"https:\/\/eistudymaterial.s3.amazonaws.com\/Tan-60-Degrees-05-300x296.png\")
<\/span><\/p>\n\n
Trigonometric approach<\/b><\/h3>\n<\/li>\n<\/ul>\n
Example<\/b><\/h2>\n
So, tan (60\u00b0) + sin (60\u00b0)
= \u221a3 + \u221a3\/2
= (3\u221a3)\/2
<\/b><\/p>\n
We know that tan (60\u00b0) = \u221a3 and cos (30\u00b0) = \u221a3\/2
So, 2 tan (60\u00b0) \u2013 2 cos (30\u00b0)
= 2\u221a3 – 2(\u221a3\/2)
= 2\u221a3 – \u221a3
= \u221a3<\/p>\n
\n” _builder_version=”4.9.11″ _module_preset=”default” header_font=”|700|||||||” header_font_size=”25px” text_orientation=”center” custom_margin=”0px||0px||true|false” custom_padding=”8px|15px|0px|15px|false|true” locked=”off” global_colors_info=”{}”]<\/p>\nExplore Other Topics<\/h1>\n
\n” _builder_version=”4.9.11″ _module_preset=”default” header_font=”|700|||||||” header_font_size=”25px” text_orientation=”center” custom_margin=”0px||0px||true|false” custom_padding=”8px|15px|0px|15px|false|true” locked=”off” global_colors_info=”{}”]<\/p>\nRelated Concepts<\/h1>\n